Chemistry 9701/41 — October/November 2017

Background Concept

The rate equation (or rate law) expresses the relationship between the rate of a reaction and the concentrations of the reactants. It has the general form $\text{rate} = k[\text{A}]^m[\text{B}]^n$, where $k$ is the rate constant, and $m$ and $n$ are the orders of reaction with respect to reactants A and B. The overall order is $m + n$.

Understanding the Question

The question asks for the definition of 'order of reaction' with respect to a particular reagent. This is a straightforward recall question.

Approach

State the standard definition: the exponent (power) to which the concentration term is raised in the rate equation.

Step-by-Step Reasoning

• The rate equation is $\text{rate} = k[\text{NO}]^a[\text{Br}_2]^b$.
• The term $[\text{NO}]^a$ shows that the concentration of $\text{NO}$ is raised to the power $a$.
• Therefore, $a$ is the order of reaction with respect to $\text{NO}$.
• Definition: The order of reaction with respect to a reactant is the power to which its concentration is raised in the rate equation.

Key Takeaways

The order of reaction is an experimental quantity determined from the rate equation, not necessarily related to the stoichiometric coefficients in the balanced equation.

Common Mistakes

• Confusing 'order of reaction' with 'molecularity' (which is a theoretical concept for elementary steps).
• Saying 'the number of molecules' or 'the coefficient in the equation'.

Things to Be Careful About

Use precise terminology: 'power' or 'exponent', not 'number' or 'coefficient'. Must mention 'rate equation'.

Background Concept

To determine the order of reaction with respect to each reactant, we use the method of initial rates. We compare two experiments where the concentration of one reactant changes while the concentration of the other(s) remains constant. The ratio of the rates equals the ratio of the concentrations raised to the power of their respective orders.

Understanding the Question

We are given a table of initial rates for different concentrations of $\text{NO}$ and $\text{Br}_2$. We need to deduce the values of $a$ and $b$ in the rate equation $\text{rate} = k[\text{NO}]^a[\text{Br}_2]^b$.

Approach

1. Find $b$ (order w.r.t $\text{Br}_2$): Look for two experiments where $[\text{NO}]$ is the same. Experiments 1 and 2 have $[\text{NO}] = 0.03$. $[\text{Br}_2]$ doubles ($0.02 \to 0.04$). Rate doubles ($3.4 \to 6.8$). So order is 1.
2. Find $a$ (order w.r.t $\text{NO}$): Look for two experiments where $[\text{Br}_2]$ is the same. Experiments 2 and 3 have $[\text{Br}_2] = 0.04$. $[\text{NO}]$ triples ($0.03 \to 0.09$). Rate increases by a factor of $9$ ($6.8 \times 10^{-3} \to 6.1 \times 10^{-2}$). So order is 2.

Step-by-Step Reasoning

• Finding $b$:
  $\frac{\text{rate}_2}{\text{rate}_1} = \left(\frac{[\text{Br}_2]_2}{[\text{Br}_2]_1}\right)^b$
  $\frac{6.8 \times 10^{-3}}{3.4 \times 10^{-3}} = \left(\frac{0.04}{0.02}\right)^b$
  $2 = 2^b \implies b = 1$

• Finding $a$:
  $\frac{\text{rate}_3}{\text{rate}_2} = \left(\frac{[\text{NO}]_3}{[\text{NO}]_2}\right)^a$
  $\frac{6.1 \times 10^{-2}}{6.8 \times 10^{-3}} = \left(\frac{0.09}{0.03}\right)^a$
  $8.97 \approx 9 = 3^a \implies a = 2$

• Conclusion: The rate equation is $\text{rate} = k[\text{NO}]^2[\text{Br}_2]^1$.

Key Takeaways

Always isolate the variable you want to study by keeping other concentrations constant. Use ratios to eliminate the rate constant $k$.

Common Mistakes

• Comparing experiments where both concentrations change (e.g., 1 and 3), which makes it impossible to isolate the effect of one reactant.
• Arithmetic errors in calculating the ratio of rates or concentrations.

Things to Be Careful About

The rate in experiment 3 is $6.1 \times 10^{-2}$, which is $0.061$. $0.061 / 0.0068 = 8.97$, which is approximately $9$. Recognise that $9 = 3^2$, so the order is exactly $2$. Don't get bogged down in the slight discrepancy due to experimental error; use the nearest integer power.

Background Concept

Once the rate equation and the rate constant $k$ are known, we can calculate the rate for any set of concentrations by simple substitution.

Understanding the Question

We need to calculate the initial rate for experiment 4, where $[\text{NO}] = 0.12$ and $[\text{Br}_2] = 0.06$. We can use the rate constant $k$ calculated in part (iv), or we can use a ratio method comparing experiment 4 to another experiment (e.g., experiment 2).

Approach

Method 1: Use $k = 188.9$ and substitute into $\text{rate} = k[\text{NO}]^2[\text{Br}_2]$.
Method 2: Compare experiment 4 to experiment 2. $[\text{NO}]$ increases by factor of $4$ ($0.03 \to 0.12$), so rate increases by $4^2 = 16$. $[\text{Br}_2]$ increases by factor of $1.5$ ($0.04 \to 0.06$), so rate increases by $1.5$. Total factor = $16 \times 1.5 = 24$. New rate = $24 \times 6.8 \times 10^{-3}$.

Step-by-Step Reasoning

• Using $k$:
  $\text{rate} = 188.9 \times (0.12)^2 \times 0.06$
  $= 188.9 \times 0.0144 \times 0.06$
  $= 188.9 \times 0.000864$
  $= 0.1632 \text{ mol dm}^{-3}\text{ s}^{-1}$

• Using ratios:
  $\text{rate}_4 = \text{rate}_2 \times \left(\frac{[\text{NO}]_4}{[\text{NO}]_2}\right)^2 \times \left(\frac{[\text{Br}_2]_4}{[\text{Br}_2]_2}\right)$
  $= 6.8 \times 10^{-3} \times \left(\frac{0.12}{0.03}\right)^2 \times \left(\frac{0.06}{0.04}\right)$
  $= 6.8 \times 10^{-3} \times 16 \times 1.5$
  $= 6.8 \times 10^{-3} \times 24$
  $= 0.1632 \text{ mol dm}^{-3}\text{ s}^{-1}$

• Rounding to 2 significant figures (consistent with the data): $0.16 \text{ mol dm}^{-3}\text{ s}^{-1}$.

Key Takeaways

You can calculate rates either by direct substitution using $k$, or by using proportional reasoning from a known experiment. Both methods should give the same result.

Common Mistakes

• Forgetting to square the $[\text{NO}]$ term.
• Using the wrong concentration values.
• Not rounding to the correct number of significant figures.

Things to Be Careful About

The mark scheme gives $0.16(32)$, meaning $0.1632$ is acceptable, but $0.16$ is the expected answer to 2 s.f. Ensure units are correct: $\text{mol dm}^{-3}\text{ s}^{-1}$.

Background Concept

The rate constant $k$ is specific to a reaction at a given temperature. It can be calculated by rearranging the rate equation: $k = \frac{\text{rate}}{[\text{A}]^m[\text{B}]^n}$. The units of $k$ depend on the overall order of the reaction. For an overall order $n$, the units are $(\text{concentration})^{1-n} \text{ time}^{-1}$.

Understanding the Question

We need to calculate the numerical value and units of $k$ using data from experiment 1. The rate equation is $\text{rate} = k[\text{NO}]^2[\text{Br}_2]^1$ (overall order = 3).

Approach

1. Rearrange the rate equation to solve for $k$.
2. Substitute the values from experiment 1.
3. Calculate the numerical value.
4. Determine the units by substituting the units into the rearranged equation.

Step-by-Step Reasoning

• Numerical value:
  $k = \frac{3.4 \times 10^{-3}}{(0.03)^2 \times 0.02} = \frac{3.4 \times 10^{-3}}{0.0009 \times 0.02} = \frac{3.4 \times 10^{-3}}{1.8 \times 10^{-5}} = 188.88... \approx 189$

• Units:
  $k = \frac{\text{rate}}{[\text{NO}]^2[\text{Br}_2]}$
  Units of $k = \frac{\text{mol dm}^{-3}\text{ s}^{-1}}{(\text{mol dm}^{-3})^2 \times (\text{mol dm}^{-3})}$
  $= \frac{\text{mol dm}^{-3}\text{ s}^{-1}}{\text{mol}^3 \text{ dm}^{-9}}$
  $= \text{mol}^{1-3} \text{ dm}^{-3-(-9)} \text{ s}^{-1}$
  $= \text{mol}^{-2} \text{ dm}^6 \text{ s}^{-1}$

Key Takeaways

Always include units for $k$. Derive them systematically by substituting the units of rate and concentration into the rearranged rate equation.

Common Mistakes

• Forgetting to square the $[\text{NO}]$ term in the denominator.
• Incorrect unit algebra (e.g., $\text{dm}^{-3} / \text{dm}^{-9} = \text{dm}^{-6}$ instead of $\text{dm}^6$).
• Rounding too early or to the wrong number of significant figures.

Things to Be Careful About

The mark scheme accepts $188.9$ or $189$. Use at least 3 significant figures. The units must be exactly $\text{mol}^{-2} \text{ dm}^6 \text{ s}^{-1}$.

Background Concept

The rate constant $k$ is temperature-dependent. According to the Arrhenius equation, $k$ increases exponentially with increasing temperature. Conversely, $k$ decreases as temperature decreases. The rate of reaction is directly proportional to $k$ (for a given set of concentrations).

Understanding the Question

We are asked to explain why the rate decreases with decreasing temperature, specifically by considering the rate equation $\text{rate} = k[\text{NO}]^2[\text{Br}_2]$.

Approach

1. State the effect of temperature on $k$.
2. Relate $k$ to the rate using the rate equation.

Step-by-Step Reasoning

• As temperature decreases, the rate constant $k$ decreases (fewer molecules have energy $\geq$ activation energy).
• In the rate equation $\text{rate} = k[\text{NO}]^2[\text{Br}_2]$, if $k$ decreases (and concentrations are held constant), the rate must also decrease.

Key Takeaways

Temperature affects the rate constant $k$, not the concentrations (unless the system is not at equilibrium and concentrations change over time, but here we are talking about initial rates). The rate is directly proportional to $k$.

Common Mistakes

• Saying 'the rate constant is constant' (it's only constant at a constant temperature).
• Explaining in terms of collision frequency or activation energy without linking back to $k$ and the rate equation as requested.

Things to Be Careful About

The question specifically says 'By considering the rate equation'. You must mention that $k$ decreases. Don't just say 'molecules have less energy' without connecting it to $k$ and then to the rate.