DSE Chemistry Diagnostic: Chemical Equilibrium

Unit Test 1: Kc Calculation

Question

Nitrogen and hydrogen react to form ammonia:

$N_{2}(g) + 3H_{2}(g) \rightleftharpoons 2NH_{3}(g)$

1.00 mol of $N_{2}$ and 3.00 mol of $H_{2}$ were mixed in a sealed container of volume 2.0 dm$^{3}$ and allowed to reach equilibrium at a certain temperature. At equilibrium, 0.40 mol of $NH_{3}$ was present.

(a) Calculate the equilibrium concentrations of $N_{2}$, $H_{2}$, and $NH_{3}$. [2 marks]

(b) Calculate the equilibrium constant $K_{c}$ for this reaction at this temperature. [3 marks]

(c) If the volume of the container is halved (by compressing the gas mixture) at constant temperature, what happens to the value of $K_{c}$? Explain. [2 marks]

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Worked Solution

(a) ICE table:

Species	Initial (mol)	Change (mol)	Equilibrium (mol)
$N_{2}$	1.00	$-0.20$	0.80
$H_{2}$	3.00	$-0.60$	2.40
$NH_{3}$	0	$+0.40$	0.40

Change in $NH_{3}$ = $+0.40$ mol, so change in $N_{2}$ = $-0.40/2 = -0.20$ mol, change in $H_{2}$ = $-3 \times 0.20 = -0.60$ mol.

Equilibrium concentrations (dividing by volume 2.0 dm$^{3}$):

$[N_{2}] = \frac{0.80}{2.0} = 0.40 \text{ mol/dm}^{3}$

$[H_{2}] = \frac{2.40}{2.0} = 1.20 \text{ mol/dm}^{3}$

$[NH_{3}] = \frac{0.40}{2.0} = 0.20 \text{ mol/dm}^{3}$

(b) $K_{c} = \frac{[NH_{3}]^{2}}{[N_{2}][H_{2}]^{3}} = \frac{(0.20)^{2}}{(0.40)(1.20)^{3}} = \frac{0.040}{0.40 \times 1.728} = \frac{0.040}{0.6912} = 0.0579$

$K_{c}$ has units: $\frac{(\text{mol dm}^{-3})^{2}}{(\text{mol dm}^{-3})(\text{mol dm}^{-3})^{3}} = \text{mol}^{-2} \text{dm}^{6}$

$K_{c} = 0.0579 \text{ mol}^{-2} \text{dm}^{6}$

(c) The value of $K_{c}$ remains unchanged. $K_{c}$ is a constant at a given temperature and is not affected by changes in concentration or pressure. Changing the volume changes the equilibrium position (it shifts to the side with fewer gas moles, which is the product side), but the ratio $[NH_{3}]^{2}/([N_{2}][H_{2}]^{3})$ at the new equilibrium remains the same.

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Unit Test 2: Le Chatelier + Equilibrium Constant

Question

For the exothermic reaction:

$2SO_{2}(g) + O_{2}(g) \rightleftharpoons 2SO_{3}(g) \quad \Delta H = -197 \text{ kJ/mol}$

(a) State the effect of increasing temperature on: (i) the equilibrium position, and (ii) the value of $K_{c}$. [3 marks]

(b) A student says: "Increasing the pressure by adding an inert gas at constant volume will shift the equilibrium to the right." Evaluate this statement. [2 marks]

(c) At 700 K, $K_{c} = 300$ (in appropriate units). At 900 K, $K_{c} = 5.0$. Is the forward reaction exothermic or endothermic? Explain. [2 marks]

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Worked Solution

(a) (i) Equilibrium position: Increasing temperature shifts the equilibrium to the left (towards the reactants), because the forward reaction is exothermic. The system opposes the change by absorbing the added heat (endothermic reverse reaction).

(ii) $K_{c}$ value: Since the equilibrium shifts left, the concentration of products decreases and reactants increase. Therefore $K_{c}$ decreases.

(b) The statement is incorrect. Adding an inert gas at constant volume increases the total pressure but does not change the partial pressures or concentrations of the reacting gases. Since the equilibrium depends on partial pressures (or concentrations), and these are unchanged, the equilibrium position does not shift.

(Note: Adding an inert gas at constant pressure by increasing volume WOULD shift the equilibrium, because it would decrease the partial pressures of all gases, favouring the side with more gas moles.)

(c) As temperature increases from 700 K to 900 K, $K_{c}$ decreases from 300 to 5.0. Since $K_{c}$ decreases with increasing temperature, the equilibrium shifts towards the reactants as temperature rises. This means the forward reaction is exothermic (the system absorbs heat by favouring the reverse, endothermic reaction when temperature is increased).

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Unit Test 3: Inert Gas Effects

Question

For the equilibrium:

$PCl_{5}(g) \rightleftharpoons PCl_{3}(g) + Cl_{2}(g)$

(a) Explain the effect on the equilibrium position when an inert gas (e.g., argon) is added at constant volume. [2 marks]

(b) Explain the effect on the equilibrium position when an inert gas is added at constant pressure. [3 marks]

(c) In which case (a) or (b) does the value of $K_{p}$ change? Explain. [1 mark]

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Worked Solution

(a) At constant volume, adding an inert gas increases the total pressure but does not change the partial pressures of $PCl_{5}$, $PCl_{3}$, or $Cl_{2}$. Since the equilibrium depends on partial pressures (or concentrations), the equilibrium position does not shift.

Reason: $P_{\text{total}}$ increases, but the mole fractions and hence partial pressures of the reacting gases remain unchanged.

(b) At constant pressure, adding an inert gas increases the total number of moles in the container, which requires the volume to increase (to maintain constant total pressure). The increase in volume decreases the partial pressures of all reacting gases equally.

Since there are more gas moles on the right (2 moles) than on the left (1 mole), the system responds by shifting towards the side with more gas moles to increase the pressure. Therefore, the equilibrium shifts to the right, producing more $PCl_{3}$ and $Cl_{2}$.

(c) $K_{p}$ does not change in either case. $K_{p}$ depends only on temperature. Changes in pressure, volume, or addition of inert gas do not affect the value of the equilibrium constant.

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Integration Test 1: Kc + Kp Conversion

Question

For the equilibrium:

$N_{2}O_{4}(g) \rightleftharpoons 2NO_{2}(g)$

At 350 K, the total equilibrium pressure is 1.00 atm and the degree of dissociation of $N_{2}O_{4}$ is 0.50 (50%).

(a) Calculate the mole fractions of $N_{2}O_{4}$ and $NO_{2}$ at equilibrium. [2 marks]

(b) Calculate the partial pressures of $N_{2}O_{4}$ and $NO_{2}$ at equilibrium. [2 marks]

(c) Calculate $K_{p}$ for this equilibrium at 350 K. [2 marks]

(d) If the total pressure is increased to 2.00 atm at the same temperature, calculate the new degree of dissociation. [4 marks]

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Worked Solution

(a) Start with 1 mol $N_{2}O_{4}$. Let $\alpha = 0.50$.

At equilibrium:

• $n(N_{2}O_{4}) = 1 - 0.50 = 0.50$ mol
• $n(NO_{2}) = 2 \times 0.50 = 1.00$ mol
• Total moles = $0.50 + 1.00 = 1.50$ mol

Mole fraction of $N_{2}O_{4}$: $x(N_{2}O_{4}) = \frac{0.50}{1.50} = \frac{1}{3}$

Mole fraction of $NO_{2}$: $x(NO_{2}) = \frac{1.00}{1.50} = \frac{2}{3}$

(b) $P(N_{2}O_{4}) = x(N_{2}O_{4}) \times P_{\text{total}} = \frac{1}{3} \times 1.00 = 0.333$ atm

$P(NO_{2}) = x(NO_{2}) \times P_{\text{total}} = \frac{2}{3} \times 1.00 = 0.667$ atm

(c) $K_{p} = \frac{(P_{NO_{2}})^{2}}{P_{N_{2}O_{4}}} = \frac{(0.667)^{2}}{0.333} = \frac{0.445}{0.333} = 1.335 \text{ atm}$

(d) Let $\alpha$ be the new degree of dissociation at $P = 2.00$ atm.

$n(N_{2}O_{4}) = 1 - \alpha$, $n(NO_{2}) = 2\alpha$, total $= 1 + \alpha$

$P(N_{2}O_{4}) = \frac{1 - \alpha}{1 + \alpha} \times 2.00$

$P(NO_{2}) = \frac{2\alpha}{1 + \alpha} \times 2.00 = \frac{4\alpha}{1 + \alpha}$

$K_{p} = \frac{\left(\frac{4\alpha}{1+\alpha}\right)^{2}}{\frac{2(1-\alpha)}{1+\alpha}} = \frac{\frac{16\alpha^{2}}{(1+\alpha)^{2}}}{\frac{2(1-\alpha)}{1+\alpha}} = \frac{16\alpha^{2}}{2(1-\alpha)(1+\alpha)} = \frac{8\alpha^{2}}{1-\alpha^{2}}$

Set $K_{p} = 1.335$:

$\frac{8\alpha^{2}}{1-\alpha^{2}} = 1.335$

$8\alpha^{2} = 1.335(1-\alpha^{2})$

$8\alpha^{2} = 1.335 - 1.335\alpha^{2}$

$9.335\alpha^{2} = 1.335$

$\alpha^{2} = \frac{1.335}{9.335} = 0.1430$

$\alpha = 0.378$

The new degree of dissociation is 0.378 (37.8%), which is less than 0.50, consistent with Le Chatelier's principle (increasing pressure favours the side with fewer gas moles).

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Integration Test 2: Industrial Process Optimisation

Question

The Haber process for ammonia synthesis:

$N_{2}(g) + 3H_{2}(g) \rightleftharpoons 2NH_{3}(g) \quad \Delta H = -92 \text{ kJ/mol}$

(a) Explain why industrial conditions use a temperature of approximately 450$^{\circ}$C rather than a lower temperature, even though a lower temperature would give a higher equilibrium yield. [3 marks]

(b) Explain why industrial conditions use a high pressure (200 atm) rather than atmospheric pressure. [2 marks]

(c) At 450$^{\circ}$C and 200 atm, the equilibrium yield of ammonia is approximately 15%. An engineer suggests using 500 atm to increase the yield. Calculate the approximate equilibrium yield at 500 atm, assuming ideal gas behaviour and that only pressure changes. [3 marks]

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Worked Solution

(a) Although a lower temperature would shift the equilibrium to the right (exothermic forward reaction) and increase the equilibrium yield, it would also decrease the rate of reaction significantly. At very low temperatures, the reaction would be too slow to be economically viable. A compromise temperature of 450$^{\circ}$C provides a reasonable rate (via the iron catalyst) while still achieving an acceptable equilibrium yield. Additionally, at 450$^{\circ}$C the iron catalyst is most effective.

(b) High pressure shifts the equilibrium to the right (towards fewer gas moles: 4 mol of gas on the left vs 2 mol on the right), increasing the equilibrium yield of ammonia. This is in accordance with Le Chatelier's principle.

(c) Assuming ideal gas behaviour, the mole fraction of $NH_{3}$ at equilibrium scales approximately with pressure for this reaction. For the Haber process at a given temperature, increasing pressure by a factor of $500/200 = 2.5$ increases the yield. However, the relationship is not linear.

A reasonable estimate: if the yield is 15% at 200 atm, at 500 atm it would be approximately 25--30%. The exact calculation requires solving the equilibrium expression, but qualitatively, increasing pressure favours ammonia production.

For a rough estimate using the simplified relationship: yield $\propto P^{0.5}$ (approximately), yield at 500 atm $\approx 15\% \times \sqrt{500/200} = 15\% \times 1.58 = 23.7\%$.

A more accurate estimate would give approximately 25%.

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Integration Test 3: Equilibrium Position + Quantitative Prediction

Question

For the equilibrium:

$H_{2}(g) + I_{2}(g) \rightleftharpoons 2HI(g)$

$K_{c} = 49.0$ at a certain temperature. 2.00 mol of $HI$ is placed in a 1.0 dm$^{3}$ container and allowed to reach equilibrium.

(a) Calculate the equilibrium concentrations of $H_{2}$, $I_{2}$, and $HI$. [5 marks]

(b) At equilibrium, an additional 1.00 mol of $I_{2}$ is added to the container. Calculate the new equilibrium concentrations after the system re-equilibrates. [4 marks]

(c) After re-equilibration, is the value of $K_{c}$ the same as before? Explain. [1 mark]

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Worked Solution

(a) Initial: $[HI] = 2.00$ mol/dm$^{3}$, $[H_{2}] = [I_{2}] = 0$.

Let $x$ mol/dm$^{3}$ of $HI$ dissociate.

At equilibrium: $[HI] = 2.00 - 2x$, $[H_{2}] = x$, $[I_{2}] = x$.

$K_{c} = \frac{[HI]^{2}}{[H_{2}][I_{2}]} = \frac{(2.00 - 2x)^{2}}{x^{2}} = 49.0$

$\frac{2.00 - 2x}{x} = 7.00$

$2.00 - 2x = 7x$

$2.00 = 9x$

$x = 0.222 \text{ mol/dm}^{3}$

At equilibrium:

• $[H_{2}] = [I_{2}] = 0.222$ mol/dm$^{3}$
• $[HI] = 2.00 - 2(0.222) = 1.556$ mol/dm$^{3}$

(b) After adding 1.00 mol $I_{2}$ to 1.0 dm$^{3}$: $[I_{2}]_{\text{new initial}} = 0.222 + 1.00 = 1.222$ mol/dm$^{3}$.

The equilibrium shifts left to partially consume the added $I_{2}$. Let $y$ mol/dm$^{3}$ of $H_{2}$ react with $I_{2}$.

New equilibrium: $[H_{2}] = 0.222 - y$, $[I_{2}] = 1.222 - y$, $[HI] = 1.556 + 2y$.

$K_{c} = \frac{(1.556 + 2y)^{2}}{(0.222 - y)(1.222 - y)} = 49.0$

$1.556 + 2y = 7.00\sqrt{(0.222 - y)(1.222 - y)}$

Expanding: $(1.556 + 2y)^{2} = 49.0(0.222 - y)(1.222 - y)$

$2.421 + 6.224y + 4y^{2} = 49.0(0.2713 - 1.444y + y^{2})$

$2.421 + 6.224y + 4y^{2} = 13.29 - 70.76y + 49.0y^{2}$

$0 = 10.87 - 76.98y + 45y^{2}$

Using the quadratic formula: $y = \frac{76.98 \pm \sqrt{76.98^{2} - 4 \times 45 \times 10.87}}{2 \times 45}$

$= \frac{76.98 \pm \sqrt{5926 - 1957}}{90} = \frac{76.98 \pm \sqrt{3969}}{90} = \frac{76.98 \pm 63.0}{90}$

$y = \frac{76.98 - 63.0}{90} = \frac{13.98}{90} = 0.155$ (taking the smaller root, since $y \lt 0.222$)

New equilibrium concentrations:

• $[H_{2}] = 0.222 - 0.155 = 0.067$ mol/dm$^{3}$
• $[I_{2}] = 1.222 - 0.155 = 1.067$ mol/dm$^{3}$
• $[HI] = 1.556 + 2(0.155) = 1.866$ mol/dm$^{3}$

(c) Yes, $K_{c}$ remains the same ($49.0$). $K_{c}$ depends only on temperature, and the temperature has not changed. The equilibrium position shifts, but the ratio of product concentrations to reactant concentrations at equilibrium remains constant.