Chemistry 9701/14 — May/June 2025

Background Concept

Ionisation energy is the energy required to remove an electron from a gaseous atom or ion. Successive ionisation energies increase because each subsequent electron is removed from an increasingly positive ion, experiencing a stronger electrostatic attraction to the nucleus. However, a very large increase (a "jump") in ionisation energy occurs when an electron is removed from a new, inner principal quantum shell (energy level). This inner electron is closer to the nucleus, experiences less shielding from inner electrons, and is therefore held much more strongly. By identifying where these large jumps occur in a graph of successive ionisation energies, we can deduce the number of electrons in each shell and thus the full electronic configuration of the element.

Understanding the Question

The question provides a graph of the logarithm of the first 13 ionisation energies of an unknown element plotted against the number of electrons removed (from 1 to 13). We must use the pattern of the graph to identify the element or a correct property of it from the four given options. The logarithmic scale is used to compress the large range of ionisation energy values onto a readable graph, but the relative positions of the jumps remain the same.

Approach

We will locate the large jumps in the ionisation energy values on the graph. The position of the first large jump tells us the number of outer-shell electrons. The position of subsequent large jumps tells us the number of electrons in the inner shells. This allows us to deduce the full electronic configuration and identify the element. Then we evaluate each option to find the correct statement.

Step-by-Step Reasoning

1. First jump (outer shell): The graph shows a gradual increase from the 1st to the 4th ionisation energy, followed by a larger jump to the 5th. This indicates that the first 4 electrons are removed from the outermost shell, and the 5th electron is removed from a shell closer to the nucleus. Therefore, the element has 4 electrons in its outermost shell. This immediately eliminates option B (aluminium has 3 outer electrons, configuration 2, 8, 3, which would show a jump after the 3rd IE) and option D (the element does not have only one outer electron).
2. Second jump (inner shell): The graph shows a gradual increase from the 5th to the 12th ionisation energy, followed by a larger jump to the 13th. This indicates that electrons 5 through 12 (a total of 8 electrons) are removed from the next inner shell, and the 13th electron is removed from an even deeper shell. Thus, the second shell contains 8 electrons.
3. Innermost shell: Since the 13th electron is being removed from a new shell, and the first shell can hold a maximum of 2 electrons, the element must have 2 electrons in its first shell. (The graph only goes up to 13, so we only see the removal of 1 electron from the first shell, but the jump at 13 confirms it is the start of the n=1 shell).
4. Full configuration and identification: Combining these findings, the electronic configuration is 2, 8, 4. The total number of electrons is 14, which corresponds to silicon (atomic number 14). This confirms option C is correct.
5. Evaluating option A: Ionisation is the process of removing electrons from atoms to form positive ions. Protons are located in the nucleus and are not removed during chemical ionisation processes. Thus, option A is fundamentally incorrect.

Key Takeaways

• A large jump in successive ionisation energies indicates the removal of an electron from a new, inner principal quantum shell.
• The number of electrons in the outermost shell equals the ionisation number just before the first large jump.
• The capacity of each shell (2, 8, 8, ...) can be deduced from the number of electrons removed between successive jumps.

Common Mistakes

• Confusing protons and electrons: Option A is a distractor for students who do not recall that ionisation involves the loss of electrons, not protons.
• Misreading the graph: Students might count the points incorrectly or fail to identify the "jumps" versus the "gradual increases". Remember to look for the steepest segments of the graph.
• Assuming the total number of electrons: The graph only shows the first 13 ionisation energies. Students must deduce that the first shell holds 2 electrons even though only 1 is shown being removed (the jump at 13 implies the start of the n=1 shell, which holds 2).

Things to Be Careful About

• Logarithmic scale: The y-axis is logarithmic. This does not change the relative positions of the jumps, but students should not try to read the actual ionisation energy values from the axis unless specifically asked.
• State symbols and terminology: When writing about ionisation, always specify that electrons are removed, not protons or neutrons.
• Aluminium vs Silicon: Aluminium (Group 13) has 3 outer electrons and shows a jump after IE3. Silicon (Group 14) has 4 outer electrons and shows a jump after IE4. Phosphorus (Group 15) has 5 and shows a jump after IE5. Matching the jump position to the group number is a reliable check.

Background Concept

The empirical formula gives the simplest whole-number ratio of atoms of each element in a compound. The molecular formula gives the actual number of atoms of each element in one molecule. For many simple compounds the two are identical, but when the molecular formula has subscripts that share a common factor, the empirical formula is obtained by dividing them all by that factor.

Understanding the Question

The question asks directly for the empirical formula of butanoic acid. This requires two things: knowing the molecular formula of butanoic acid, and then simplifying it to the smallest whole-number ratio.

Approach

1. Recall the molecular formula of butanoic acid.
2. Find the highest common factor of the subscripts.
3. Divide each subscript by that factor to obtain the empirical formula.
4. Match the result to one of the options.

Step-by-Step Reasoning

• Butanoic acid is a saturated carboxylic acid with four carbon atoms: CH₃CH₂CH₂COOH.
• Counting atoms: 4 C, 8 H, 2 O, giving the molecular formula C₄H₈O₂.
• The subscripts 4, 8 and 2 all share a highest common factor of 2.
• Dividing each subscript by 2: C₂H₄O.
• This matches option A.

Key Takeaways

• The empirical formula is always the simplest whole-number ratio, not necessarily the molecular formula.
• Carboxylic acids with even carbon numbers often have empirical formulas different from their molecular formulas because of the two oxygen atoms.
• When the molecular formula is given (or recalled), always check whether the subscripts can be divided down.

Common Mistakes

• Choosing C₄H₈O (option C) by forgetting the second oxygen in the carboxylic acid group — butanoic acid has two oxygen atoms, not one.
• Assuming the molecular formula is already the empirical formula without checking for a common factor.
• Confusing butanoic acid (C₄H₈O₂) with butanal (C₄H₈O), which has only one oxygen.

Things to Be Careful About

• The —COOH group contributes two oxygen atoms and one hydrogen atom; do not forget them when counting.
• Always reduce the formula fully — an answer like C₄H₈O₂ would be the molecular formula, not the empirical formula.
• Read the options carefully: several are plausible one-carbon-off distractors, so the exact ratio of C:H:O (2:4:1) must match.

Background Concept

VSEPR theory predicts molecular shapes from the number of electron pairs around the central atom. Both bonding pairs and lone pairs repel each other, but lone pairs occupy more space and compress the bond angles between bonding pairs.

Common shapes and approximate bond angles include:

• linear: 180°
• trigonal planar: 120°
• tetrahedral: 109.5°
• octahedral: 90°
• trigonal pyramidal: about 107°
• bent: about 104.5°

Understanding the Question

The question asks for the pair in which the bond angle in the first species is smaller than the smallest bond angle in the second species. This is important because some species, such as SF₆, have more than one bond angle, so we must use the smallest one when comparing.

Approach

Recall the shape and bond angle of each species, identify the smallest bond angle in the second species, and compare it with the bond angle of the first species.

Step-by-Step Reasoning

• A: CH₄ is tetrahedral, bond angle 109.5°. SF₆ is octahedral, with bond angles of 90° and 180°. The smallest is 90°. Since 109.5° > 90°, this pair does not work.
• B: CO₂ is linear, bond angle 180°. BF₃ is trigonal planar, bond angle 120°. Since 180° > 120°, this pair does not work.
• C: H₂O is bent, bond angle about 104.5°. H₃O⁺ is trigonal pyramidal, bond angle about 107°. Since 104.5° < 107°, this pair satisfies the condition.
• D: NH₄⁺ is tetrahedral, bond angle 109.5°. NH₃ is trigonal pyramidal, bond angle about 107°. Since 109.5° > 107°, this pair does not work.

Therefore, the correct answer is C.

Key Takeaways

• Lone pairs reduce bond angles: H₂O has two lone pairs and a smaller angle than H₃O⁺, which has only one lone pair.
• Octahedral molecules such as SF₆ have 90° bond angles, which are often the smallest.
• NH₄⁺ has a larger bond angle than NH₃ because NH₄⁺ has no lone pairs on the central atom.

Common Mistakes

• Using the largest or only bond angle in the second species instead of the smallest.
• Forgetting that SF₆ has 90° angles.
• Assuming NH₃ has a larger bond angle than NH₄⁺; the opposite is true because NH₃ has a lone pair.

Things to Be Careful About

• Read “smallest bond angle” carefully.
• Approximate values such as 104.5° and 107° are sufficient for comparison.
• Remember that a positive charge on the central atom, as in H₃O⁺, reduces the lone-pair effect compared with the neutral H₂O.

Background Concept

The behaviour of an ideal gas is described by the ideal gas equation $pV = nRT$. When the amount of gas ($n$) is constant, this can be rearranged into the combined gas law: $\frac{p_1 V_1}{T_1} = \frac{p_2 V_2}{T_2}$. This equation is valid provided temperature is measured in Kelvin. When a gas expands into an evacuated vessel, the total volume available to the gas is the sum of the volumes of all connected vessels.

Understanding the Question

We have a fixed amount of helium gas initially confined to vessel M at 20 °C and $1 \times 10^5 \text{ Pa}$. Vessel N is evacuated and has three times the volume of M. When the valve is opened, the gas expands to fill both vessels, so the new volume is the sum of the two. Then the temperature of the whole apparatus is raised to 100 °C. We need to find the final pressure in the system.

Approach

1. Identify the initial state of the gas (pressure, volume, temperature in Kelvin).
2. Identify the final state of the gas (final volume is $V_M + V_N = 4V_M$, final temperature in Kelvin).
3. Apply the combined gas law to find the final pressure.

Step-by-Step Reasoning

Let the volume of vessel M be $V$. Since N has three times the volume of M, $V_N = 3V$.

Initial state (gas in M only):

• $P_1 = 1 \times 10^5 \text{ Pa}$
• $V_1 = V$
• $T_1 = 20 + 273 = 293 \text{ K}$

Final state (gas in both M and N):

• $V_2 = V + 3V = 4V$
• $T_2 = 100 + 273 = 373 \text{ K}$
• $P_2 = ?$

Substitute into $\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$ and rearrange for $P_2$:

$$P_2 = P_1 \left(\frac{V_1}{V_2}\right) \left(\frac{T_2}{T_1}\right)$$ $$P_2 = 1 \times 10^5 \times \frac{V}{4V} \times \frac{373}{293}$$ $$P_2 = 1 \times 10^5 \times 0.25 \times 1.273$$ $$P_2 = 3.18 \times 10^4 \text{ Pa}$$

This matches option A.

Why the other options are wrong:

• B ($4.24 \times 10^4 \text{ Pa}$): This is the result if you forget to add the volume of M and use $V_2 = 3V$ instead of $4V$. ($10^5 \times \frac{1}{3} \times \frac{373}{293} = 4.24 \times 10^4$)
• C ($1.25 \times 10^5 \text{ Pa}$): This is the result if you ignore the volume change and only account for the temperature change, perhaps using 25 °C (298 K) as the initial temperature. ($10^5 \times \frac{373}{298} = 1.25 \times 10^5$)
• D ($5.09 \times 10^5 \text{ Pa}$): This is the result if you multiply by the volume ratio instead of dividing (i.e., using $\frac{V_2}{V_1}$ instead of $\frac{V_1}{V_2}$). ($10^5 \times 4 \times \frac{373}{293} = 5.09 \times 10^5$)

Key Takeaways

• Always convert temperatures to Kelvin before using any gas law equation.
• When a gas expands into connected evacuated vessels, the final volume is the sum of all the volumes, not just the volume of the evacuated vessel.
• The combined gas law $\frac{p_1 V_1}{T_1} = \frac{p_2 V_2}{T_2}$ is the most direct tool when the amount of gas is constant and both pressure, volume, and temperature change.

Common Mistakes

• Forgetting to add the initial volume to the evacuated volume (using $3V$ instead of $4V$).
• Using Celsius instead of Kelvin in the gas law calculation.
• Inverting the volume ratio (multiplying by $\frac{V_2}{V_1}$ instead of $\frac{V_1}{V_2}$), which gives a pressure higher than the initial pressure when the volume actually increased.

Things to Be Careful About

• The temperature must always be in Kelvin for gas law calculations: $T(\text{K}) = T(°\text{C}) + 273$.
• Pay close attention to the wording: "three times the volume of M" means $V_N = 3V_M$, so the total volume is $V_M + 3V_M = 4V_M$.
• Significant figures: the given pressure $1 \times 10^5$ has at least one significant figure, but the options are given to 3 significant figures, so carry at least 3 sig figs through the calculation.

Background Concept

Bond enthalpy (bond dissociation enthalpy) is the energy required to break one mole of a specific covalent bond in the gaseous state. The enthalpy change of a reaction can be estimated from bond enthalpies using:

$$\Delta H = \sum \text{(bonds broken)} - \sum \text{(bonds formed)}$$

Breaking bonds requires energy (endothermic, positive contribution), while forming bonds releases energy (exothermic, negative contribution). The net enthalpy change is the difference between the two.

Hess's law states that the enthalpy change of a reaction depends only on the initial and final states, not on the route taken. This allows us to construct energy cycles linking reactants, intermediates, and products.

Understanding the Question

We are given:

• $\Delta H_c^\ominus[\text{S}_8(\text{s})] = -2376 \text{ kJ mol}^{-1}$ (enthalpy of combustion of $\text{S}_8$)
• Energy to break 1 mol $\text{S}_8(\text{s})$ into gaseous atoms = 2232 kJ mol⁻¹
• O=O bond enthalpy = 496 kJ mol⁻¹

We need to find the S=O bond enthalpy in $\text{SO}_2$ (which is O=S=O, so two S=O bonds per molecule).

The combustion reaction is $\text{S}_8(\text{s}) + 8\text{O}_2(\text{g}) \rightarrow 8\text{SO}_2(\text{g})$. The enthalpy change of this reaction is −2376 kJ mol⁻¹.

Approach

Write the balanced equation, identify all bonds broken and formed, apply $\Delta H = \sum \text{(bonds broken)} - \sum \text{(bonds formed)}$, and solve for the unknown S=O bond enthalpy.

Step-by-Step Reasoning

1. Write the balanced combustion equation: $\text{S}_8(\text{s}) + 8\text{O}_2(\text{g}) \rightarrow 8\text{SO}_2(\text{g})$.

2. Bonds broken:

   • $\text{S}_8(\text{s}) \rightarrow 8\text{S}(\text{g})$: 2232 kJ (given directly — this is the energy to atomise the whole $\text{S}_8$ molecule).
   • 8 O=O bonds: $8 \times 496 = 3968$ kJ.
   • Total: $2232 + 3968 = 6200$ kJ.

3. Bonds formed: 8 $\text{SO}_2$ molecules, each with 2 S=O bonds, so 16 S=O bonds formed.

4. Apply the bond enthalpy relation:

$$\Delta H = 6200 - 16 \times (\text{S=O}) = -2376$$

1. Solve:

$$16 \times (\text{S=O}) = 6200 + 2376 = 8576$$ $$\text{S=O} = \frac{8576}{16} = 536 \text{ kJ mol}^{-1}$$

Distractor analysis:

• A (239): Obtained if the sign of $\Delta H$ is mishandled: $(6200 - 2376)/16 = 239$. This treats the enthalpy change as if less energy is released on bond formation than is consumed on bond breaking, forgetting that the combustion is exothermic.
• B (257): A plausible but incorrect value with no clean derivation from the given data.
• C (319): Obtained by miscounting the number of S=O bonds formed: $(6200 - 2376)/12 = 318.7 \approx 319$. This treats only 12 S=O bonds (e.g., forgetting that each $\text{SO}_2$ has two S=O bonds).

Key Takeaways

• Bond enthalpy calculations require careful bookkeeping of every bond broken and formed, including stoichiometric coefficients.
• The sign of $\Delta H$ matters: exothermic reactions release more energy on bond formation than is consumed on bond breaking.
• Each $\text{SO}_2$ molecule (O=S=O) contains two S=O double bonds — 8 molecules give 16 S=O bonds.

Common Mistakes

• Forgetting the sign of $\Delta H$: Treating −2376 as +2376, or subtracting it instead of adding it when rearranging, gives the wrong answer (option A).
• Miscounting the number of S=O bonds: 8 $\text{SO}_2$ molecules × 2 S=O bonds each = 16, not 8 or 12.
• Forgetting that the energy to break $\text{S}_8$ into atoms (2232 kJ) is for the whole molecule, not per S atom.
• Not including the 8 $\text{O}_2$ molecules in the bond-breaking total — each contributes one O=O bond worth 496 kJ.

Things to Be Careful About

• The energy to break $\text{S}_8$ into gaseous atoms is given per mole of $\text{S}_8$, not per mole of S atoms — use it as-is.
• The O=O bond enthalpy is per mole of O=O bonds; with 8 $\text{O}_2$ molecules, that is $8 \times 496$ kJ.
• State symbols matter: $\text{S}_8$ is solid, $\text{O}_2$ and $\text{SO}_2$ are gases — the energy to atomise $\text{S}_8$ is already included in the bond-breaking term.
• Units: all energies are in kJ mol⁻¹, and the final answer must carry the same unit.

Background Concept

Oxidation state (oxidation number) is a bookkeeping number assigned to an atom by a set of rules:

• The oxidation state of an uncombined element is $0$.
• For a monatomic ion, the oxidation state equals its charge.
• Oxygen is usually $-2$ (except in peroxides, where it is $-1$).
• Hydrogen is usually $+1$.
• The sum of oxidation states in a neutral compound is $0$; in a polyatomic ion it equals the ion charge.

A positive increase in oxidation state is oxidation; a decrease is reduction. In some reactions an element is simultaneously oxidised and reduced — this is disproportionation — and then different atoms of that element show different oxidation-state changes. The question only asks for the largest increase, so we look at the biggest positive difference, not the highest final state.

The question also specifically tells us to use the average oxidation state of sulfur in $\text{S}_2\text{O}_3^{2-}$ and $\text{S}_2\text{O}_4^{2-}$. This is because the two sulfur atoms are not equivalent in these ions, but for this comparison we treat them as if they share the oxidation state equally.

Understanding the Question

Underlined elements are Cr in option A, N in option B, and S in options C and D. For each reaction we must find the largest increase in oxidation state of that element, not the overall oxidation state change of all atoms. For example, an element might be partly oxidised and partly reduced; the largest increase could come from the atoms that are oxidised. Since all four options are set up so that several atoms of the underlined element change oxidation state, we compare the maximum positive change in each option.

Approach

Assign oxidation states to the underlined element in every reactant and product species. Then, for each option, find the highest product oxidation state and subtract the reactant oxidation state. Compare these differences across the four options and select the largest.

Step-by-Step Reasoning

Option A — Cr

$\text{CrO}_4^{3-}$:

$$x + 4(-2) = -3 \Rightarrow x = +5$$

$\text{CrO}_4^{2-}$:

$$x + 4(-2) = -2 \Rightarrow x = +6$$

And $\text{Cr}^{3+}$ has oxidation state $+3$. Some Cr goes from $+5$ to $+6$; that is an increase of $+1$. Another Cr atom goes from $+5$ to $+3$, a decrease of $-2$. The largest increase is therefore $+1$.

Option B — N

In $\text{NO}_2$:

$$x + 2(-2) = 0 \Rightarrow x = +4$$

In $\text{HNO}_3$, with H $= +1$ and three O $= -6$:

$$+1 + x + 3(-2) = 0 \Rightarrow x = +5$$

In $\text{HNO}_2$, with H $= +1$ and two O $= -4$:

$$+1 + x + 2(-2) = 0 \Rightarrow x = +3$$

The largest increase is $+5 - (+4) = +1$.

Option C — S

In $\text{S}_2\text{O}_3^{2-}$, using the average oxidation state $x$ for S:n

$$2x + 3(-2) = -2 \Rightarrow 2x = +4 \Rightarrow x = +2$$

The products are $\text{S}$ and $\text{SO}_2$. Elemental sulfur has oxidation state $0$. In $\text{SO}_2$:

$$x + 2(-2) = 0 \Rightarrow x = +4$$

So one sulfur atom changes from average $+2$ to $0$ (decrease), while another changes from average $+2$ to $+4$ (increase $+2$). The largest increase is therefore $+2$.

Option D — S

In $\text{S}_2\text{O}_4^{2-}$, using average oxidation state $x$ for S:

$$2x + 4(-2) = -2 \Rightarrow 2x = +6 \Rightarrow x = +3$$

In $\text{S}_2\text{O}_3^{2-}$, average S is $+2$. In $\text{HSO}_3^-$, with H $= +1$ and three O $= -6$:

$$+1 + x + 3(-2) = -1 \Rightarrow x = +4$$

So some sulfur increases from $+3$ to $+4$: an increase of $+1$. Other sulfur decreases to $+2$. The largest increase is $+1$.

Comparing all options, A, B and D each have a largest increase of $+1$, while option C has a largest increase of $+2$. Therefore the correct answer is C.

Key Takeaways

• Oxidation state is a bookkeeping tool; changes in oxidation state identify oxidation and reduction.
• In a polyatomic ion, the oxidation states must sum to the charge of the ion, not to zero.
• When asked for the largest increase, subtract the reactant oxidation state from the highest product oxidation state.
• When the question says to use an average oxidation state, follow that instruction exactly; it removes ambiguity in ions such as thiosulfate and dithionite.

Common Mistakes

• Picking the product with the highest oxidation state, such as $+6$ in $\text{CrO}_4^{2-}$ or $+5$ in $\text{HNO}_3$, instead of calculating the increase. Here this might seem attractive, but the increase is only $+1$.
• Forgetting that the sum of oxidation states in an ion equals the ion charge, not zero. For $\text{CrO}_4^{3-}$, if the charge is ignored, Cr would incorrectly be given $+8$.
• Treating the two sulfur atoms in $\text{S}_2\text{O}_3^{2-}$ and $\text{S}_2\text{O}_4^{2-}$ as having different oxidation states without being told to. The question explicitly asks for the average, so use $+2$ and $+3$.
• Ignoring the word "increase" and looking for the option with the most redox change or the largest decrease.

Things to Be Careful About

• Always include the charge of a polyatomic ion when finding oxidation states: e.g. $\text{HSO}_3^-$ has total charge $-1$, not $0$.
• Watch the sign of the change: oxidation is a positive increase, reduction is a negative change.
• In disproportionation reactions, some atoms increase and some decrease; only the positive increase is relevant when the question asks for the largest increase.
• Compare all four options before selecting the answer; several options may have the same increase, and the largest one is the correct choice.

Background Concept

This is a homogeneous gas-phase equilibrium:

$$\text{CO(g)} + 2\text{H}_2\text{(g)} \rightleftharpoons \text{CH}_3\text{OH(g)} \quad \Delta H = -100 \text{ kJ mol}^{-1}$$

For a general equilibrium $a\text{A} + b\text{B} \rightleftharpoons c\text{C} + d\text{D}$, the equilibrium constant in terms of partial pressures is

$$K_p = \frac{p_{\text{C}}^c \, p_{\text{D}}^d}{p_{\text{A}}^a \, p_{\text{B}}^b}$$

Each partial pressure is raised to the stoichiometric coefficient of that species. Le Chatelier's principle states that if a stress is applied to a system at equilibrium, the position of equilibrium shifts to counteract that stress. An increase in pressure favours the side with fewer moles of gas; a change in temperature depends on the sign of $\Delta H$; a catalyst does not change the position of equilibrium.

Understanding the Question

The question asks which single statement about this equilibrium is correct. Four possible statements are given: one about the form of the $K_p$ expression, one about the effect of pressure, one about the effect of temperature, and one about the effect of a catalyst. To answer, you need to use the stoichiometric coefficients, count the number of gas moles on each side, use the sign of $\Delta H$, and recall what a catalyst does to an equilibrium.

Approach

Check each statement systematically.

• For A, write the correct $K_p$ expression and compare it with the one given.
• For B, count gas moles on each side and apply Le Chatelier's principle to an increase in pressure.
• For C, use the sign of $\Delta H$: an exothermic forward reaction is favoured by a decrease in temperature.
• For D, recall that a catalyst lowers the activation energy for both directions equally, so it does not alter the equilibrium composition.

Only one statement will be chemically correct.

Step-by-Step Reasoning

Statement A

The correct expression is

$$K_p = \frac{p_{\text{CH}_3\text{OH}}}{p_{\text{CO}} \times p_{\text{H}_2}^2}$$

because the coefficient of hydrogen in the balanced equation is 2. The option shows $p_{\text{H}_2}$ without the square, so it is incorrect.

Statement B

On the left-hand side there are $1 + 2 = 3$ moles of gas. On the right-hand side there is 1 mole of gas. Increasing the pressure favours the side with fewer gas molecules, which is the right-hand side. Therefore the equilibrium yield of methanol increases. This statement is correct.

Statement C

The forward reaction has $\Delta H = -100 \text{ kJ mol}^{-1}$, so it is exothermic. Increasing the temperature adds heat, and the equilibrium shifts in the endothermic direction, which is the reverse reaction. This decreases the yield of methanol, so the statement is incorrect.

Statement D

A catalyst provides an alternative pathway with lower activation energy for both the forward and reverse reactions. It speeds up the attainment of equilibrium but does not change the position of equilibrium or the equilibrium yield. Therefore this statement is incorrect.

Hence the correct answer is B.

Key Takeaways

• The $K_p$ expression must include each partial pressure raised to its stoichiometric coefficient.
• The effect of pressure on a gas-phase equilibrium depends on the change in the number of moles of gas: higher pressure favours fewer gas moles.
• The effect of temperature depends on the sign of $\Delta H$: increasing temperature favours the endothermic direction.
• A catalyst affects the rate of approach to equilibrium, not the equilibrium position.

Common Mistakes

• Forgetting to square $p_{\text{H}_2}$ in the $K_p$ expression.
• Assuming that increasing pressure always increases yield without counting the number of gas moles on each side.
• Thinking that an exothermic reaction is favoured by heating; in fact, heating favours the endothermic reverse reaction.
• Stating that a catalyst increases the equilibrium yield; it only helps equilibrium be reached faster.

Things to Be Careful About

• Use partial pressures, not concentrations, when writing $K_p$, and raise each to the correct power.
• Count only gaseous species when determining the change in moles of gas.
• Note that $\Delta H$ is negative, so the forward reaction is exothermic.
• A catalyst does not change $K_p$ or the equilibrium composition, only the rate at which equilibrium is reached.

Background Concept

The Boltzmann distribution shows the spread of molecular energies in a gas at a given temperature. The curve is not symmetric: it starts at zero, rises sharply to a peak (the most probable energy), then falls gradually to a long tail at high energy. The total area under the curve is proportional to the total number of molecules. At a higher temperature, the distribution broadens: the peak moves to a higher energy and becomes lower, because the same number of molecules is spread over a wider range of energies. At a lower temperature, the distribution narrows: the peak moves to a lower energy and becomes higher, because more molecules cluster near the average energy.

Understanding the Question

The question asks which change in conditions increases the number of molecules that have the most probable energy — i.e., raises the peak of the distribution. It offers four options that vary temperature and pressure. We need to recall that the shape of the Boltzmann distribution depends only on temperature (for a fixed amount of ideal gas), not on pressure.

Approach

Think of the peak height of the Boltzmann curve. Temperature is the key variable: lowering temperature raises the peak; raising temperature lowers it. Pressure changes at constant temperature do not change the distribution of molecular energies, so options A and C cannot be correct. Among the temperature-changing options, we need the one that lowers temperature, which is B.

Step-by-Step Reasoning

• The "most probable energy" is the energy at the maximum of the distribution curve.
• At constant temperature, the distribution of molecular energies is fixed regardless of pressure (for an ideal gas). Therefore, changing pressure alone (A or C) does not increase the number of molecules at the most probable energy.
• Decreasing temperature at constant pressure (B) narrows the distribution and raises the peak, so more molecules have the most probable (lower) energy.
• Increasing temperature at constant pressure (D) broadens the distribution and lowers the peak, so fewer molecules have the most probable energy.
• Therefore, the correct answer is B.

Key Takeaways

• The Boltzmann distribution is a temperature-dependent curve; its peak height and position change with temperature.
• Lower temperature → higher, narrower peak; higher temperature → lower, broader peak.
• Pressure does not affect the shape of the molecular-energy distribution for an ideal gas at a given temperature.

Common Mistakes

• Confusing "most probable energy" with "average energy": the most probable energy is the peak, not the mean.
• Thinking that increasing pressure increases the number of molecules at a given energy: pressure changes the number of molecules per unit volume but not the distribution shape.
• Choosing D because higher temperature gives molecules more energy: but the peak height decreases, so the number at the most probable energy is smaller.

Things to Be Careful About

• The question asks for the number of molecules with the most probable energy, not the value of that energy.
• At higher temperature, the most probable energy increases, but the number of molecules at that energy decreases.
• The distribution is for a fixed amount of gas; if the amount changes, the total area changes, but the question implies a fixed sample.

Background Concept

The electronic configuration of atoms follows the Aufbau principle, the Pauli exclusion principle, and Hund's rule. Hund's rule states that electrons fill degenerate orbitals (orbitals of the same energy, such as the three 3p orbitals) singly first, with parallel spins, before pairing up. This maximises the number of unpaired electrons in a subshell until it is half-filled.

Understanding the Question

The question asks to identify the correct graph showing the number of unpaired electrons for the six Period 3 elements with proton numbers 13 to 18 (Al, Si, P, S, Cl, Ar). The x-axis is the proton number, and the y-axis is the number of unpaired electrons.

Approach

Determine the valence electron configuration for each element from proton number 13 to 18. Apply Hund's rule to the 3p subshell to count the number of unpaired electrons for each element. Plot these values mentally and match the resulting trend to one of the given graphs.

Step-by-Step Reasoning

1. Elements 13 to 18 are in Period 3. Their inner electron configuration is that of neon, $1\text{s}^2 2\text{s}^2 2\text{p}^6$. All inner electrons are paired, so we only need to consider the valence electrons in the 3s and 3p subshells.
2. The 3s subshell is always fully occupied ($3\text{s}^2$) for all these elements, contributing 0 unpaired electrons.
3. The 3p subshell has three degenerate orbitals. As we add electrons from proton number 13 to 18:
   • Al (13): $3\text{p}^1$. One electron in the first 3p orbital. Unpaired = 1.
   • Si (14): $3\text{p}^2$. Two electrons in separate 3p orbitals (Hund's rule). Unpaired = 2.
   • P (15): $3\text{p}^3$. Three electrons, one in each 3p orbital. Unpaired = 3. This is the half-filled subshell.
   • S (16): $3\text{p}^4$. Four electrons. One orbital must now contain a pair. Unpaired = 2.
   • Cl (17): $3\text{p}^5$. Five electrons. Two orbitals contain pairs, one contains a single electron. Unpaired = 1.
   • Ar (18): $3\text{p}^6$. Six electrons. All three 3p orbitals are fully paired. Unpaired = 0.
4. The sequence of unpaired electrons is 1, 2, 3, 2, 1, 0.
5. Graph D shows exactly this trend: rising from 1 at proton number 13 to a peak of 3 at proton number 15, then falling to 0 at proton number 18.

Key Takeaways

Hund's rule governs the filling of degenerate orbitals. The number of unpaired electrons in a p-subshell increases from 1 to 3 as electrons are added, then decreases from 2 to 0 as pairing begins. This creates a symmetric peak at the half-filled subshell.

Common Mistakes

• Forgetting Hund's rule and assuming electrons pair up immediately (e.g., thinking $3\text{p}^2$ has 0 unpaired electrons).
• Ignoring that the 3s electrons are paired and only counting 3p electrons (though for these elements, 3s is always $3\text{s}^2$, so it doesn't change the unpaired count, but it is good practice).
• Assuming the trend continues to increase linearly across the period.

Things to Be Careful About

• State that the inner shells are fully paired and contribute zero unpaired electrons.
• Remember that the 3p subshell has 3 orbitals, not 1.
• Graph C shows a plateau, which would be incorrect because after the half-filled subshell (P), electrons begin to pair up, reducing the number of unpaired electrons.

Background Concept

Nitrogen dioxide, $\text{NO}_2$, is a covalent oxide of nitrogen. When it 'decomposes into its elements', the products are nitrogen gas and oxygen gas, both diatomic molecules: $\text{N}_2$ and $\text{O}_2$. The balanced equation for the decomposition is

$$2\text{NO}_2(\text{g}) \rightarrow \text{N}_2(\text{g}) + 2\text{O}_2(\text{g})$$

For a gas or mixture of gases at constant volume and temperature, $pV = nRT$ means that pressure is directly proportional to the total number of moles of gas: $p \propto n$. This lets us turn a pressure ratio into a mole ratio.

Understanding the Question

We are given 69.0 g of $\text{NO}_2$ and told that the initial pressure is $P$. After partial decomposition, the total pressure is $1.1P$. The question asks for the mass of oxygen molecules present. The key point is that $1.1P$ is the total pressure of all gases in the vessel, not the partial pressure of oxygen, and not simply '10% of the NO2 has decomposed'. The catalyst is only there to make the decomposition fast enough; it does not change the stoichiometry or the equilibrium position.

Approach

1. Convert the initial mass of NO2 into moles using its molar mass.
2. Use the pressure ratio to find the final total number of moles of gas.
3. Write and balance the decomposition equation.
4. Relate the increase in total moles to the amount of oxygen formed.
5. Convert moles of oxygen to mass and select the option.

Step-by-Step Reasoning

Step 1: initial moles of NO2.

$M_r(\text{NO}_2) = 14.0 + 2(16.0) = 46.0$

$n(\text{NO}_2) = \frac{69.0}{46.0} = 1.50 \text{ mol}$

Step 2: final total moles.

At constant $V$ and $T$, $n \propto p$, so

$n_{\text{total, final}} = 1.1 \times 1.50 = 1.65 \text{ mol}$

The increase in total moles is $1.65 - 1.50 = 0.15 \text{ mol}$.

Step 3: balanced equation and stoichiometry.

$$2\text{NO}_2(\text{g}) \rightarrow \text{N}_2(\text{g}) + 2\text{O}_2(\text{g})$$

Let $x$ be the amount of NO2 decomposed. Then remaining NO2 = $1.50 - x$, N2 formed = $x/2$, O2 formed = $x$. Total moles = $(1.50 - x) + x/2 + x = 1.50 + x/2$. Set this equal to 1.65:

$1.50 + x/2 = 1.65$

$x/2 = 0.15$, so $x = 0.30 \text{ mol}$.

Therefore O2 formed = $0.30 \text{ mol}$.

Step 4: mass of oxygen.

$m(\text{O}_2) = 0.30 \times 32.0 = 9.60 \text{ g}$

So the correct option is B.

Key Takeaways

• In a gas mixture at constant V and T, pressure is a direct measure of total moles.
• 'Elements' for nitrogen and oxygen means N2 and O2, so balance the equation accordingly.
• The increase in total moles per reaction is the key stoichiometric link.
• A catalyst affects rate, not the equilibrium amounts or the stoichiometry.

Common Mistakes

• Treating $1.1P$ as '10% of NO2 decomposed'. The pressure increase is only 0.1P, but because 2 mol of gas become 3 mol, this corresponds to 20% decomposition, giving 0.30 mol O2 (9.60 g), not 0.15 mol O2 (4.80 g, option A).
• Assuming complete decomposition: that would give 1.5 mol O2 and 48.0 g (option D).
• Writing the products as N and O atoms instead of N2 and O2, which gives an unbalanced equation and wrong mole ratios.
• Using the total pressure as the partial pressure of oxygen.
• Forgetting to multiply by the molar mass of O2 when converting moles to mass.

Things to Be Careful About

• Keep state symbols in the balanced equation; they are not essential for the calculation but show the physical states.
• The pressure–mole proportionality only holds if volume and temperature are constant; this is implied by a reaction vessel under the same conditions.
• The question asks for oxygen molecules, O2, so use $M_r = 32.0$.
• The catalyst is a distractor: it only speeds up the reaction.
• Watch significant figures: the data include 69.0 and 1.1; the intended option is 9.60 g.

Background Concept

PF5 is a covalent molecule of phosphorus pentafluoride. Phosphorus has 5 valence electrons, and each fluorine contributes one electron to a single P–F bond. This gives 5 bonding pairs and no lone pairs on the central atom. According to VSEPR theory, 5 electron pairs arrange themselves as far apart as possible in a trigonal bipyramidal shape.

Understanding the Question

The question asks which statement about PF5 is correct. We need to check the shape, bond angles, lone pairs, and polarity of the molecule.

Approach

Apply VSEPR theory: count the electron pairs around the central atom, predict the shape, then check each option against that shape and the symmetry of the molecule.

Step-by-Step Reasoning

1. Count valence electrons of P: 5.
2. Each F forms one single bond with P, using one electron from P and one from F. So there are 5 bonding pairs and 0 lone pairs around P.
3. Five electron pairs give a trigonal bipyramidal shape: 3 equatorial positions at 120° to each other, and 2 axial positions at 90° to the equatorial plane.
4. Option A is false because equatorial F–P–F angles are 120°, not all 90°.
5. Option B is true because the central atom has no lone pair.
6. Option C is false because the identical P–F bonds are arranged symmetrically, so the individual bond dipoles cancel and the molecule has no overall dipole moment.
7. Option D is false because octahedral geometry requires 6 electron pairs; PF5 has 5.

Key Takeaways

• PF5 is trigonal bipyramidal, not octahedral.
• The central phosphorus atom has 5 bonding pairs and no lone pairs.
• A molecule can have polar bonds but still be non-polar if the bond dipoles cancel by symmetry.

Common Mistakes

• Assuming that because P–F bonds are polar, the whole molecule is polar.
• Confusing 5 electron pairs with octahedral geometry, which needs 6.
• Thinking all bond angles in a trigonal bipyramid are 90°; the equatorial angles are 120°.

Things to Be Careful About

• Count lone pairs as well as bonding pairs when applying VSEPR.
• Remember that PF5 is an example of an expanded octet: phosphorus has 10 electrons around it.
• Check each option individually rather than stopping at the first plausible statement.

Background Concept

Ionic bonding arises from the transfer of electrons between a metal and a non-metal, producing oppositely charged ions held by electrostatic attraction. Covalent bonding involves the sharing of electron pairs between atoms. A coordinate (dative) bond is a special covalent bond in which both electrons of the shared pair come from a single atom (a lone-pair donor).

Understanding the Question

The question asks for a solid compound that has both ionic and covalent bonding, but not coordinate bonding. This means we need a compound that is ionic overall (metal + polyatomic ion, or ammonium salt) and also contains covalent bonds within a polyatomic ion, while having no dative bonds.

Approach

Evaluate each option systematically:

1. Decide whether the compound is ionic (metal + non-metal, or ammonium salt).
2. Check whether it contains covalent bonds within a polyatomic ion.
3. Check whether any bond is a coordinate (dative) bond.

Step-by-Step Reasoning

• A — Al₂Cl₆: Aluminium chloride is a covalent compound. In the gas/solid phase it exists as a dimer with bridging chlorine atoms. Each bridging Cl donates a lone pair to an Al atom, forming coordinate bonds. So it has covalent and coordinate bonding, but no ionic bonding. Eliminated.
• B — CH₃COONa (sodium ethanoate): Na⁺ and CH₃COO⁻ form an ionic lattice. Within the ethanoate ion, the C–C, C–H, C=O and C–O bonds are all ordinary covalent bonds formed by sharing one electron from each atom. There is no lone-pair donation, so no coordinate bond. This compound has both ionic and covalent bonding but no coordinate bonding. Answer.
• C — MgCl₂: Magnesium chloride is an ionic compound of Mg²⁺ and Cl⁻. There are no covalent bonds. Eliminated.
• D — NH₄Cl (ammonium chloride): Ionic between NH₄⁺ and Cl⁻, and covalent within the ammonium ion. However, one of the four N–H bonds is a coordinate bond: the H⁺ ion has no electrons, so the lone pair on NH₃ is donated to form that bond. Thus it contains coordinate bonding. Eliminated.

Key Takeaways

• Ionic compounds can contain covalent bonds within polyatomic ions (e.g. ethanoate, nitrate, sulfate, ammonium).
• A coordinate bond is a covalent bond where both electrons come from one atom; it is common in NH₄⁺, H₃O⁺ and Al₂Cl₆.
• The ammonium ion always contains one coordinate bond, even though all four N–H bonds look identical.

Common Mistakes

• Assuming NH₄Cl has no coordinate bond because all N–H bonds appear the same. In fact, one is a dative bond.
• Thinking MgCl₂ has significant covalent character; it is essentially ionic.
• Forgetting that Al₂Cl₆ is covalent, not ionic.

Things to Be Careful About

• Distinguish between ionic + covalent (CH₃COONa) and ionic + covalent + coordinate (NH₄Cl).
• The phrase "solid compound" is a distractor here — all four options are solids, so it does not affect the answer.

Background Concept

The standard enthalpy change of formation, $\Delta H_f^\ominus$, is defined as the enthalpy change when one mole of a compound is formed from its elements, all in their standard states, under standard conditions (usually 298 K and 1 atm). The standard state of a substance is its most stable physical form at these conditions. For carbon, the standard state is solid graphite, $\text{C(s)}$; for hydrogen, it is the diatomic gas $\text{H}_2\text{(g)}$; for oxygen, it is $\text{O}_2\text{(g)}$. For ethanol, the standard state at 298 K and 1 atm is a liquid, $\text{C}_2\text{H}_5\text{OH(l)}$.

This question tests whether you can translate that definition into a correctly balanced chemical equation.

Understanding the Question

The question asks you to identify which of four equations correctly represents the standard enthalpy change of formation of ethanol. The command is essentially "identify", so you must check each option against the definition. Three conditions must all be satisfied:

1. The reactants are the elements in their standard states.
2. The equation is balanced so that exactly one mole of ethanol is formed.
3. The ethanol product is in its standard physical state (liquid).

Any option that fails even one of these conditions is incorrect.

Approach

Work through the definition methodically:

• Write the formula of ethanol and count the atoms: $\text{C}_2\text{H}_5\text{OH}$ contains 2 carbon atoms, 6 hydrogen atoms, and 1 oxygen atom.
• Convert this into the formation equation: 2 mol $\text{C(s)} + 3$ mol $\text{H}_2\text{(g)} + \frac{1}{2}$ mol $\text{O}_2\text{(g)} \rightarrow 1$ mol $\text{C}_2\text{H}_5\text{OH}$.
• Check the physical state of each substance at standard conditions.
• Eliminate options that fail any condition.

Step-by-Step Reasoning

1. Identify the elements and their standard states.

   • Carbon: $\text{C(s)}$ (graphite).
   • Hydrogen: $\text{H}_2\text{(g)}$.
   • Oxygen: $\text{O}_2\text{(g)}$.
     All four options use these correct reactant forms.

2. Balance the equation for one mole of ethanol.

   • Ethanol has 2 C atoms, so we need 2 mol $\text{C(s)}$.
   • Ethanol has 6 H atoms, so we need 3 mol $\text{H}_2\text{(g)}$ (since each $\text{H}_2$ provides 2 H atoms).
   • Ethanol has 1 O atom, so we need $\frac{1}{2}$ mol $\text{O}_2\text{(g)}$.
   • Options A and B use $2\frac{1}{2}\text{H}_2\text{(g)}$, which provides only 5 hydrogen atoms — the equation would not balance (5 H vs 6 H). These are eliminated.

3. Check the physical state of the product.

   • At 298 K and 1 atm, ethanol is a liquid. Therefore the product must be written as $\text{C}_2\text{H}_5\text{OH(l)}$.
   • Option C has the correct $3\text{H}_2$ stoichiometry but writes ethanol as a gas, $\text{C}_2\text{H}_5\text{OH(g)}$, which is not its standard state. Eliminated.

4. Select the correct option.

   • Option D has the correct stoichiometry (2 C, 3 H2, ½ O2) and the correct product state (liquid). It satisfies all three conditions of the definition.

Key Takeaways

• The standard enthalpy change of formation always involves one mole of product formed from its elements in their standard states.
• "Standard state" is a physical-state requirement: you must know whether the compound is solid, liquid, or gas at 298 K and 1 atm.
• When checking a formation equation, always verify: (a) reactants are elements in standard states, (b) the equation balances to exactly one mole of product, and (c) the product is in its standard state.

Common Mistakes

• Choosing the gas state for ethanol (options A and C): ethanol is a liquid at standard conditions, so $\text{C}_2\text{H}_5\text{OH(g)}$ cannot represent $\Delta H_f^\ominus$.
• Miscounting hydrogen atoms: $2\frac{1}{2}\text{H}_2$ provides only 5 H atoms, not the 6 required for ethanol. Some students see "2½" and think it is just a fraction without checking the atom count.
• Confusing formation with combustion: a combustion equation would have $\text{O}_2$ as a reactant and $\text{CO}_2$ and $\text{H}_2\text{O}$ as products. Formation is the opposite direction — building the compound from its elements.

Things to Be Careful About

• State symbols are part of the definition of $\Delta H_f^\ominus$ — omitting or miswriting them changes the meaning of the equation.
• The equation must produce exactly one mole of the compound; coefficients must be scaled accordingly (e.g. $\frac{1}{2}\text{O}_2$ is allowed, not $\text{O}_2$).
• The elements must be in their standard states — for carbon that is solid graphite, not $\text{C(g)}$ or diamond.
• Always count atoms carefully when a coefficient is a fraction; a fraction like $2\frac{1}{2}$ can hide an unbalanced equation.

Background Concept

Redox reactions involve the transfer of electrons. Oxidation number is a bookkeeping number assigned to each atom. An increase in oxidation number means oxidation (electron loss); a decrease means reduction (electron gain). In any redox reaction, the total number of electrons lost by the reducing agent equals the total number gained by the oxidising agent. This electron balance lets us find mole ratios even when the full balanced equation is not given.

For $\text{K}_2\text{MnO}_4$, it is important to recognise that this is potassium manganate(VI), not potassium permanganate. The two $\text{K}^+$ ions and four $\text{O}^{2-}$ ions fix Mn at $+6$. In contrast, $\text{KMnO}_4$ has Mn at $+7$. This distinction is the key to the calculation.

Understanding the Question

The reaction is between $\text{K}_2\text{MnO}_4$ and excess concentrated hydrochloric acid. The products include $\text{Cl}_2$ and $\text{MnCl}_2$. All Mn ends up as $\text{MnCl}_2$, so every Mn atom is reduced from $+6$ to $+2$. Chlorine in HCl is oxidised from $-1$ to $0$ in $\text{Cl}_2$. The statement that no other element is oxidised or reduced tells us that K and O keep their oxidation numbers, so the only electron transfer is from $\text{Cl}^-$ to Mn. We need the moles of $\text{Cl}_2$ produced by 1 mol of $\text{K}_2\text{MnO}_4$. Because HCl is in excess, it is not the limiting reagent; the amount of $\text{Cl}_2$ is fixed by how many electrons 1 mol of $\text{K}_2\text{MnO}_4$ can accept.

Approach

1. Find the oxidation number of Mn in $\text{K}_2\text{MnO}_4$ and in $\text{MnCl}_2$.
2. Determine the electrons gained per mole of Mn.
3. Determine the electrons released per mole of $\text{Cl}_2$ formed.
4. Equate electrons gained and lost to calculate the moles of $\text{Cl}_2$.

Step-by-Step Reasoning

Let $x$ be the oxidation number of Mn in $\text{K}_2\text{MnO}_4$. K is $+1$ and O is $-2$:

$$2(+1) + x + 4(-2) = 0 \Rightarrow x = +6$$

In $\text{MnCl}_2$, Cl is $-1$:

$$x + 2(-1) = 0 \Rightarrow x = +2$$

So Mn changes from $+6$ to $+2$, a decrease of 4. Each mole of Mn gains 4 mol of electrons:

$$\text{Mn}^{+6} + 4\text{e}^- \rightarrow \text{Mn}^{+2}$$

In HCl, Cl is $-1$. In $\text{Cl}_2$, each Cl atom is $0$. Each $\text{Cl}^-$ loses 1 electron, but $\text{Cl}_2$ is diatomic, so forming 1 mol of $\text{Cl}_2$ releases 2 mol of electrons:

$$2\text{Cl}^- \rightarrow \text{Cl}_2 + 2\text{e}^-$$

1 mol of $\text{K}_2\text{MnO}_4$ contains 1 mol of Mn, so it can accept 4 mol of electrons. These 4 mol of electrons oxidise $\frac{4}{2} = 2$ mol of $\text{Cl}_2$. Therefore the answer is A.

The full balanced equation is a useful check:

$$\text{K}_2\text{MnO}_4 + 8\text{HCl} \rightarrow \text{MnCl}_2 + 2\text{KCl} + 2\text{Cl}_2 + 4\text{H}_2\text{O}$$

It also shows 2 mol of $\text{Cl}_2$ per mol of $\text{K}_2\text{MnO}_4$.

Distractors: Option B (2.5 mol) comes from treating $\text{K}_2\text{MnO}_4$ as $\text{KMnO}_4$ with Mn at $+7$, giving 5 electrons per Mn and 2.5 mol of $\text{Cl}_2$. Option D (4.0 mol) comes from forgetting that $\text{Cl}_2$ is diatomic and counting 4 mol of Cl atoms as 4 mol of $\text{Cl}_2$. Option C (3.0 mol) has no simple redox basis; it might come from a miscount of the electron change.

Key Takeaways

• Oxidation number changes give electron transfer directly.
• Always compare the oxidation numbers of the element in the reactant and product.
• For diatomic molecules, account for two atoms per molecule when converting electron loss to moles of product.
• The statement that no other element is oxidised or reduced lets us ignore spectator elements.

Common Mistakes

• Assuming $\text{K}_2\text{MnO}_4$ has Mn at $+7$. It has two $\text{K}^+$ and four $\text{O}^{2-}$, so Mn is $+6$. This leads to option B.
• Forgetting to divide by 2 for $\text{Cl}_2$. Each mole of $\text{Cl}_2$ needs two $\text{Cl}^-$ and releases two electrons; 4 mol of electrons give 2 mol of $\text{Cl}_2$, not 4 mol.
• Trying to balance the full equation first; this is not necessary and can waste time. Use electron transfer.
• Ignoring the phrase "excess hydrochloric acid" and treating HCl as limiting; the amount of $\text{Cl}_2$ is determined by $\text{K}_2\text{MnO}_4$.

Things to Be Careful About

• Use oxidation number rules consistently: K is $+1$, O is $-2$ in normal compounds, and Cl is $-1$ in chlorides.
• The oxidation number of an element in its elemental form is $0$, but $\text{Cl}_2$ contains two Cl atoms, so the electron change per mole of $\text{Cl}_2$ is 2 mol of electrons.
• Keep units as mol of electrons; do not confuse mol of atoms with mol of molecules.
• If a full equation is used as a check, it must be balanced in atoms and charge; here the balanced equation confirms the 2:1 ratio.

Background Concept

For a gaseous mixture at equilibrium, the partial pressure of a gas is the pressure it would exert if it alone occupied the whole container. It is given by:

$$P_A = x_A P_{\text{total}}$$

where $x_A$ is the mole fraction of gas A:

$$x_A = \frac{\text{moles of A}}{\text{total moles of gas}}$$

So, to find the partial pressure of ammonia, we need the equilibrium amounts of all gases, not just the amount of ammonia produced.

Understanding the Question

We are told that nitrogen and hydrogen are mixed in the mole ratio 1 : 3 and reach equilibrium according to:

$$\text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3$$

50% of the nitrogen reacts. The total final pressure is $p$. We are asked for the partial pressure of ammonia in the equilibrium mixture. Notice that the question does not ask for $K_p$, so we only need mole fractions and the total pressure.

Approach

Because only ratios matter, choose convenient initial amounts: 1 mol of $\text{N}_2$ and 3 mol of $\text{H}_2$. Use the stoichiometry of the reaction to calculate how much $\text{H}_2$ is consumed and how much $\text{NH}_3$ is formed when 0.5 mol of $\text{N}_2$ reacts. Then add the equilibrium moles to find the total, calculate the mole fraction of ammonia, and multiply by the total pressure $p$.

Step-by-Step Reasoning

1. Initial amounts
   Let the initial amounts be:

   • $\text{N}_2$: 1 mol
   • $\text{H}_2$: 3 mol
   • $\text{NH}_3$: 0 mol

2. Amount of nitrogen reacted
   50% of 1 mol = 0.5 mol of $\text{N}_2$ reacts.

3. Use stoichiometry
   The balanced equation shows that 1 mol of $\text{N}_2$ reacts with 3 mol of $\text{H}_2$ to form 2 mol of $\text{NH}_3$.

   So when 0.5 mol of $\text{N}_2$ reacts:

   • $\text{H}_2$ consumed $= 3 \times 0.5 = 1.5$ mol
   • $\text{NH}_3$ formed $= 2 \times 0.5 = 1.0$ mol

4. Equilibrium amounts

   • $\text{N}_2$: $1 - 0.5 = 0.5$ mol
   • $\text{H}_2$: $3 - 1.5 = 1.5$ mol
   • $\text{NH}_3$: $0 + 1.0 = 1.0$ mol

5. Total moles at equilibrium

   $$0.5 + 1.5 + 1.0 = 3.0 \text{ mol}$$

6. Mole fraction of ammonia

   $$x_{\text{NH}_3} = \frac{1.0}{3.0} = \frac{1}{3}$$

7. Partial pressure of ammonia

   $$P_{\text{NH}_3} = x_{\text{NH}_3} \times p = \frac{1}{3}p = \frac{p}{3}$$

   Therefore, the correct option is C.

Key Takeaways

• Partial pressure is mole fraction multiplied by total pressure.
• Always use equilibrium amounts, not initial amounts, when finding mole fractions at equilibrium.
• Stoichiometry tells us the exact amounts of reactants consumed and products formed.
• Choosing convenient initial amounts, such as 1 mol and 3 mol, makes the calculation simple when only ratios are given.

Common Mistakes

• Using the initial total moles instead of the equilibrium total moles. The initial total is 4 mol, but the equilibrium total is 3 mol. Using 4 mol would give $p/4$, which is option B and is incorrect.
• Forgetting that 3 mol of $\text{H}_2$ are consumed for every 1 mol of $\text{N}_2$. Some might subtract only 0.5 mol from $\text{H}_2$ instead of 1.5 mol.
• Confusing mole fraction with partial pressure. Mole fraction is a dimensionless ratio; partial pressure has pressure units and requires multiplying by $p$.
• Reading "50% of nitrogen reacted" as "50% of nitrogen remains". 50% reacted means half of the initial nitrogen is used up, so 0.5 mol remains from 1 mol.

Things to Be Careful About

• The initial ratio 1 : 3 matches the stoichiometric ratio, so both reactants are consumed in the same proportion, but the total number of moles changes because 4 moles of reactants form 2 moles of product.
• The answer is expressed as a fraction of $p$, so no numerical pressure value is needed.
• Keep the amounts in moles consistent throughout; do not mix moles with masses or volumes.