NESA Chemistry Calculating the Equilibrium Constant

15 sample questions with marking guides and sample answers · Avg. score: 61.8%

Q5
2023
QCAA
Paper 1
1 mark
Q5
1 mark

The question refers to the decomposition of hydrogen iodide gas (HI) to produce hydrogen gas (H2\text{H}_2) and iodine gas (I2\text{I}_2) in a sealed 1-litre container.

2HI(g)H2(g)+I2(g)ΔH=+53.6 kJ mol12\text{HI(g)} \rightleftharpoons \text{H}_2\text{(g)} + \text{I}_2\text{(g)} \quad \Delta H = +53.6 \text{ kJ mol}^{-1}
Colourless Colourless Purple

Determine the equilibrium expression (KcK_c) for the reaction.

A

Kc=[H2][I2]2[HI]K_c = \frac{[\text{H}_2][\text{I}_2]}{2[\text{HI}]}

B

Kc=[H2][I2][HI]2K_c = \frac{[\text{H}_2][\text{I}_2]}{[\text{HI}]^2}

C

Kc=2[H]2[I]2[HI]K_c = \frac{2[\text{H}]2[\text{I}]}{2[\text{HI}]}

D

Kc=2[H]2[I][HI]2K_c = \frac{2[\text{H}]2[\text{I}]}{[\text{HI}]^2}

Reveal Answer
A

Kc=[H2][I2]2[HI]K_c = \frac{[\text{H}_2][\text{I}_2]}{2[\text{HI}]}

This is incorrect because the stoichiometric coefficient of the reactant (2) must be used as an exponent, not a multiplier. The correct term is [HI]2[\text{HI}]^2, not 2[HI]2[\text{HI}].

B

Kc=[H2][I2][HI]2K_c = \frac{[\text{H}_2][\text{I}_2]}{[\text{HI}]^2}

Correct Answer

This is correct based on the law of mass action for the reaction 2HIH2+I22\text{HI} \rightleftharpoons \text{H}_2 + \text{I}_2. The concentration of products is in the numerator, the reactant is in the denominator, and the coefficient 2 for HI becomes an exponent.

C

Kc=2[H]2[I]2[HI]K_c = \frac{2[\text{H}]2[\text{I}]}{2[\text{HI}]}

This is incorrect because it uses atomic species (H, I) instead of the molecular species (H2\text{H}_2, I2\text{I}_2) actually present in the reaction, and it treats coefficients as multipliers rather than exponents.

D

Kc=2[H]2[I][HI]2K_c = \frac{2[\text{H}]2[\text{I}]}{[\text{HI}]^2}

This is incorrect because it substitutes atomic concentrations for molecular products and uses coefficients as multipliers in the numerator.

Q15
2025
VCAA
1 mark
Q15
1 mark

Consider the following two reactions that are at equilibrium at 500 °C.

N2(g)+3H2(g)2NH3(g)Kc=2.86×101 M2N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g) \qquad K_c = 2.86 \times 10^{-1}\ \mathrm{M^{-2}}

4NH3(g)2N2(g)+6H2(g)4NH_3(g) \rightleftharpoons 2N_2(g) + 6H_2(g)

The magnitude of the value of KcK_c for the second reaction is

A

8.18×1028.18 \times 10^{-2}

B

5.72×1015.72 \times 10^{-1}

C

1.75×1001.75 \times 10^0

D

1.22×1011.22 \times 10^1

Reveal Answer
A

8.18×1028.18 \times 10^{-2}

This value is (Kc)2(K_c)^2, which would be the equilibrium constant if the first reaction were multiplied by 2 but not reversed.

B

5.72×1015.72 \times 10^{-1}

This value is 2×Kc2 \times K_c. When a reaction is multiplied by a coefficient, the equilibrium constant must be raised to that power, not multiplied by it.

C

1.75×1001.75 \times 10^0

This value results from incorrect mathematical manipulation of the equilibrium constant. The correct operation is to take the inverse square of the original KcK_c.

D

1.22×1011.22 \times 10^1

Correct Answer

The second reaction is the reverse of the first reaction multiplied by 2. Therefore, its equilibrium constant is (Kc)2=(2.86×101)2=1.22×101(K_c)^{-2} = (2.86 \times 10^{-1})^{-2} = 1.22 \times 10^1.

Q23
2024
NESA
3 marks
Q23
3 marks

Consider the following equilibrium system.

[Co(H2O)6]2+(aq)+4Cl(aq)[CoCl4]2(aq)+6H2O(l)[\text{Co(H}_2\text{O)}_6]^{2+}(aq) + 4\text{Cl}^-(aq) \rightleftharpoons [\text{CoCl}_4]^{2-}(aq) + 6\text{H}_2\text{O}(l)

[Co(H2O)6]2+(aq)[\text{Co(H}_2\text{O)}_6]^{2+}(aq) is pink and [CoCl4]2(aq)[\text{CoCl}_4]^{2-}(aq) is blue. When a solution of these ions and chloride ions is heated, the mixture becomes more blue.

Relate the observed colour change to the change in KeqK_{eq}.

Reveal Answer

The mixture becomes more blue as temperature is increased, which suggests that the concentration of [CoCl4]2[\text{CoCl}_4]^{2-} is increasing. This means that an increase in temperature favours the forward reaction. This response will result in a larger value of KeqK_{eq} as

Keq=[[CoCl4]2][[Co(H2O)6]2+][Cl]4K_{eq} = \frac{[[\text{CoCl}_4]^{2-}]}{[[\text{Co}(\text{H}_2\text{O})_6]^{2+}][\text{Cl}^-]^4}
Marking Criteria
DescriptorMarks

Provides a sound relationship between the observed colour change and KeqK_{eq}

3

Provides some information about changes in concentration of reactants and/or products

2

Provides some relevant information

1

None of the above

0
Q5
2020
SCSA
1 mark
Q5
1 mark

Which of the following statements about pure water are correct?

(i) Pure water is a weak electrolyte that undergoes self-ionisation.
(ii) The equilibrium constant for the ionisation of pure water at 25 °C is 1.00×10141.00 \times 10^{-14}.
(iii) Pure water ionises completely at 25 °C, hence [H⁺] = [OH⁻].
(iv) The ionisation of pure water produces twice as many hydrogen ions as hydroxide ions.

A

i and ii only

B

ii and iii only

C

iii and iv only

D

i, ii, iii and iv

Reveal Answer
A

i and ii only

Correct Answer

Statements (i) and (ii) are correct. Pure water is a weak electrolyte that slightly self-ionises, and its ion-product constant (KwK_w) at 25 °C is 1.00×10141.00 \times 10^{-14}.

B

ii and iii only

Statement (iii) is incorrect. While it is true that [H+]=[OH][H^+] = [OH^-] in pure water, water only ionises to a very small extent, not completely.

C

iii and iv only

Statements (iii) and (iv) are both incorrect. Water does not ionise completely, and its self-ionisation produces equal amounts of hydrogen and hydroxide ions, not twice as many.

D

i, ii, iii and iv

This option is incorrect because statements (iii) and (iv) are false. Water only partially ionises and produces a 1:1 ratio of hydrogen to hydroxide ions.

Q8
2023
QCAA
Paper 2
7 marks
Q8

Two experiments were conducted to investigate the effect of temperature on the equilibrium formed during the decomposition of hydrogen iodide (HI).

2HI(g)H2(g)+I2(g)ΔH=+53.6 kJ mol12\text{HI(g)} \rightleftharpoons \text{H}_2\text{(g)} + \text{I}_2\text{(g)} \quad \Delta H = +53.6 \text{ kJ mol}^{-1}

ExperimentInitial [HI]Initial [H2_2]Initial [I2_2]Eq [HI]Eq [H2_2]Eq [I2_2]KcK_c
10.080.000.00 0.01 2.78 ×\times 102^{-2}
20.000.060.060.060.030.03 

(Note: Concentrations are in mol L1^{-1})

Q8a
2 marks

Determine the concentration of HI(g) and I2_2(g) at equilibrium for experiment 1.

[HI]:

[I2_2]:

Reveal Answer

[HI]:0.06 mol L1[HI]: 0.06 \text{ mol L}^{-1}
[I2]:0.01 mol L1[I_2]: 0.01 \text{ mol L}^{-1}

Marking Criteria
DescriptorMarks

determines [HI] is 0.06

1

determines [I₂] is 0.01

1
Q8b
2 marks

Calculate the equilibrium constant (KcK_c) for experiment 2. Show your working.

Reveal Answer

Kc=[H2][I2][HI]2K_c = \frac{[H_2][I_2]}{[HI]^2}

Kc=(0.03)2(0.06)2=0.25K_c = \frac{(0.03)^2}{(0.06)^2} = 0.25

Marking Criteria
DescriptorMarks

uses appropriate substitution

1

calculates KcK_c is 0.25

1
Q8c
3 marks

Determine which experiment was conducted at a higher temperature. Explain your reasoning.

Reveal Answer

Temperature was higher for experiment 2. The KcK_c value was larger for experiment 2, indicating that the equilibrium shifted towards the products (endothermic direction) to compensate for the increase in temperature.

Marking Criteria
DescriptorMarks

determines experiment 2 was conducted at a higher temperature

1

explains that KcK_c value for experiment 2 is larger

1

explains higher KcK_c value indicates equilibrium shifts to the right

1
Q20
2021
SCSA
1 mark
Q20
1 mark

Consider the following reversible reaction:

2NO(g)+Cl2(g)2NOCl(g)Kc=6.5×104 at 35 °C2 NO(g) + Cl_2(g) \rightleftharpoons 2 NOCl(g) \qquad K_c = 6.5 \times 10^4 \text{ at 35 °C}

Which of the following statements describes the relative concentrations of reactants and products in this system when equilibrium is established in a closed vessel at 35 °C?

A

The concentrations of reactants and products will be equal.

B

There will be a greater concentration of products than reactants.

C

The reactant concentration will be greater than that of the products.

D

The concentrations of NOCl and NO will be double the concentration of Cl2Cl_2.

Reveal Answer
A

The concentrations of reactants and products will be equal.

Equal concentrations of reactants and products do not generally occur at equilibrium, especially when the equilibrium constant KcK_c is significantly different from 1.

B

There will be a greater concentration of products than reactants.

Correct Answer

Because the equilibrium constant KcK_c is much greater than 1 (6.5×1046.5 \times 10^4), the equilibrium lies far to the right, meaning the formation of products is strongly favored.

C

The reactant concentration will be greater than that of the products.

A greater concentration of reactants than products would only be expected if the equilibrium constant KcK_c was much less than 1, indicating a reactant-favored system.

D

The concentrations of NOCl and NO will be double the concentration of Cl2Cl_2.

The coefficients in the balanced chemical equation dictate the stoichiometric ratio in which substances react and form, not their absolute concentrations at equilibrium.

Q24
2020
QCAA
Paper 1
5 marks
Q24

This table shows the effect of temperature on the pH of pure water.

Temperature (°C)pH
107.27
157.17
207.08
257.00
306.92
506.63
Q24a
3 marks

Analyse the data to explain whether the self-ionisation of water is endothermic or exothermic. Explain your reasoning.

Reveal Answer

As the temperature increases, the [H3_3O+^+] increases.
2H2_2O(l) \rightleftharpoons H3_3O+^+(aq) + OH^-(aq)
Therefore, the equilibrium shifts towards the products.
Increasing temperature shifts equilibrium in the endothermic direction, therefore the self-ionisation of water is endothermic.

Marking Criteria
DescriptorMarks

Identifies [H3_3O+^+] increases as temperature increases

1

Identifies equilibrium shifts towards the products and the endothermic direction

1

Determines self-ionisation of water is endothermic

1
Q24b
2 marks

Calculate the KwK_w of pure water at 50 °C. Show your working. (to three significant figures)

Reveal Answer

pH = log[H+]=6.63-\log [H^+] = 6.63
[H+]=106.63[H^+] = 10^{-6.63}
=2.34×107= 2.34 \times 10^{-7}
Kw=[H+][OH]K_w = [H^+] [OH^-]
=(2.34×107)2= (2.34 \times 10^{-7})^2
Kw=5.48×1014K_w = 5.48 \times 10^{-14} (to 3 significant figures)

Marking Criteria
DescriptorMarks

Determines [H+]=2.34×107[H^+] = 2.34 \times 10^{-7}

1

Determines consequentially correct KwK_w

1
Q18
2022
QCAA
Paper 1
1 mark
Q18
1 mark

Determine the KbK_b expression for the weak base shown in the equilibrium equation.

NH3(aq)+H2O(l)NH4+(aq)+OH(aq)\text{NH}_3\text{(aq)} + \text{H}_2\text{O(l)} \rightleftharpoons \text{NH}_4^+\text{(aq)} + \text{OH}^-\text{(aq)}

A

Kb=[NH3][H2O][NH4+]K_b = \frac{[\text{NH}_3][\text{H}_2\text{O}]}{[\text{NH}_4^+]}

B

Kb=[NH3][H2O][OH]K_b = \frac{[\text{NH}_3][\text{H}_2\text{O}]}{[\text{OH}^-]}

C

Kb=[NH4+][OH][NH3]K_b = \frac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_3]}

D

Kb=[NH4+][OH][H2O]K_b = \frac{[\text{NH}_4^+][\text{OH}^-]}{[\text{H}_2\text{O}]}

Reveal Answer
A

Kb=[NH3][H2O][NH4+]K_b = \frac{[\text{NH}_3][\text{H}_2\text{O}]}{[\text{NH}_4^+]}

This expression incorrectly places the reactants in the numerator and the products in the denominator. Additionally, pure liquids like H2O\text{H}_2\text{O} are excluded from equilibrium expressions.

B

Kb=[NH3][H2O][OH]K_b = \frac{[\text{NH}_3][\text{H}_2\text{O}]}{[\text{OH}^-]}

This option incorrectly places reactants in the numerator and includes liquid water. Equilibrium expressions follow the ratio [products][reactants]\frac{[\text{products}]}{[\text{reactants}]}.

C

Kb=[NH4+][OH][NH3]K_b = \frac{[\text{NH}_4^+][\text{OH}^-]}{[\text{NH}_3]}

Correct Answer

The KbK_b expression is the product of the concentrations of the products divided by the reactant concentration (K=[products][reactants]K = \frac{[\text{products}]}{[\text{reactants}]}). Pure liquids, such as H2O\text{H}_2\text{O}, are omitted from the expression.

D

Kb=[NH4+][OH][H2O]K_b = \frac{[\text{NH}_4^+][\text{OH}^-]}{[\text{H}_2\text{O}]}

This expression incorrectly includes water in the denominator. Pure liquids are omitted from equilibrium constants, and the reactant NH3\text{NH}_3 is missing.

Q8
2023
VCAA
1 mark
Q8
1 mark

At 327 °C, the equilibrium constant for the reaction N2(g)+3H2(g)2NH3(g)\text{N}_2\text{(g)} + 3\text{H}_2\text{(g)} \rightleftharpoons 2\text{NH}_3\text{(g)} is 4.10 M24.10 \text{ M}^{-2}.

What is the equilibrium constant at 327 °C for the reaction 2N2(g)+6H2(g)4NH3(g)2\text{N}_2\text{(g)} + 6\text{H}_2\text{(g)} \rightleftharpoons 4\text{NH}_3\text{(g)}?

A

8.20 M28.20 \text{ M}^{-2}

B

8.20 M48.20 \text{ M}^{-4}

C

16.8 M216.8 \text{ M}^{-2}

D

16.8 M416.8 \text{ M}^{-4}

Reveal Answer
A

8.20 M28.20 \text{ M}^{-2}

Incorrect. When a reaction equation is multiplied by a factor of 2, the equilibrium constant must be squared, not multiplied by 2.

B

8.20 M48.20 \text{ M}^{-4}

Incorrect. Although the units are correctly squared, the numerical value was incorrectly multiplied by 2 instead of being squared.

C

16.8 M216.8 \text{ M}^{-2}

Incorrect. The numerical value was correctly squared, but the units must also be squared from M2\text{M}^{-2} to M4\text{M}^{-4}.

D

16.8 M416.8 \text{ M}^{-4}

Correct Answer

Correct. When a reaction equation is multiplied by a factor of 2, the new equilibrium constant is the square of the original. Therefore, Knew=(4.10 M2)2=16.8 M4K_{new} = (4.10 \text{ M}^{-2})^2 = 16.8 \text{ M}^{-4}.

Q39
2024
NESA
4 marks
Q39
4 marks

Water and octan-1-ol do not mix. When an aqueous solution of bromoacetic acid (BrCH2COOH\text{BrCH}_2\text{COOH}) is shaken with octan-1-ol, an equilibrium system is established between bromoacetic acid dissolved in the octan-1-ol and in the water.

BrCH2COOH(aq)BrCH2COOH(octan-1-ol)\text{BrCH}_2\text{COOH}(aq) \rightleftharpoons \text{BrCH}_2\text{COOH}(octan\text{-}1\text{-}ol)

The equilibrium constant expression for this system is

Keq=[BrCH2COOH(octan-1-ol)][BrCH2COOH(aq)].K_{eq} = \frac{[\text{BrCH}_2\text{COOH}(octan\text{-}1\text{-}ol)]}{[\text{BrCH}_2\text{COOH}(aq)]}.

An aqueous solution of bromoacetic acid with an initial concentration of 0.1000 mol L10.1000 \text{ mol L}^{-1} is shaken with an equal volume of octan-1-ol. Bromoacetic acid does not dissociate in octan-1-ol but does dissociate in water, with Ka=1.29×103K_a = 1.29 \times 10^{-3}. When the system has reached equilibrium, the [H+][\text{H}^+] is 9.18×103 mol L19.18 \times 10^{-3} \text{ mol L}^{-1}.

Calculate the equilibrium concentration of aqueous bromoacetic acid and hence, or otherwise, calculate the KeqK_{eq} for the octan-1-ol and water system.

Reveal Answer

Ka=[H+][BrCH2COO][BrCH2COOH]K_a = \frac{[\text{H}^+][\text{BrCH}_2\text{COO}^-]}{[\text{BrCH}_2\text{COOH}]}

[BrCH2COOH](aq)=[H+][BrCH2COO]Ka=[9.18×103][9.18×103]1.29×103=0.0654 mol L1\begin{align*} [\text{BrCH}_2\text{COOH}](aq) &= \frac{[\text{H}^+][\text{BrCH}_2\text{COO}^-]}{K_a}\\ &= \frac{[9.18 \times 10^{-3}][9.18 \times 10^{-3}]}{1.29 \times 10^{-3}}\\ &= 0.0654 \text{ mol L}^{-1} \end{align*}

[BrCH2COOH](octan-1-ol)=0.10000.06549.18×103=0.0254 mol L1[\text{BrCH}_2\text{COOH}](\text{octan-1-ol}) = 0.1000 - 0.0654 - 9.18 \times 10^{-3} = 0.0254 \text{ mol L}^{-1}

Keq=[BrCH2COOH(octan-1-ol)][BrCH2COOH(aq)]=0.0254 mol L10.0654 mol L1=0.390K_{eq} = \frac{[\text{BrCH}_2\text{COOH}(\text{octan-1-ol})]}{[\text{BrCH}_2\text{COOH}(aq)]} = \frac{0.0254 \text{ mol L}^{-1}}{0.0654 \text{ mol L}^{-1}} = 0.390
Marking Criteria
DescriptorMarks

Calculates the KeqK_{eq}

4

Provides most steps of the calculation

3

Provides some steps of the calculation

2

Provides some relevant information

1

None of the above

0
Q17
2023
SCSA
1 mark
Q17
1 mark

In the conversion between sulfur dioxide and sulfur trioxide in a sealed vessel, the following equilibrium is established:

2 SO3(g)2 SO2(g)+O2(g)ΔH=198 kJ mol12\text{ SO}_3\text{(g)} \rightleftharpoons 2\text{ SO}_2\text{(g)} + \text{O}_2\text{(g)} \quad \Delta\text{H} = 198\text{ kJ mol}^{-1}

For this system, which of the following statements about the equilibrium constant K is correct? K will

A

increase if the temperature of the system is decreased.

B

decrease if the partial pressure of SO2(g)\text{SO}_2\text{(g)} is increased.

C

increase if the temperature of the system is increased.

D

increase if the pressure of the system is increased.

Reveal Answer
A

increase if the temperature of the system is decreased.

The reaction is endothermic (ΔH>0\Delta\text{H} > 0). Decreasing the temperature shifts the equilibrium to the left, which decreases the equilibrium constant KK.

B

decrease if the partial pressure of SO2(g)\text{SO}_2\text{(g)} is increased.

Changing the partial pressure of a reactant or product shifts the equilibrium position, but it does not change the value of the equilibrium constant KK, which is only temperature-dependent.

C

increase if the temperature of the system is increased.

Correct Answer

Because the reaction is endothermic (ΔH=198 kJ mol1\Delta\text{H} = 198\text{ kJ mol}^{-1}), increasing the temperature shifts the equilibrium to the right to absorb the added heat, resulting in a higher equilibrium constant KK.

D

increase if the pressure of the system is increased.

Changing the total pressure of the system will shift the equilibrium position (in this case, to the left where there are fewer moles of gas), but it will not change the equilibrium constant KK.

Q1
2025
SCSA
1 mark
Q1
1 mark

Which of the following gives the equilibrium constant expression for the following equilibrium?

Cu(OH)2(s)+4NH3(aq)[Cu(NH3)4]2+(aq)+2OH(aq)Cu(OH)_2(s) + 4 NH_3(aq) \leftrightharpoons [Cu(NH_3)_4]^{2+}(aq) + 2 OH^-(aq)

A

K=[[Cu(NH3)4]2+][OH]2[NH3]4K = \frac{[[Cu(NH_3)_4]^{2+}][OH^-]^2}{[NH_3]^4}

B

K=[NH3]4[[Cu(NH3)4]2+][OH]2K = \frac{[NH_3]^4}{[[Cu(NH_3)_4]^{2+}][OH^-]^2}

C

K=[Cu(OH)2][NH3]4[[Cu(NH3)4]2+][OH]2K = \frac{[Cu(OH)_2][NH_3]^4}{[[Cu(NH_3)_4]^{2+}][OH^-]^2}

D

K=[[Cu(NH3)4]2+][OH]2[Cu(OH)2][NH3]4K = \frac{[[Cu(NH_3)_4]^{2+}][OH^-]^2}{[Cu(OH)_2] [NH_3]^4}

Reveal Answer
A

K=[[Cu(NH3)4]2+][OH]2[NH3]4K = \frac{[[Cu(NH_3)_4]^{2+}][OH^-]^2}{[NH_3]^4}

Correct Answer

This is the correct expression. It follows the law of mass action by placing product concentrations in the numerator and reactant concentrations in the denominator, raising them to the power of their coefficients, and omitting the solid Cu(OH)2Cu(OH)_2.

B

K=[NH3]4[[Cu(NH3)4]2+][OH]2K = \frac{[NH_3]^4}{[[Cu(NH_3)_4]^{2+}][OH^-]^2}

This expression is inverted. The equilibrium constant KK is defined as the ratio of product concentrations to reactant concentrations, not the other way around.

C

K=[Cu(OH)2][NH3]4[[Cu(NH3)4]2+][OH]2K = \frac{[Cu(OH)_2][NH_3]^4}{[[Cu(NH_3)_4]^{2+}][OH^-]^2}

This option is both inverted (reactants over products) and incorrectly includes the solid Cu(OH)2Cu(OH)_2. Pure solids are excluded from equilibrium expressions because their concentration remains constant.

D

K=[[Cu(NH3)4]2+][OH]2[Cu(OH)2][NH3]4K = \frac{[[Cu(NH_3)_4]^{2+}][OH^-]^2}{[Cu(OH)_2] [NH_3]^4}

This option incorrectly includes the concentration of the solid reactant, Cu(OH)2Cu(OH)_2. In heterogeneous equilibria, pure solids and liquids are omitted from the KK expression.

Q14
2020
QCAA
Paper 1
1 mark
Q14
1 mark

Ammonia gas reacts with oxygen gas in the following equilibrium reaction.

4NH3(g)+3O2(g)2N2(g)+6H2O(g)4NH_3(g) + 3O_2(g) \rightleftharpoons 2N_2(g) + 6H_2O(g)

The equilibrium expression for the reaction is

A

[NH3][O2][N2][H2O]\frac{[NH_3][O_2]}{[N_2][H_2O]}

B

[N2][H2O][NH3][O2]\frac{[N_2][H_2O]}{[NH_3][O_2]}

C

[NH3]4[O2]3[N2]2[H2O]6\frac{[NH_3]^4[O_2]^3}{[N_2]^2[H_2O]^6}

D

[N2]2[H2O]6[NH3]4[O2]3\frac{[N_2]^2[H_2O]^6}{[NH_3]^4[O_2]^3}

Reveal Answer
A

[NH3][O2][N2][H2O]\frac{[NH_3][O_2]}{[N_2][H_2O]}

This option incorrectly places reactants in the numerator and products in the denominator, and it fails to raise the concentrations to the power of their stoichiometric coefficients.

B

[N2][H2O][NH3][O2]\frac{[N_2][H_2O]}{[NH_3][O_2]}

While this option correctly places products over reactants, it fails to raise the concentrations to the power of their stoichiometric coefficients found in the balanced equation.

C

[NH3]4[O2]3[N2]2[H2O]6\frac{[NH_3]^4[O_2]^3}{[N_2]^2[H_2O]^6}

This expression is the inverse of the correct equilibrium constant; it incorrectly places the reactants in the numerator and the products in the denominator.

D

[N2]2[H2O]6[NH3]4[O2]3\frac{[N_2]^2[H_2O]^6}{[NH_3]^4[O_2]^3}

Correct Answer

The equilibrium expression is calculated as the product of product concentrations divided by the product of reactant concentrations, with each raised to the power of its stoichiometric coefficient: [N2]2[H2O]6[NH3]4[O2]3\frac{[N_2]^2[H_2O]^6}{[NH_3]^4[O_2]^3}.

Q15
2024
NESA
1 mark
Q15
1 mark

The thermal decomposition of lithium peroxide (Li2O2\text{Li}_2\text{O}_2) is given by the equation shown.

2Li2O2(s)2Li2O(s)+O2(g)2\text{Li}_2\text{O}_2(s) \rightleftharpoons 2\text{Li}_2\text{O}(s) + \text{O}_2(g)

Mixtures of Li2O2\text{Li}_2\text{O}_2, Li2O\text{Li}_2\text{O} and O2\text{O}_2 were allowed to reach equilibrium in two identical, closed containers, P and Q, at the same temperature. The amount of Li2O2(s)\text{Li}_2\text{O}_2(s) in container P is double that in container Q. The amount of Li2O(s)\text{Li}_2\text{O}(s) is the same in each container.

What is the ratio of [O2(g)][\text{O}_2(g)] in container P to [O2(g)][\text{O}_2(g)] in container Q?

A

1 : 1

B

2 : 1

C

3 : 2

D

5 : 4

Reveal Answer
A

1 : 1

Correct Answer

The equilibrium constant expression is Kc=[O2(g)]K_c = [\text{O}_2(g)] because solids are excluded. Since the temperature is the same in both containers, the equilibrium concentration of O2(g)\text{O}_2(g) is identical, making the ratio 1:1.

B

2 : 1

This incorrectly assumes the concentration of O2(g)\text{O}_2(g) is proportional to the amount of solid Li2O2\text{Li}_2\text{O}_2. The amounts of pure solids do not affect the equilibrium position.

C

3 : 2

This ratio has no basis in the equilibrium expression. The amounts of solid reactants and products do not change the equilibrium concentration of the gas.

D

5 : 4

This is incorrect because the equilibrium concentration of O2(g)\text{O}_2(g) depends only on the temperature-dependent equilibrium constant, not on the initial amounts of solids.

Q6
2022
QCAA
Paper 1
1 mark
Q6
1 mark

The equilibrium concentration of A is 2.8×1042.8 \times 10^{-4} M and B is 1.2×1041.2 \times 10^{-4} M.

A(g)B(g)ΔH>0\text{A(g)} \rightleftharpoons \text{B(g)} \quad \Delta H > 0

Which option represents the ratio of molecules present in a sample of the gaseous mixture when the temperature is decreased and a new equilibrium established?

A

8 molecules of A and 2 molecules of B

B

5 molecules of A and 5 molecules of B

C

3 molecules of A and 7 molecules of B

D

2 molecules of A and 8 molecules of B

Reveal Answer
A

8 molecules of A and 2 molecules of B

Correct Answer

The reaction is endothermic (ΔH>0\Delta H > 0), so decreasing the temperature shifts the equilibrium to the left (reactants) to generate heat. Since the initial concentration of A was already higher than B (2.8>1.22.8 > 1.2), shifting left increases the ratio of A to B further, making the 8:2 distribution the correct representation.

B

5 molecules of A and 5 molecules of B

This option represents a 1:1 ratio of A to B. Since the initial mixture already contained more A than B and the equilibrium shifts further toward A upon cooling, the final mixture must have a ratio of A to B significantly greater than 1.

C

3 molecules of A and 7 molecules of B

This option shows a higher amount of product B than reactant A. Decreasing the temperature of an endothermic reaction favors the reverse reaction, which would decrease the amount of B and increase the amount of A.

D

2 molecules of A and 8 molecules of B

This option implies a shift toward the product B. According to Le Chatelier's principle, removing heat (lowering temperature) from an endothermic reaction drives the system toward the reactants, so A should be the predominant species.

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