How many NESA Chemistry questions cover Calculating the Equilibrium Constant?

AusGrader has 51 NESA Chemistry questions on Calculating the Equilibrium Constant, all with instant AI grading and detailed marking feedback.

NESA (HSC) Chemistry — Calculating the Equilibrium Constant Questions & Answers | AusGrader

Q15

2024

NESA

1 mark

Q15

1 mark

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

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

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

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

A

1 : 1

B

2 : 1

C

3 : 2

D

5 : 4

Q20

2023

NESA

1 mark

Q20

1 mark

Nitrogen monoxide and oxygen combine to form nitrogen dioxide, according to the following equation.

2\text{NO}(g) + \text{O}_2(g) \rightleftharpoons 2\text{NO}_2(g) \quad \quad K_{eq} = 2.47 \times 10^{12}

A 2.00 L vessel is filled with 1.80 mol of $\text{NO}_2(g)$ and the system is allowed to reach equilibrium.

What is the equilibrium concentration of $\text{NO}(g)$ ?

A

$0.00 \text{ mol L}^{-1}$

B

$4.34 \times 10^{-5} \text{ mol L}^{-1}$

C

$6.90 \times 10^{-5} \text{ mol L}^{-1}$

D

$8.69 \times 10^{-5} \text{ mol L}^{-1}$

Correct Answer

This is correct. Using an ICE table, the initial concentration of $\text{NO}_2$ is $0.900 \text{ M}$ . At equilibrium, $[\text{NO}] = 2x$ and $[\text{O}_2] = x$ . Solving $2.47 \times 10^{12} = (0.900)^2 / ((2x)^2(x))$ yields $x = 4.34 \times 10^{-5}$ , making $[\text{NO}] = 2x = 8.69 \times 10^{-5} \text{ M}$ .

Q39

2024

NESA

4 marks

Q39

4 marks

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

\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

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 $0.1000 \text{ mol L}^{-1}$ 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 $K_a = 1.29 \times 10^{-3}$ . When the system has reached equilibrium, the $[\text{H}^+]$ is $9.18 \times 10^{-3} \text{ mol L}^{-1}$ .

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

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

\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*}

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

K_{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

Descriptor	Marks
Calculates the $K_{eq}$	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

Q23

2024

NESA

3 marks

Q23

3 marks

Consider the following equilibrium system.

[\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)

$[\text{Co(H}_2\text{O)}_6]^{2+}(aq)$ is pink and $[\text{CoCl}_4]^{2-}(aq)$ 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 $K_{eq}$ .

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

K_{eq} = \frac{[[\text{CoCl}_4]^{2-}]}{[[\text{Co}(\text{H}_2\text{O})_6]^{2+}][\text{Cl}^-]^4}

Marking Criteria

Descriptor	Marks
Provides a sound relationship between the observed colour change and $K_{eq}$	3
Provides some information about changes in concentration of reactants and/or products	2
Provides some relevant information	1
None of the above	0

Q21

2024

QCAA

Paper 1

3 marks

Q21

Reactants A and B are placed in a 1.00 L container and react to form product C. The reaction then reaches equilibrium.

$\text{A(g)} + \text{B(g)} \rightleftharpoons \text{C(g)} \quad (K_c = 9.9 \times 10^2)$

Q21a

1 mark

Explain why a reversible arrow ( $\rightleftharpoons$ ) is used to symbolise this reaction.

The reversible arrow indicates that the reaction occurs in both directions (forward and reverse) simultaneously.

Marking Criteria

Descriptor	Marks
Explains that forward and reverse reactions occur simultaneously	1

Q21b

2 marks

Deduce whether the equilibrium for the reaction lies towards the reactants or products. Explain your reasoning.

$K_c$ is greater than 1, therefore the equilibrium for the reaction lies towards the products.

Marking Criteria

Descriptor	Marks
Deduces the equilibrium lies towards the products	1
Explains that $K_c$ is greater than 1	1

NESA Chemistry Calculating the Equilibrium Constant

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