Chemistry 9701/52 — May/June 2024

Cambridge A-Level · Planning, Analysis and Evaluation · worked solutions for every part, with the mark scheme

questions

marks

minutes

Topics Planning · Analysis, Conclusions and Evaluation

Q1MediumPlanningAnalysis, Conclusions and Evaluation

Calcium carbonate, $\text{CaCO}_3(\text{s})$ , decomposes when heated, as shown.

\text{CaCO}_3(\text{s}) \rightarrow \text{CaO}(\text{s}) + \text{CO}_2(\text{g})

The enthalpy change of reaction, $\Delta H_r$ , for the thermal decomposition of $\text{CaCO}_3(\text{s})$ cannot be measured directly. Instead, a procedure involving two experiments is used. In each experiment, the enthalpy change of a different reaction is determined.

The equation for the reaction in experiment 1 is shown. The enthalpy change for this reaction is $\Delta H_1$ .

\text{experiment 1} \quad \text{CaCO}_3(\text{s}) + 2\text{HCl}(\text{aq}) \rightarrow \text{CaCl}_2(\text{aq}) + \text{H}_2\text{O}(\text{l}) + \text{CO}_2(\text{g})

The equation for the reaction in experiment 2 is shown. The enthalpy change for this reaction is $\Delta H_2$ .

\text{experiment 2} \quad \text{CaO}(\text{s}) + 2\text{HCl}(\text{aq}) \rightarrow \text{CaCl}_2(\text{aq}) + \text{H}_2\text{O}(\text{l})

Experiment 1

step 1 Weigh a $0.0500\text{ mol}$ sample of powdered $\text{CaCO}_3(\text{s})$ .

step 2 Transfer $50.00\text{ cm}^3$ , an excess, of $2.00\text{ mol dm}^{-3}$ hydrochloric acid, $\text{HCl}(\text{aq})$ , into a small glass beaker.

step 3 Start a timer and measure the temperature of the $\text{HCl}(\text{aq})$ in the beaker every 30 seconds for $2\frac{1}{2}$ minutes.

step 4 After 3 minutes add the sample of $\text{CaCO}_3(\text{s})$ to the $\text{HCl}(\text{aq})$ in the beaker. Continue measuring the temperature of the reaction mixture every 30 seconds for a further 5 minutes.

Experiment 2

Repeat experiment 1 using calcium oxide, $\text{CaO}(\text{s})$ , instead of $\text{CaCO}_3(\text{s})$ .

(a)

Suggest why the enthalpy change of reaction for the thermal decomposition of calcium carbonate cannot be measured directly.

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DifficultyEasy

Worked solution

Answer

The thermal decomposition requires continuous heating, making it impossible to accurately measure the temperature change of the system while heat is being applied from an external source.

Final answer

Cannot easily measure temperature change while applying heat

Detailed explanation

Background Concept

Enthalpy changes are typically measured using calorimetry, where the temperature change of a solution is monitored as a reaction occurs in an insulated container. For this to work, the reaction must proceed spontaneously or be initiated easily, and the heat exchange must occur primarily between the reaction and the solution.

Understanding the Question

The question asks why the enthalpy change for the thermal decomposition of calcium carbonate cannot be measured directly. Thermal decomposition requires sustained, high-temperature heating (typically above 800 °C) to break the chemical bonds.

Approach

Consider the practical requirements of direct calorimetry versus the conditions required for thermal decomposition. If you apply heat directly to the reaction vessel, the heat source itself will raise the temperature of the surroundings, making it impossible to isolate the temperature change caused solely by the chemical reaction.

Step-by-Step Reasoning

Direct measurement requirement: To measure $\Delta H$ directly, you would heat the $\text{CaCO}_3$ and measure the temperature change of the system.
The problem: The reaction requires an external heat source to proceed. This external heat will continuously raise the temperature of the reaction mixture and the apparatus.
Conclusion: You cannot distinguish between the temperature increase caused by the chemical reaction and the temperature increase caused by the external heating. Therefore, the enthalpy change cannot be measured directly.

Key Takeaways

Some reactions require conditions (like high heat) that are incompatible with standard calorimetry techniques. In such cases, indirect methods like Hess's law must be used.

Common Mistakes

Stating "the reaction is too slow" or "the reaction does not go to completion." While true, the primary reason is the interference of the external heat source with temperature measurement.
Saying "you cannot measure temperature." Students can measure temperature; the issue is attributing the change to the reaction.

Things to Be Careful About

Focus on the interference of the heat source. The mark scheme specifically looks for the inability to measure temperature change while applying heat.

Techniques used

evaluate direct calorimetry limitations

(b)

(i)

Calculate the mass, in g, of $\text{CaCO}_3(\text{s})$ to be weighed using a two-decimal-place balance in step 1.

1M Similar

DifficultyEasy

Worked solution

Working

M_r(\text{CaCO}_3) = 40.1 + 12.0 + (3 \times 16.0) = 100.1

\text{mass} = 0.0500 \text{ mol} \times 100.1 \text{ g mol}^{-1} = 5.005 \text{ g}

Rounding to two decimal places gives $5.01 \text{ g}$ .

Answer

$5.01 \text{ g}$

Final answer

5.01 g

Detailed explanation

Background Concept

The mass of a substance can be calculated from the number of moles and its molar mass using the equation $m = n \times M_r$ . The molar mass is the sum of the atomic masses of all atoms in the formula. In experimental chemistry, the precision of the balance used dictates the number of decimal places required for the measured mass.

Understanding the Question

The student is asked to calculate the mass of $0.0500 \text{ mol}$ of $\text{CaCO}_3$ to be weighed using a balance that reads to two decimal places.

Approach

Calculate the molar mass of $\text{CaCO}_3$ .
Multiply by the number of moles to find the theoretical mass.
Round the result to two decimal places to match the balance's precision.

Step-by-Step Reasoning

Molar mass calculation: $M_r = 40.1 (\text{Ca}) + 12.0 (\text{C}) + 3 \times 16.0 (\text{O}) = 100.1 \text{ g mol}^{-1}$ .
Mass calculation: $m = 0.0500 \times 100.1 = 5.005 \text{ g}$ .
Rounding: The balance measures to $0.01 \text{ g}$ . $5.005$ rounds to $5.01 \text{ g}$ (using standard rounding rules, or $5.00$ if using round-to-even, but CIE typically accepts standard rounding where 5 rounds up, giving $5.01$ ). The mark scheme gives $5.01 \text{ g}$ .

Key Takeaways

Always calculate the exact mass first, then round to the appropriate number of decimal places based on the apparatus specified in the question.

Common Mistakes

Forgetting to round to two decimal places.
Using incorrect atomic masses (e.g., $\text{Ca} = 40$ instead of $40.1$ ). Using $40$ gives $5.00 \text{ g}$ , which may not match the mark scheme's expected $5.01 \text{ g}$ .

Things to Be Careful About

Check the atomic masses provided in the data booklet. $\text{Ca}$ is $40.1$ , not $40$ . This small difference affects the final rounded value.

Techniques used

calculate mass from moles and molar mass

(ii)

Outline how a student should weigh by difference using a weighing boat in order to determine the exact mass of $\text{CaCO}_3(\text{s})$ added to $\text{HCl}(\text{aq})$ in the beaker. Draw a results table, with appropriate headings, ready for the student to complete.

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DifficultyMedium-Easy

Worked solution

Answer

Weighing by difference:

Weigh the weighing boat containing the $\text{CaCO}_3$ sample.
Transfer the $\text{CaCO}_3$ to the beaker containing the $\text{HCl}$ .
Weigh the empty weighing boat (or boat with any residue) again.
The difference in mass is the exact mass of $\text{CaCO}_3$ added.

Results table:

mass of weighing boat + $\text{CaCO}_3$ (before transfer) / g
mass of weighing boat (+ residue) (after addition) / g
mass of $\text{CaCO}_3$ added / g

Final answer

See working

Detailed explanation

Background Concept

Weighing by difference is a precise method for transferring a specific mass of solid into a container. Instead of trying to add solid to a balance until it reads exactly $5.01 \text{ g}$ (which is difficult and risks spilling), you weigh the container + solid, transfer the solid, and weigh the container again. The difference is the exact mass transferred.

Understanding the Question

The student must outline the procedure for weighing by difference and draw a results table to record the measurements.

Approach

Describe the sequence of weighing: initial mass (boat + solid), transfer, final mass (boat + residue).
Design a table with clear headings, including units, to capture these values and the calculated difference.

Step-by-Step Reasoning

Procedure: The mark scheme requires two key steps: weighing the boat with the sample before transfer, and weighing the boat after the sample is added to the beaker. The difference gives the mass added.
Table design: A results table needs column headings with quantities and units. Since this is a single measurement sequence, a single column with rows for each measurement is appropriate.
- Row 1: Mass of boat + $\text{CaCO}_3$ before transfer / g
- Row 2: Mass of boat (with residue) after addition / g
- Row 3: Mass of $\text{CaCO}_3$ added / g (calculated as Row 1 - Row 2)

Key Takeaways

When asked to outline a technique, be precise about the sequence of actions. For tables, always include units in the headings.

Common Mistakes

Describing weighing the empty boat first, then adding solid to make $5.01 \text{ g}$ . This is not weighing by difference.
Forgetting units in the table headings. The mark scheme explicitly requires units (/ g).
Including unnecessary rows like "mass of beaker" or "mass of HCl". The table should only record what is needed to find the mass of $\text{CaCO}_3$ added.

Things to Be Careful About

The mark scheme accepts "mass of boat (+ residue)" because some solid may stick to the boat. Ensure your table reflects the actual physical steps.

Techniques used

outline weighing by difference techniqueconstruct a results table

(c)

Identify which piece of apparatus should be used to measure the volume of $\text{HCl}(\text{aq})$ in step 2 and give a reason for your choice.

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DifficultyMedium-Easy

Worked solution

Answer

Apparatus: Burette
Reason: It measures volume to the nearest $0.05 \text{ cm}^3$ , providing the necessary precision for a measurement of $50.00 \text{ cm}^3$ .

Working

A measuring cylinder typically measures to $1 \text{ cm}^3$ or $0.5 \text{ cm}^3$ . A volumetric pipette measures a fixed volume (e.g., $25.0 \text{ cm}^3$ or $50.0 \text{ cm}^3$ ) but is less flexible for adding an "excess" where exact precision of the excess volume is less critical than the precision of the measurement itself. However, a burette can deliver $50.00 \text{ cm}^3$ with precision to $0.05 \text{ cm}^3$ , matching the four significant figures in $50.00$ .

Answer

Burette; measures to nearest $0.05 \text{ cm}^3$

Final answer

Burette; measures to nearest 0.05 cm^3

Detailed explanation

Background Concept

Volumetric apparatus comes in different precisions. A measuring cylinder is approximate ( $\pm 1 \text{ cm}^3$ ). A volumetric pipette is highly precise for a fixed volume ( $\pm 0.05 \text{ cm}^3$ for $25 \text{ cm}^3$ ). A burette is precise for variable volumes ( $\pm 0.05 \text{ cm}^3$ ).

Understanding the Question

The student needs to measure $50.00 \text{ cm}^3$ of $\text{HCl}$ . The value $50.00$ implies a precision to two decimal places ( $0.01 \text{ cm}^3$ or better).

Approach

Identify the apparatus that can measure $50.00 \text{ cm}^3$ with the required precision. A burette reads to $0.05 \text{ cm}^3$ (or $0.01$ if estimated), which is sufficient. A measuring cylinder is not precise enough.

Step-by-Step Reasoning

Required precision: $50.00 \text{ cm}^3$ has 4 significant figures, implying precision to $0.01$ or $0.05 \text{ cm}^3$ .
Apparatus choice: A burette is designed to deliver variable volumes with high precision. Its scale is typically read to $0.05 \text{ cm}^3$ (with estimation to $0.01$ ).
Reason: The burette measures to the nearest $0.05 \text{ cm}^3$ , which justifies the $50.00$ recording.

Key Takeaways

Match the number of decimal places in the volume to the precision of the apparatus. $50.00$ requires a burette or volumetric pipette.

Common Mistakes

Suggesting a measuring cylinder. It cannot measure to $0.01$ or $0.05 \text{ cm}^3$ .
Suggesting a volumetric pipette. While precise, it is usually for fixed volumes (like $25.00 \text{ cm}^3$ ). A $50.00 \text{ cm}^3$ pipette exists but is less common; a burette is the standard answer for delivering an excess volume with this precision.

Things to Be Careful About

The mark scheme specifically looks for "burette" AND "measures to nearest $0.05 \text{ cm}^3$ ". Both are required for the mark.

Techniques used

select appropriate volumetric apparatus

(d)

Without making any changes to the apparatus, suggest an instruction to be added to step 3 and step 4 to make the experiment more accurate.

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Worked solution

Answer

Stir the solution / mixture continuously.

Working

In calorimetry experiments, heat is generated or absorbed at the site of reaction. Without stirring, temperature gradients can form, leading to inaccurate maximum/minimum temperature readings. Stirring ensures the heat is distributed evenly throughout the solution, allowing the thermometer to record the true average temperature.

Answer

Stir the solution / mixture

Final answer

Stir the solution

Detailed explanation

Background Concept

In simple solution calorimetry, the assumption is that the temperature measured by the thermometer represents the temperature of the entire solution. This is only true if the heat from the reaction is distributed uniformly.

Understanding the Question

The procedure involves adding a solid to an acid and measuring temperature. The question asks for an instruction to add to steps 3 and 4 to improve accuracy, without changing apparatus.

Approach

Think about what could cause inaccurate temperature readings in a beaker reaction. Localized heating or cooling. The solution is to mix the contents.

Step-by-Step Reasoning

Problem: When $\text{CaCO}_3$ is added to $\text{HCl}$ , the reaction occurs at the surface of the solid. The solution near the solid gets hot, but the rest may not.
Effect: The thermometer might record a lower temperature than the actual maximum if it's not in the hot spot, or it might record a fluctuating temperature.
Solution: Stirring ensures convection and uniform temperature distribution. This allows the thermometer to record the true maximum temperature reached by the reaction mixture.

Key Takeaways

Stirring is a fundamental technique in solution calorimetry to ensure thermal equilibrium.

Common Mistakes

Suggesting "use a lid" or "insulate the beaker". The question says "without making any changes to the apparatus". A lid is an additional piece of apparatus.
Suggesting "use a digital thermometer". Again, apparatus change.

Things to Be Careful About

Read the constraint: "Without making any changes to the apparatus". Only procedural instructions are allowed.

Techniques used

improve calorimetry technique

(e)

A student carries out experiment 1 and obtains the results given in Table 1.1.

Table 1.1

time / minutes	0.5	1.0	1.5	2.0	2.5	3.0	3.5	4.0	4.5	5.0	5.5	6.0
temperature / $^{\circ}\text{C}$	19.0	19.0	19.0	19.0	19.0	—	27.5	36.0	34.5	32.5	32.0	31.0

time / minutes	6.5	7.0	7.5	8.0
temperature / $^{\circ}\text{C}$	29.0	28.0	26.0	25.5

(i)

Plot a graph on the grid in Fig. 1.1 to show the relationship between temperature and time. Use a cross ( $\times$ ) to plot each data point.

The points and line of best fit for the data before 3 minutes have been drawn for you.

Draw a line of best fit for the data after 3 minutes that will enable you to determine the theoretical temperature increase at 3.0 minutes.

Fig. 1.1

2M Similar

DifficultyMedium

Worked solution

Answer

Graph plotting:
Plot the following points using crosses ( $\times$ ):

$(3.5, 27.5)$
$(4.0, 36.0)$
$(4.5, 34.5)$
$(5.0, 32.5)$
$(5.5, 32.0)$
$(6.0, 31.0)$
$(6.5, 29.0)$
$(7.0, 28.0)$
$(7.5, 26.0)$
$(8.0, 25.5)$

Line of best fit:
Draw a smooth curve or line of best fit through these points from $t = 4.0$ to $t = 8.0$ . Extend (extrapolate) this line back to $t = 3.0$ minutes. The line should show a peak around $4.0$ minutes and then a gradual decline due to heat loss.

Answer

See graph description

Final answer

See working

Detailed explanation

Background Concept

In exothermic reactions measured in a simple calorimeter, heat is lost to the surroundings as the temperature rises. This means the recorded maximum temperature is lower than the theoretical maximum (the temperature if no heat was lost). To find the theoretical temperature change, we use an extrapolation method.

Understanding the Question

The student must plot the temperature-time data for the reaction phase (after 3 minutes) and draw a line of best fit that can be extrapolated back to $t = 3.0$ minutes to find the theoretical maximum temperature.

Approach

Plot all 10 data points from $t = 3.5$ to $t = 8.0$ .
Draw a line/curve of best fit. The reaction is fastest at $4.0$ minutes (temp $36.0$ ), then slows down as reactants are used up and heat is lost.
Extrapolate the cooling curve (the part after the peak) back to $t = 3.0$ minutes. This gives the theoretical temperature at the moment of mixing, assuming no heat loss had occurred.

Step-by-Step Reasoning

Plotting: Ensure all 10 points are plotted accurately. The grid has major lines every $5^\circ\text{C}$ and minor lines every $0.5^\circ\text{C}$ . X-axis has major lines every $1$ min, minor every $0.1$ min.
Line of best fit: The points from $4.0$ to $8.0$ show a cooling trend (mostly). Draw a curve that fits these points. The point at $3.5$ ( $27.5$ ) is on the rising part, but the reaction is still ongoing. The mark scheme says "line of best fit drawn involving points from 4 to 8 minutes AND extrapolated back to 3 minutes".
Extrapolation: Extend the line from the cooling phase (e.g., from $5.0$ to $8.0$ ) back to $t = 3.0$ . Read the temperature at $t = 3.0$ on this extrapolated line. This value is typically around $38.5^\circ\text{C}$ .

Key Takeaways

Extrapolation is used to correct for heat loss in calorimetry. The line of best fit for the cooling phase is extended back to the start of the reaction.

Common Mistakes

Plotting points incorrectly (e.g., misreading the axes).
Drawing a straight line through all points from $3.5$ to $8.0$ . The reaction has a peak; the line should curve or at least reflect the cooling trend.
Not extrapolating back to $t = 3.0$ . The extrapolation is essential to find the theoretical max temp.

Things to Be Careful About

The mark scheme requires 10 points plotted correctly and the line extrapolated to 3 minutes. Check your grid reading carefully.

Techniques used

plot temperature-time graphextrapolate line of best fit

(ii)

Use your graph to determine the theoretical temperature increase at 3.0 minutes.

1M Similar

DifficultyMedium-Easy

Worked solution

Working

Theoretical temperature at $t = 3.0$ minutes (from extrapolation) $\approx 38.5 ^\circ\text{C}$ .
Initial temperature at $t = 3.0$ minutes $= 19.0 ^\circ\text{C}$ .

\Delta T = 38.5 - 19.0 = 19.5 ^\circ\text{C}

Answer

$19.5 ^\circ\text{C}$

Final answer

19.5 °C

Detailed explanation

Background Concept

The temperature increase ( $\Delta T$ ) is the difference between the theoretical maximum temperature (corrected for heat loss) and the initial temperature before the reaction started.

Understanding the Question

The student must use the graph from part (e)(i) to find the theoretical temperature at $t = 3.0$ minutes and calculate the increase.

Approach

Read the temperature on the extrapolated line at $t = 3.0$ minutes.
Subtract the initial temperature ( $19.0 ^\circ\text{C}$ ) from this value.

Step-by-Step Reasoning

Reading the graph: The extrapolated line from the cooling curve (points 4.0 to 8.0) back to $t = 3.0$ intersects at approximately $38.5 ^\circ\text{C}$ . (Acceptable range is usually $38.0 - 39.0 ^\circ\text{C}$ ).
Calculation: $\Delta T = T_{\text{final}} - T_{\text{initial}} = 38.5 - 19.0 = 19.5 ^\circ\text{C}$ .

Key Takeaways

The theoretical $\Delta T$ is always higher than the observed maximum $\Delta T$ ( $36.0 - 19.0 = 17.0 ^\circ\text{C}$ ) because it corrects for heat lost during the reaction.

Common Mistakes

Using the observed maximum temperature ( $36.0 ^\circ\text{C}$ ) instead of the extrapolated value.
Calculating $36.0 - 19.0 = 17.0 ^\circ\text{C}$ , which is the observed increase, not the theoretical one.

Things to Be Careful About

The mark scheme gives $38.5 - 19.0 = 19.5 ^\circ\text{C}$ . Ensure your reading from the graph is consistent with this. If you read $38.0$ , your answer would be $19.0 ^\circ\text{C}$ , which might still be accepted depending on the mark scheme tolerance, but $19.5$ is the expected value.

Techniques used

read extrapolated temperature from graph

(f)

Suggest why the temperature measured at 3.5 minutes is lower than the temperature measured at 4.0 minutes.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

The reaction between $\text{CaCO}_3$ and $\text{HCl}$ is not complete at 3.5 minutes; it is still proceeding and releasing heat, so the temperature continues to rise until the reaction finishes.

Working

At $t = 3.0$ min, solid is added. At $t = 3.5$ min, the temperature is $27.5 ^\circ\text{C}$ . At $t = 4.0$ min, it is $36.0 ^\circ\text{C}$ . The temperature is still increasing, meaning the exothermic reaction is still occurring.

Answer

The reaction is not complete (at 3.5 minutes)

Final answer

The reaction is not complete at 3.5 minutes

Detailed explanation

Background Concept

When a solid is added to a liquid, the reaction rate depends on the surface area and concentration. It takes time for the reaction to reach completion. During this time, heat is continuously released.

Understanding the Question

The temperature at 3.5 minutes ( $27.5 ^\circ\text{C}$ ) is lower than at 4.0 minutes ( $36.0 ^\circ\text{C}$ ). Why?

Approach

The temperature is still rising, which means heat is still being produced by the reaction. Therefore, the reaction has not finished.

Step-by-Step Reasoning

Observation: Temp at 3.5 min is $27.5 ^\circ\text{C}$ , at 4.0 min is $36.0 ^\circ\text{C}$ .
Explanation: The reaction $\text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{CO}_2$ is exothermic. As long as reactants are present, heat is released.
Conclusion: At 3.5 minutes, there is still unreacted $\text{CaCO}_3$ or $\text{HCl}$ , so the reaction is ongoing and the temperature is still increasing.

Key Takeaways

A rising temperature in an exothermic reaction indicates the reaction is still proceeding.

Common Mistakes

Saying "heat loss is less at 4.0 minutes." Heat loss is always occurring, but the rate of heat production is greater than the rate of heat loss.
Saying "the thermometer is slow." While thermometers have a response time, the primary reason is the ongoing reaction.

Things to Be Careful About

Keep the explanation simple: the reaction is still happening / not complete.

Techniques used

explain reaction rate and completion

(g)

A student carries out experiment 2 and determines a temperature increase of $62.0\,^{\circ}\text{C}$ . The heat released by the reaction, $q$ , is given by:

q = mc\Delta T

where $m$ is the mass of $\text{HCl}(\text{aq})$ . Assume that $1.00\text{ cm}^3$ of $\text{HCl}(\text{aq})$ has a mass of $1.00\text{ g}$ and that the specific heat capacity of the solution, $c$ , is $4.18\text{ J g}^{-1}\text{ K}^{-1}$ .

Calculate $q$ , in J, for experiment 2 and hence determine $\Delta H_2$ in $\text{kJ mol}^{-1}$ .

2M Similar

DifficultyMedium

Worked solution

Working

Calculate $q$ :

q = m c \Delta T

q = 50.0 \text{ g} \times 4.18 \text{ J g}^{-1} \text{ K}^{-1} \times 62.0 \text{ K} = 12958 \text{ J}

Calculate $\Delta H_2$ :

\Delta H_2 = \frac{-q}{n} = \frac{-12958 \text{ J}}{0.0500 \text{ mol}} = -259160 \text{ J mol}^{-1} = -259 \text{ kJ mol}^{-1}

(Note: The reaction is exothermic, so $\Delta H$ is negative. The heat released $q$ is positive, but enthalpy change is negative.)

Answer

$q = 12958 \text{ J}$ ; $\Delta H_2 = -259 \text{ kJ mol}^{-1}$

Final answer

q = 12958 J; ΔH2 = -259 kJ mol^-1

Detailed explanation

Background Concept

The heat energy transferred ( $q$ ) in a solution is calculated using $q = mc\Delta T$ , where $m$ is the mass of the solution (assuming density $1.00 \text{ g cm}^{-3}$ , so $50.00 \text{ cm}^3 = 50.0 \text{ g}$ ), $c$ is the specific heat capacity ( $4.18 \text{ J g}^{-1} \text{ K}^{-1}$ ), and $\Delta T$ is the temperature change.
The enthalpy change per mole ( $\Delta H$ ) is $q / n$ , with a negative sign for exothermic reactions.

Understanding the Question

Given $\Delta T = 62.0 ^\circ\text{C}$ (which is $62.0 \text{ K}$ ), calculate $q$ and then $\Delta H_2$ for the reaction of $\text{CaO}$ with $\text{HCl}$ .

Approach

Calculate $q$ using the given formula and values.
Divide $q$ by the moles of $\text{CaO}$ ( $0.0500 \text{ mol}$ ) to get $\Delta H$ in $\text{J mol}^{-1}$ .
Convert to $\text{kJ mol}^{-1}$ and apply the correct sign (exothermic = negative).

Step-by-Step Reasoning

Mass: $50.00 \text{ cm}^3$ of $\text{HCl}$ has a mass of $50.0 \text{ g}$ (given $1.00 \text{ cm}^3 = 1.00 \text{ g}$ ). Note: we only consider the mass of the solution, not the solid, as the solid's heat capacity is usually negligible and not given.
$q$ calculation: $q = 50.0 \times 4.18 \times 62.0 = 12958 \text{ J}$ .
Moles: $n = 0.0500 \text{ mol}$ (same as experiment 1, as the procedure is repeated with the same amount of solid).
$\Delta H_2$ calculation: $\Delta H_2 = -12958 / 0.0500 = -259160 \text{ J mol}^{-1} = -259 \text{ kJ mol}^{-1}$ (to 3 s.f.).

Key Takeaways

Always check the sign of $\Delta H$ . Exothermic reactions have negative $\Delta H$ . The mass $m$ in $q=mc\Delta T$ is the mass of the solution, not the total mass of reactants (unless specified).

Common Mistakes

Forgetting to convert J to kJ ( $259160$ instead of $259$ ).
Getting the sign wrong ( $+259$ instead of $-259$ ). The question says "heat released", so $q$ is positive, but $\Delta H$ is negative.
Using the wrong mass (e.g., $50 + 5.01 = 55.01 \text{ g}$ ). The question says "assume that $1.00 \text{ cm}^3$ of $\text{HCl}(\text{aq})$ has a mass of $1.00 \text{ g}$ ", implying we only use the mass of the acid.

Things to Be Careful About

Significant figures: $62.0$ has 3 s.f., $4.18$ has 3 s.f., $50.0$ has 3 s.f. The answer should be to 3 s.f. ( $-259 \text{ kJ mol}^{-1}$ ).

Techniques used

calculate heat released using q=mcΔTcalculate enthalpy change per mole

(h)

Use the energy cycle below, your answer to (g) and the information given to determine $\Delta H_r$ for the thermal decomposition of $\text{CaCO}_3$ .

Enthalpy change for experiment 1, $\Delta H_1 = -84\text{ kJ mol}^{-1}$ .

(If you were unable to calculate a final answer in (g), assume a value of $179\text{ kJ mol}^{-1}$ . This is not the correct answer and the sign has been omitted.)

Fig. 1.2

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Worked solution

Answer

Using Hess's law from the energy cycle:

\Delta H_r = \Delta H_1 - \Delta H_2

\Delta H_r = -84 - (-259) = +175 \text{ kJ mol}^{-1}

Working

The cycle shows:
Path 1: $\text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O} + \text{CO}_2$ ( $\Delta H_1 = -84 \text{ kJ mol}^{-1}$ )
Path 2: $\text{CaCO}_3 \rightarrow \text{CaO} + \text{CO}_2$ ( $\Delta H_r$ ) then $\text{CaO} + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O}$ ( $\Delta H_2 = -259 \text{ kJ mol}^{-1}$ )

By Hess's law, the total enthalpy change is the same for both paths:

\Delta H_1 = \Delta H_r + \Delta H_2

\Delta H_r = \Delta H_1 - \Delta H_2 = -84 - (-259) = +175 \text{ kJ mol}^{-1}

Answer

$+175 \text{ kJ mol}^{-1}$

Final answer

+175 kJ mol^-1

Detailed explanation

Background Concept

Hess's law states that the total enthalpy change for a reaction is independent of the route taken. An energy cycle (or Hess's law cycle) can be used to calculate an unknown $\Delta H$ by combining known $\Delta H$ values.

Understanding the Question

The student must use the given energy cycle and the calculated $\Delta H_2$ to find $\Delta H_r$ for the decomposition of $\text{CaCO}_3$ .

Approach

Identify the paths in the cycle.
Set up the equation: Left path = Right path (or sum of arrows = 0 if going around the cycle).
Solve for $\Delta H_r$ .

Step-by-Step Reasoning

Cycle analysis:
- Top left: $\text{CaCO}_3(\text{s}) + 2\text{HCl}(\text{aq})$
- Top right: $\text{CaO}(\text{s}) + 2\text{HCl}(\text{aq}) + \text{CO}_2(\text{g})$ (this is $\text{CaCO}_3 + 2\text{HCl} - \text{H}_2\text{O} + \text{H}_2\text{O}$ , effectively the decomposition products plus HCl)
- Bottom: $\text{CaCl}_2(\text{aq}) + \text{H}_2\text{O}(\text{l}) + \text{CO}_2(\text{g})$
- Arrow 1 (left): $\text{CaCO}_3 + 2\text{HCl} \rightarrow \text{products}$ ( $\Delta H_1 = -84$ )
- Arrow 2 (top): $\text{CaCO}_3 + 2\text{HCl} \rightarrow \text{CaO} + 2\text{HCl} + \text{CO}_2$ ( $\Delta H_r$ )
- Arrow 3 (right): $\text{CaO} + 2\text{HCl} \rightarrow \text{CaCl}_2 + \text{H}_2\text{O}$ ( $\Delta H_2 = -259$ )
Equation: Going from top-left to bottom via the left arrow is the same as going top-left to top-right ( $\Delta H_r$ ) then top-right to bottom ( $\Delta H_2$ ). $\Delta H_1 = \Delta H_r + \Delta H_2$
Solve: $\Delta H_r = \Delta H_1 - \Delta H_2 = -84 - (-259) = +175 \text{ kJ mol}^{-1}$ .

Key Takeaways

In a Hess's law cycle, ensure you follow the direction of the arrows. If an arrow points the opposite way, you change the sign of $\Delta H$ .

Common Mistakes

Adding instead of subtracting: $\Delta H_r = \Delta H_1 + \Delta H_2 = -84 + (-259) = -343$ . This is wrong because the cycle requires subtraction.
Forgetting the negative sign for $\Delta H_2$ : $-84 - 259 = -343$ . This is also wrong.

Things to Be Careful About

Check the direction of the arrows in the energy cycle. The mark scheme gives the formula $\Delta H_r = \Delta H_1 - \Delta H_2$ .

Techniques used

apply Hess's law using energy cycle

(i)

Identify the main weakness of the experimental procedure and suggest one improvement to overcome this weakness. The main weakness is not the type of thermometer used.

2M Similar

DifficultyMedium

Worked solution

Answer

Main weakness: Heat loss to the surroundings (from the beaker and the solution).
Improvement: Use a lid (or insulate the beaker with polystyrene/foam cup) to reduce heat loss.

Working

In simple calorimetry using a glass beaker, heat is lost to the air, the bench, and the thermometer. This causes the measured temperature change to be lower than the theoretical value, leading to an underestimation of the enthalpy change.

Answer

See above

Final answer

Heat loss to surroundings; use a lid

Detailed explanation

Background Concept

Simple solution calorimetry (using a beaker and thermometer) is prone to heat loss. The system is not perfectly insulated. Heat escapes to the surroundings via convection, conduction, and radiation.

Understanding the Question

The student must identify the main weakness of the procedure (excluding the thermometer type) and suggest one improvement.

Approach

Think about what is lost in a beaker calorimetry experiment. Heat.
How can heat loss be reduced? Insulation, lids, better apparatus.

Step-by-Step Reasoning

Weakness: The reaction is exothermic. As the temperature rises, the rate of heat loss to the surroundings increases (Newton's law of cooling). This means the maximum temperature recorded is lower than it should be.
Improvement: To reduce heat loss, we can insulate the beaker or cover it. A lid prevents heat loss by convection (hot air rising). Polystyrene cups are better insulators than glass beakers.
Constraint: The question says "not the type of thermometer used". So don't suggest "use a more accurate thermometer".

Key Takeaways

Heat loss is the biggest source of error in simple calorimetry. Improvements focus on insulation and minimizing exposure to air.

Common Mistakes

Suggesting "repeat the experiment and take a mean." This reduces random error, but heat loss is a systematic error (it always makes $\Delta T$ too small). Repeating won't fix it.
Suggesting "use a digital thermometer." The question excludes this.
Suggesting "use more concentrated acid." This doesn't reduce heat loss; it might increase the temperature change, but the proportion of heat lost remains similar or worse.

Things to Be Careful About

The improvement must directly address the weakness. Weakness: heat loss. Improvement: lid/insulation. Pair them logically.

Techniques used

identify sources of error in calorimetrypropose improvements to reduce error

Q2Medium-HardPlanningAnalysis, Conclusions and Evaluation

This question is about an experiment to investigate the effect of temperature on the equilibrium constant, $K_1$ , of the reaction shown.

\text{Fe}^{3+}(\text{aq}) + \text{SCN}^-(\text{aq}) \rightleftharpoons \text{FeSCN}^{2+}(\text{aq})

The data collected is used to determine the value of the enthalpy change of the reaction.

To set up the equilibrium, aqueous iron(III) nitrate, $\text{Fe}(\text{NO}_3)_3(\text{aq})$ , is mixed with aqueous potassium thiocyanate, $\text{KSCN}(\text{aq})$ . Aqueous iron thiocyanate ions, $\text{FeSCN}^{2+}(\text{aq})$ , have a red colour.

A colorimeter is used to measure the absorbance of the reaction mixture. A calibration graph can then be used to determine the concentration of $\text{FeSCN}^{2+}(\text{aq})$ in the reaction mixture.

Table 2.1 shows the solutions for the experiments.

Table 2.1

solution	ion	concentration / $\text{mol dm}^{-3}$
A	$\text{SCN}^-(\text{aq})$	0.00920
B	$\text{SCN}^-(\text{aq})$	0.00200
C	$\text{Fe}^{3+}(\text{aq})$	0.00200
D	$\text{Fe}^{3+}(\text{aq})$	0.500

(a)

Describe how you would prepare $100.0\text{ cm}^3$ of solution B from solution A.

Include a calculation of the volume of solution A required for the preparation of solution B.

Give the name and capacity of any key apparatus that should be used.

Write your answer as a series of numbered steps.

3M Similar

DifficultyMedium-Easy

Worked solution

Working

C_1 V_1 = C_2 V_2

0.00920 \times V_1 = 0.00200 \times 100.0

V_1 = 21.739... \text{ cm}^3 \approx 21.75 \text{ cm}^3

Answer

Calculate the required volume of solution A as $21.75 \text{ cm}^3$ .
Use a burette to transfer $21.75 \text{ cm}^3$ of solution A into a $100 \text{ cm}^3$ volumetric flask.
Top up the volumetric flask to the $100 \text{ cm}^3$ mark with distilled water and mix thoroughly.

Final answer

Volume of A = 21.75 cm^3; use burette to transfer to 100 cm^3 volumetric flask and top up with distilled water.

Detailed explanation

Background Concept

Preparing a solution of a specific concentration from a more concentrated stock solution involves dilution. The number of moles of solute remains constant before and after dilution, so $C_1V_1 = C_2V_2$ applies. To prepare an exact volume of a diluted solution, a volumetric flask is used because it is calibrated to contain a precise volume at a specific temperature. A burette or pipette is used to transfer the concentrated solution accurately.

Understanding the Question

You need to prepare $100.0 \text{ cm}^3$ of solution B ( $0.00200 \text{ mol dm}^{-3}$ ) from solution A ( $0.00920 \text{ mol dm}^{-3}$ ). The question asks for a calculation of the volume of A needed, the name and capacity of key apparatus, and a numbered step-by-step procedure.

Approach

Use the dilution equation $C_1V_1 = C_2V_2$ to find the volume of A required.
Round the calculated volume to a value that can be measured with a burette (nearest $0.05 \text{ cm}^3$ ).
Describe the standard procedure for making a solution in a volumetric flask: transfer the measured concentrated solution, then add solvent to the mark.

Step-by-Step Reasoning

Calculation: $V_1 = \frac{C_2 V_2}{C_1} = \frac{0.00200 \times 100.0}{0.00920} = 21.739... \text{ cm}^3$ . A burette reads to $0.05 \text{ cm}^3$ , so we round to $21.75 \text{ cm}^3$ .
Apparatus: A $100 \text{ cm}^3$ volumetric flask is needed to make exactly $100.0 \text{ cm}^3$ of solution. A burette is used to measure out $21.75 \text{ cm}^3$ of the concentrated solution accurately.
Procedure: First, transfer the measured volume of A into the volumetric flask. Then, add distilled water until the bottom of the meniscus touches the $100 \text{ cm}^3$ calibration mark. Finally, stopper and invert to mix.

Key Takeaways

Dilution calculations rely on $C_1V_1 = C_2V_2$ . When preparing standard solutions, always use a volumetric flask for the final volume and a burette or pipette for the measured aliquot of the stock solution.

Common Mistakes

Forgetting to round the calculated volume to the precision of the apparatus (burette = $0.05 \text{ cm}^3$ ).
Suggesting a measuring cylinder instead of a volumetric flask; measuring cylinders are not precise enough for preparing standard solutions.
Not specifying "distilled water" for the top-up, or forgetting to mention mixing/diluting to the mark.

Things to Be Careful About

Ensure the units for volume and concentration are consistent. The final answer for volume should reflect the precision of the measuring instrument (burette), not the raw calculator output.

Techniques used

calculate dilution volume using C1V1 = C2V2describe volumetric flask preparation

(b)

Before starting the experiment, solutions B and D are used to produce a calibration graph. Known volumes of each solution are added together and the absorbance for each mixture is recorded. The calibration graph is shown in Fig. 2.1.

The concentration of solution D is much greater than the concentration of solution B in order that solution D is in excess. Suggest a reason why solution D is in excess.

Fig. 2.1

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

Solution D is in excess to ensure that all of the thiocyanate ions ( $\text{SCN}^-$ ) from solution B react completely to form $\text{FeSCN}^{2+}$ . This guarantees that the concentration of $\text{FeSCN}^{2+}$ produced is known and equal to the initial concentration of $\text{SCN}^-$ , which is necessary for constructing an accurate calibration graph.

Final answer

To ensure all SCN- reacts so [FeSCN2+] is known.

Detailed explanation

Background Concept

A calibration graph (or standard curve) relates a measurable property (like absorbance) to the concentration of an analyte. For the reaction $\text{Fe}^{3+}(\text{aq}) + \text{SCN}^-(\text{aq}) \rightleftharpoons \text{FeSCN}^{2+}(\text{aq})$ , the position of equilibrium depends on the initial concentrations. If neither reactant is in large excess, the exact concentration of the product $\text{FeSCN}^{2+}$ at equilibrium is unknown and difficult to calculate without knowing $K_1$ .

Understanding the Question

The question asks why solution D (containing $\text{Fe}^{3+}$ at $0.500 \text{ mol dm}^{-3}$ ) is used in large excess compared to solution B (containing $\text{SCN}^-$ at $0.00200 \text{ mol dm}^{-3}$ ) when making the calibration mixtures.

Approach

Consider Le Chatelier's principle. Adding a large excess of one reactant will shift the equilibrium position far to the right, effectively driving the reaction to completion. This means we can assume 100% of the limiting reactant ( $\text{SCN}^-$ ) has been converted to product ( $\text{FeSCN}^{2+}$ ).

Step-by-Step Reasoning

By making $[\text{Fe}^{3+}]$ much greater than $[\text{SCN}^-]$ , the equilibrium is forced almost entirely to the right.
Therefore, the concentration of $\text{FeSCN}^{2+}$ formed is essentially equal to the initial concentration of $\text{SCN}^-$ added.
Since we know exactly how much B was added, we know the exact concentration of $\text{FeSCN}^{2+}$ in each calibration mixture. This allows us to plot absorbance against a known concentration.

Key Takeaways

In calibration experiments involving equilibria, an excess of one reagent is used to drive the reaction to completion, establishing a known concentration of the product for the standard curve.

Common Mistakes

Stating "to make the reaction go faster"; excess reagent shifts equilibrium, it doesn't necessarily change the rate in this context.
Saying "to ensure the reaction happens"; the reaction happens anyway, but the extent depends on equilibrium position.

Things to Be Careful About

Be precise: say "all $\text{SCN}^-$ reacts" or "equilibrium is shifted to the right" so that the concentration of product is known. Avoid vague statements like "to make more product".

Techniques used

apply Le Chatelier's principle to equilibriumunderstand calibration curve requirements

(c)

The following experimental procedure is used.

step 1 Half-fill a large beaker with water at room temperature ( $25\,^{\circ}\text{C}$ ).

step 2 Transfer about $40\text{ cm}^3$ of solution B into a boiling tube and place the boiling tube in the beaker of water.

step 3 Transfer $5.00\text{ cm}^3$ of solution C into a test-tube and place the test-tube in the beaker of water.

step 4 Wait for 10 minutes.

step 5 Transfer $5.00\text{ cm}^3$ of solution B from the boiling tube to the test-tube containing solution C. Stir the mixture in the test-tube and record the temperature of the mixture.

step 6 Measure the absorbance of the mixture in the test-tube using the colorimeter.

Change the temperature of the water in the beaker and repeat steps 3 to 6 for different temperatures.

(i)

Identify the dependent variable.

1M Similar

(ii)

Describe how you would adjust the temperature of the water in the large beaker to obtain a temperature of $10\,^{\circ}\text{C}$ .

1M Similar

(d)

A student obtains the results given in Table 2.2.

Table 2.2

1	2	3	4
temperature / $^{\circ}\text{C}$	relative absorbance	$[\text{FeSCN}^{2+}] / 10^{-5}\text{ mol dm}^{-3}$	$K_1$
25	0.60
55	0.42

The value of the equilibrium constant, $K_1$ , can be determined using equation 1.

\text{equation 1} \quad K_1 = \frac{x}{(0.0010 - x)^2}

$x$ is the value of $[\text{FeSCN}^{2+}]$ in $\text{mol dm}^{-3}$ .

(i)

Use the calibration graph in Fig. 2.1 to complete column 3 in Table 2.2. Record values to one decimal place.

1M Similar

DifficultyMedium-Easy

Worked solution

Working

From Fig. 2.1 (calibration graph):

For relative absorbance = 0.60, read across to the line of best fit and down to the x-axis: $[\text{FeSCN}^{2+}] / 10^{-5} \text{ mol dm}^{-3} \approx 8.9$ .
For relative absorbance = 0.42, read across to the line of best fit and down to the x-axis: $[\text{FeSCN}^{2+}] / 10^{-5} \text{ mol dm}^{-3} \approx 6.2$ .

Answer

1	2	3
temperature / $^\circ\text{C}$	relative absorbance	$[\text{FeSCN}^{2+}] / 10^{-5} \text{ mol dm}^{-3}$
25	0.60	8.9
55	0.42	6.2

Final answer

8.9 and 6.2

Detailed explanation

Background Concept

A calibration graph plots a measurable signal (absorbance) against a known concentration. To find an unknown concentration from a measured signal, you read the signal on the y-axis, move horizontally to the line of best fit, and then read down to the x-axis to find the concentration. The x-axis label often includes a multiplier (e.g., $\times 10^{-5}$ ), which must be accounted for.

Understanding the Question

You are given absorbance values (0.60 and 0.42) and must use Fig. 2.1 to find the corresponding $[\text{FeSCN}^{2+}]$ values to fill in column 3. The values must be recorded to one decimal place.

Approach

Locate 0.60 on the y-axis (relative absorbance).
Move horizontally to the straight line of best fit.
Move vertically down to the x-axis and read the value.
Repeat for 0.42.

Step-by-Step Reasoning

For absorbance 0.60: The y-axis has major lines every 0.2 and minor lines every 0.02. 0.60 is a major line. Following across to the line and down, the x-axis value is between 8 and 10. The major lines are every 2, minor every 0.2. The reading is approximately 8.9. (Acceptable range 8.8–9.0).
For absorbance 0.42: 0.42 is one minor line above 0.40. Following across and down, the x-axis value is just above 6.0. The reading is approximately 6.2. (Acceptable range 6.1–6.3).
The column header is $[\text{FeSCN}^{2+}] / 10^{-5} \text{ mol dm}^{-3}$ , so the numbers 8.9 and 6.2 are the values to enter directly.

Key Takeaways

Reading graphs requires aligning with the line of best fit, not necessarily the data points (though here only the line is shown). Pay attention to axis multipliers and required decimal places.

Common Mistakes

Reading from the wrong axis (e.g., reading absorbance from the x-axis).
Not rounding to the required one decimal place.
Forgetting the $10^{-5}$ multiplier and writing $8.9 \times 10^{-5}$ in the box, but the column header already divides by $10^{-5}$ , so just write 8.9.

Things to Be Careful About

The graph shows a line of best fit, not individual data points for the calibration. Use the line. Ensure your reading is to one decimal place as requested.

Techniques used

read values from a calibration graph

(ii)

Use equation 1 to complete column 4 in Table 2.2. Record values to the nearest whole number.

1M Similar

DifficultyMedium

Worked solution

Working

Equation 1: $K_1 = \frac{x}{(0.0010 - x)^2}$ , where $x = [\text{FeSCN}^{2+}]$ in $\text{mol dm}^{-3}$ .

For 25 °C:
$x = 8.9 \times 10^{-5} \text{ mol dm}^{-3} = 0.000089 \text{ mol dm}^{-3}$

K_1 = \frac{0.000089}{(0.0010 - 0.000089)^2} = \frac{0.000089}{(0.000911)^2} = \frac{0.000089}{8.299 \times 10^{-7}} \approx 107.2

Rounded to nearest whole number: 107

For 55 °C:
$x = 6.2 \times 10^{-5} \text{ mol dm}^{-3} = 0.000062 \text{ mol dm}^{-3}$

K_1 = \frac{0.000062}{(0.0010 - 0.000062)^2} = \frac{0.000062}{(0.000938)^2} = \frac{0.000062}{8.798 \times 10^{-7}} \approx 70.47

Rounded to nearest whole number: 70

Answer

3	4
$[\text{FeSCN}^{2+}] / 10^{-5} \text{ mol dm}^{-3}$	$K_1$
8.9	107
6.2	70

Final answer

107 and 70

Detailed explanation

Background Concept

The equilibrium constant $K_c$ (or $K_1$ here) is calculated from the equilibrium concentrations of products and reactants. For the reaction $\text{Fe}^{3+} + \text{SCN}^- \rightleftharpoons \text{FeSCN}^{2+}$ , the expression is $K_1 = \frac{[\text{FeSCN}^{2+}]}{[\text{Fe}^{3+}][\text{SCN}^-]}$ .

In this experiment, the initial concentrations after mixing equal volumes (5.00 cm³ of B + 5.00 cm³ of C) are halved. Initial $[\text{SCN}^-] = \frac{0.00200}{2} = 0.0010 \text{ mol dm}^{-3}$ . Initial $[\text{Fe}^{3+}] = \frac{0.00200}{2} = 0.0010 \text{ mol dm}^{-3}$ .

At equilibrium, $[\text{FeSCN}^{2+}] = x$ . Then $[\text{SCN}^-] = 0.0010 - x$ and $[\text{Fe}^{3+}] = 0.0010 - x$ . Thus, $K_1 = \frac{x}{(0.0010 - x)^2}$ , which is equation 1.

Understanding the Question

You must calculate $K_1$ for two temperatures using the concentrations found in part (d)(i) and the given equation. Values must be rounded to the nearest whole number.

Approach

Convert the $[\text{FeSCN}^{2+}]$ values from column 3 (which are in units of $10^{-5} \text{ mol dm}^{-3}$ ) to $\text{mol dm}^{-3}$ by multiplying by $10^{-5}$ . This is $x$ .
Substitute $x$ into equation 1.
Calculate the result and round to the nearest whole number.

Step-by-Step Reasoning

25 °C: $x = 8.9 \times 10^{-5} = 0.000089$ . Denominator: $(0.0010 - 0.000089)^2 = (0.000911)^2 = 8.29921 \times 10^{-7}$ . $K_1 = 0.000089 / 8.29921 \times 10^{-7} = 107.24... \rightarrow 107$ .
55 °C: $x = 6.2 \times 10^{-5} = 0.000062$ . Denominator: $(0.0010 - 0.000062)^2 = (0.000938)^2 = 8.79844 \times 10^{-7}$ . $K_1 = 0.000062 / 8.79844 \times 10^{-7} = 70.466... \rightarrow 70$ .

Key Takeaways

When using equilibrium expressions, ensure all concentrations are in consistent units ( $\text{mol dm}^{-3}$ ). Be careful with scientific notation and calculator entry to avoid order-of-magnitude errors.

Common Mistakes

Forgetting to convert from $10^{-5}$ to standard decimal form before substituting into the equation.
Using the values 8.9 and 6.2 directly without the $10^{-5}$ factor.
Rounding errors during intermediate steps; keep full calculator values until the final division.

Things to Be Careful About

The equation gives $K_1$ directly. Check significant figures/rounding: "nearest whole number". 107.2 rounds to 107; 70.5 rounds to 70 (or 71 depending on rounding convention, but 70.47 is clearly 70).

Techniques used

calculate equilibrium constant from concentrations

(e)

Another student does the same experiment for seven different temperatures, plots a graph and draws the line of best fit, as shown in Fig. 2.2.

Fig. 2.2

Theory predicts that the relationship between $K_1$ and $T$ is given by equation 2.

\text{equation 2} \quad \log K_1 = \frac{-\Delta H}{2.303 R T} + \text{constant}

$\Delta H$ is the enthalpy change of reaction and $T$ is the temperature in Kelvin.

(i)

Explain why the graph supports the relationship between $K_1$ and $T$ given in equation 2.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

Equation 2 is of the form $y = mx + c$ (a linear equation), where $y = \log K_1$ and $x = 1/T$ . The graph of $\log K_1$ against $1/T$ is a straight line, which supports the relationship given in equation 2.

Final answer

The graph is a straight line.

Detailed explanation

Background Concept

The van 't Hoff equation relates the equilibrium constant to temperature: $\ln K = -\frac{\Delta H}{RT} + \frac{\Delta S}{R}$ . Converting to base-10 logarithm gives $\log K = -\frac{\Delta H}{2.303RT} + \text{constant}$ . This is a linear equation of the form $y = mx + c$ , where $y = \log K$ , $x = 1/T$ , and the gradient $m = -\frac{\Delta H}{2.303R}$ .

Understanding the Question

You are given equation 2 and a graph of $\log K_1$ against $1/T$ . You must explain why the graph supports the equation.

Approach

Compare the mathematical form of equation 2 to the standard equation of a straight line ( $y = mx + c$ ). If the variables on the graph axes correspond to $y$ and $x$ in the equation, and the plot is linear, the relationship is supported.

Step-by-Step Reasoning

Equation 2: $\log K_1 = \left(-\frac{\Delta H}{2.303R}\right) \left(\frac{1}{T}\right) + \text{constant}$ .
Let $y = \log K_1$ and $x = 1/T$ . Then $y = mx + c$ , where $m = -\frac{\Delta H}{2.303R}$ and $c = \text{constant}$ .
The graph plots $y$ against $x$ and yields a straight line of best fit.
A straight line graph confirms the linear relationship predicted by equation 2.

Key Takeaways

Linearizing a non-linear relationship (like $K$ vs $T$ ) by plotting transformed variables (like $\log K$ vs $1/T$ ) is a standard technique to verify a theoretical equation and extract parameters (like $\Delta H$ ) from the gradient.

Common Mistakes

Saying "the points lie on the line"; they don't perfectly, but the line of best fit is straight.
Not explicitly stating that the equation represents a straight line ( $y = mx + c$ form).

Things to Be Careful About

Keep the explanation concise. The key point is that the graph is a straight line, matching the linear form of the equation.

Techniques used

relate graph shape to theoretical equation

(ii)

Circle the point on the graph in Fig. 2.2 that you consider to be most anomalous.

There were no errors in the measurements in the experiment.

A student correctly suggests that the anomaly was caused because the absorbance was lower than expected by the line of best fit. Suggest why the absorbance was lower than expected.

2M Similar

DifficultyMedium

Worked solution

Answer

Anomalous point: The point at approximately $1/T = 0.0035 \text{ K}^{-1}$ , $\log K_1 \approx 2.05$ (the point below the line of best fit on the right side).

Reason for lower absorbance: Not all of solution B was transferred from the boiling tube to the test-tube in step 5 (e.g., some was left behind or spilled). This means the initial concentration of $\text{SCN}^-$ was lower than expected, resulting in a lower concentration of $\text{FeSCN}^{2+}$ produced and thus a lower absorbance.

Final answer

Point at 1/T ≈ 0.0035, log K1 ≈ 2.05; incomplete transfer of solution B.

Detailed explanation

Background Concept

An anomalous result (outlier) is a data point that does not fit the general trend of the data. In a graph of best fit, it is a point that lies far from the line. If measurements are correct, anomalies are usually due to random errors in technique or procedure, not systematic errors.

Understanding the Question

You must identify the anomalous point on Fig. 2.2 and suggest a procedural reason why the absorbance (and thus $K_1$ ) was lower than expected for that temperature. The question states there were no measurement errors (e.g., wrong thermometer reading), so it must be a technique error.

Approach

Look at Fig. 2.2 and find the point furthest from the line of best fit. The point at $1/T \approx 0.0035$ ( $T \approx 286 \text{ K} = 13^\circ\text{C}$ ) is clearly below the line.
Lower absorbance means lower $[\text{FeSCN}^{2+}]$ . Since $K_1$ is lower, the equilibrium position is further to the left, or simply less product was formed.
Think about the procedure: mixing solutions B and C. If less B was added, less reactant is available, so less product forms.

Step-by-Step Reasoning

Identifying the point: At $1/T = 0.0035$ , the line of best fit is at $\log K_1 \approx 2.12$ , but the data point is at $\approx 2.05$ . This is the most anomalous point.
Why lower absorbance? Absorbance is proportional to $[\text{FeSCN}^{2+}]$ . Lower absorbance means lower product concentration.
Procedural error: In step 5, $5.00 \text{ cm}^3$ of solution B is transferred from the boiling tube to the test-tube. If the transfer is incomplete (e.g., some solution B remains in the boiling tube or on the sides), the actual amount of $\text{SCN}^-$ added is less than $5.00 \text{ cm}^3$ .
Less $\text{SCN}^-$ means less $\text{FeSCN}^{2+}$ is formed at equilibrium, leading to lower absorbance and a lower calculated $K_1$ .

Key Takeaways

Anomalies in equilibrium experiments can arise from incomplete mixing, incomplete transfer of reactants, or failure to reach equilibrium temperature. Always link the observed error (low absorbance) to a specific procedural step.

Common Mistakes

Circling a point that is actually on the line or close to it.
Suggesting "the colorimeter was not calibrated"; the question says no measurement errors.
Saying "evaporation"; while possible, incomplete transfer is a more direct and common error in this specific step.

Things to Be Careful About

The point to circle is the one visibly below the line on the right side (around $1/T = 0.0035$ ). The reason must relate to the concentration of reactant or product being lower than expected due to technique.

Techniques used

identify anomalous data pointevaluate experimental technique errors

(iii)

Determine the gradient of the line of best fit in Fig. 2.2. State the coordinates of both points you use in your calculation. These must be selected from the line of best fit. Give the gradient to three significant figures.

2M Similar

DifficultyMedium

Worked solution

Working

Select two points on the line of best fit (far apart to minimize error):
Point 1: $(x_1, y_1) = (0.00300, 1.80)$
Point 2: $(x_2, y_2) = (0.00360, 2.16)$

\text{Gradient} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2.16 - 1.80}{0.00360 - 0.00300} = \frac{0.36}{0.00060} = 600

Using other valid points, e.g., $(0.00290, 1.74)$ and $(0.00350, 2.10)$ :

\text{Gradient} = \frac{2.10 - 1.74}{0.00350 - 0.00290} = \frac{0.36}{0.00060} = 600

Answer

Coordinates: $(0.00300, 1.80)$ and $(0.00360, 2.16)$
Gradient = 600

Final answer

Gradient = 600 (using coordinates such as (0.00300, 1.80) and (0.00360, 2.16))

Detailed explanation

Background Concept

The gradient of a line is calculated as $\frac{\Delta y}{\Delta x}$ . For a graph representing a theoretical equation, the gradient often contains physical constants or thermodynamic quantities that need to be extracted. To minimize reading errors, always use two points that are far apart on the line of best fit, not the data points themselves (which may contain anomalies).

Understanding the Question

You must determine the gradient of the line of best fit in Fig. 2.2. You must state the coordinates of the two points used, and give the gradient to 3 significant figures.

Approach

Identify two points on the straight line of best fit (not the 'x' data points).
Read their $(x, y)$ coordinates from the axes.
Calculate $\frac{\Delta y}{\Delta x}$ .

Step-by-Step Reasoning

Selecting points: Look for intersections of the line with grid lines.
- At $1/T = 0.00300$ , $\log K_1 = 1.80$ . This is a clear intersection.
- At $1/T = 0.00360$ , $\log K_1 = 2.16$ . This is also a clear intersection.
Calculation:
$\Delta y = 2.16 - 1.80 = 0.36$
$\Delta x = 0.00360 - 0.00300 = 0.00060$
$\text{Gradient} = \frac{0.36}{0.00060} = 600$
Significant figures: The coordinates are read to 3 or 4 decimal places. The gradient 600 can be written as $6.00 \times 10^2$ to show 3 sig figs, or simply 600 if unambiguous. The mark scheme accepts 600.

Key Takeaways

When calculating gradient from a graph, use points on the line of best fit, ideally at the extremes of the line, to reduce the effect of reading errors. Always state the coordinates used.

Common Mistakes

Using data points (the 'x' marks) instead of points on the line of best fit. The anomalous point will give a wrong gradient.
Not reading the x-axis correctly (e.g., reading 0.0030 as 0.030).
Calculating $\Delta x / \Delta y$ (inverse gradient).

Things to Be Careful About

The x-axis is $1/T$ in $\text{K}^{-1}$ , ranging from 0.0028 to 0.0038. Minor grid lines are every 0.00002. Ensure you read the full decimal value. The mark scheme requires coordinates to be stated.

Techniques used

calculate gradient of a line of best fit from a graph

(iv)

Use the gradient calculated in (e)(iii) and equation 2 to calculate a value for the enthalpy change of reaction, $\Delta H$ .

\text{equation 2} \quad \log K_1 = \frac{-\Delta H}{2.303 R T} + \text{constant}

(If you were unable to obtain an answer to (e)(iii), then use the value $635\text{ K}$ . This is not the correct answer.)

$\Delta H$ = .................... kJ mol $^{-1}$

2M Similar

DifficultyMedium-Hard

Worked solution

Working

Equation 2: $\log K_1 = \frac{-\Delta H}{2.303 R T} + \text{constant}$

This is in the form $y = mx + c$ , where the gradient $m = \frac{-\Delta H}{2.303 R}$ .

\text{Gradient} = \frac{-\Delta H}{2.303 \times R}

600 = \frac{-\Delta H}{2.303 \times 8.31}

-\Delta H = 600 \times 2.303 \times 8.31

-\Delta H = 11483.5 \text{ J mol}^{-1}

\Delta H = -11483.5 \text{ J mol}^{-1} = -11.5 \text{ kJ mol}^{-1}

Rounding to appropriate figures (or nearest whole number as implied by context, though 3 sig figs is standard): $\Delta H \approx -11.5 \text{ kJ mol}^{-1}$ .

Answer

$\Delta H =$ -11.5 $\text{ kJ mol}^{-1}$ (or -11.5 to 3 s.f.)

Final answer

-11.5 kJ mol^-1

Detailed explanation

Background Concept

The van 't Hoff equation in logarithmic form is $\log K = -\frac{\Delta H}{2.303 R} \cdot \frac{1}{T} + \frac{\Delta S}{2.303 R}$ . Plotting $\log K$ against $1/T$ gives a straight line with gradient $m = -\frac{\Delta H}{2.303 R}$ . The gas constant $R = 8.31 \text{ J K}^{-1} \text{ mol}^{-1}$ .

Understanding the Question

You must use the gradient calculated in part (e)(iii) and equation 2 to find $\Delta H$ . The answer must be in $\text{kJ mol}^{-1}$ .

Approach

Equate the gradient from the graph to the theoretical expression for the gradient: $\text{gradient} = -\frac{\Delta H}{2.303 R}$ .
Rearrange to solve for $\Delta H$ .
Substitute the gradient (600) and $R = 8.31$ .
Calculate $\Delta H$ in $\text{J mol}^{-1}$ and convert to $\text{kJ mol}^{-1}$ .

Step-by-Step Reasoning

Rearrangement:
$\Delta H = -\text{gradient} \times 2.303 \times R$
Substitution:
$\Delta H = -600 \times 2.303 \times 8.31$
Calculation:
$\Delta H = -600 \times 19.13853 = -11483.118 \text{ J mol}^{-1}$
Unit conversion: Divide by 1000 to get kJ:
$\Delta H = -11.483... \text{ kJ mol}^{-1}$
Rounding to 3 significant figures: $-11.5 \text{ kJ mol}^{-1}$ .

Key Takeaways

The gradient of a $\log K$ vs $1/T$ graph directly gives the enthalpy change of reaction. Remember the factor of 2.303 (which converts natural log to base-10 log) and the sign (negative gradient means exothermic if $\Delta H$ is negative, wait: gradient is positive here, so $-\Delta H$ is positive, meaning $\Delta H$ is negative, so exothermic). Let's check: gradient = 600 (positive). $600 = -\Delta H / (2.303 R) \Rightarrow -\Delta H = \text{positive} \Rightarrow \Delta H = \text{negative}$ . Yes, exothermic. This makes sense: as $1/T$ increases (T decreases), $\log K_1$ increases, so $K_1$ increases. Lower temperature favors product, so forward reaction is exothermic.

Common Mistakes

Forgetting the 2.303 factor (using $\ln$ instead of $\log$ ).
Forgetting the negative sign in the rearrangement.
Not converting from J to kJ (giving -11483 instead of -11.5).
Using the wrong value for $R$ (e.g., 8.314 is fine, but 8.31 is standard in CIE; 0.0821 is for $K_p$ with pressure in atm).

Things to Be Careful About

The question asks for the answer in $\text{kJ mol}^{-1}$ . The calculation using $R = 8.31$ gives J mol $^{-1}$ . You must divide by 1000. Also, ensure the sign is correct: a positive gradient on a $\log K$ vs $1/T$ plot indicates an exothermic reaction ( $\Delta H < 0$ ).

Techniques used

relate graph gradient to thermodynamic equationcalculate enthalpy change from gradient