How many NESA Chemistry questions cover Static and Dynamic Equilibrium?

AusGrader has 37 NESA Chemistry questions on Static and Dynamic Equilibrium, all with instant AI grading and detailed marking feedback.

NESA (HSC) Chemistry — Static and Dynamic Equilibrium Questions & Answers | AusGrader

Q1

2023

QCAA

Paper 1

1 mark

Q1

1 mark

In a chemical equation at equilibrium, a reversible arrow ( $\rightleftharpoons$ ) symbolises that

A

the forward reaction has stopped but can be reversed.

B

the moles of reactants and products present are equal.

C

half of the reactants have been converted into products.

D

the concentration of reactants and products remains constant.

Q1

2020

QCAA

Paper 1

1 mark

Q1

1 mark

A partly filled water bottle is sealed and left on a bench in a room with a constant temperature. After several minutes, it is noted that the water level in the bottle remains constant. In the water bottle, the rate of evaporation is

A

less than the rate of condensation.

B

greater than the rate of condensation.

C

equal to the rate of condensation and equal to zero.

D

equal to the rate of condensation but not equal to zero.

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

Q25

2023

QCAA

Paper 1

7 marks

Q25

During the contact process for manufacturing sulfuric acid, sulfur dioxide ( $\text{SO}_2$ ) and oxygen ( $\text{O}_2$ ) are passed over a vanadium oxide catalyst to produce sulfur trioxide ( $\text{SO}_3$ ). In the process, the vanadium oxide undergoes the following reactions.

Reaction 1: $\text{SO}_2\text{(g)} + \text{V}_2\text{O}_5\text{(s)} \rightarrow \text{SO}_3\text{(g)} + \text{V}_2\text{O}_4\text{(s)}$
Reaction 2: $2\text{V}_2\text{O}_4\text{(s)} + \text{O}_2\text{(g)} \rightarrow 2\text{V}_2\text{O}_5\text{(s)}$
Overall reaction: $2\text{SO}_2\text{(g)} + \text{O}_2\text{(g)} \xrightarrow{\text{V}_2\text{O}_5\text{(s)}} 2\text{SO}_3\text{(g)}$

Q25a

1 mark

Determine the oxidation state of vanadium in $\text{V}_2\text{O}_4\text{(s)}$ .

Descriptor	Marks
determines that the oxidation state of vanadium is +4	1

Q25b

2 marks

Determine if vanadium in $\text{V}_2\text{O}_5\text{(s)}$ in reaction 1 is acting as an oxidising or reducing agent. Explain your reasoning.

Vanadium in $V_2O_5(s)$ is acting as an oxidising agent in reaction 1 as the oxidation state decreased from +5 to +4 in $V_2O_4(s)$ .

Marking Criteria

Descriptor	Marks
determines that vanadium in $V_2O_5(s)$ is an oxidising agent	1
explains that oxidation state of vanadium decreases from +5 to +4	1

Q25c

4 marks

Use the reactions provided to explain why $\text{V}_2\text{O}_5\text{(s)}$ is a catalyst for the overall reaction.

Reaction 1: $V_2O_5(s)$ is converted to $V_2O_4(s)$ to allow $SO_2$ to be converted to $SO_3$ .
Reaction 2: $V_2O_4(s)$ is converted back to $V_2O_5(s)$ by reacting with $O_2$ .
Therefore, $V_2O_5(s)$ acts as a catalyst because it undergoes a temporary chemical change during the reaction but remains chemically unchanged at the end.

Marking Criteria

Descriptor	Marks
identifies that $V_2O_5(s)$ is involved in the chemical reaction	1
uses data from reaction 1 to support chemical change to $V_2O_4(s)$	1
uses data from reaction 2 to support chemical change to $V_2O_5(s)$	1
explains that $V_2O_5(s)$ remains unchanged at the end of the reaction	1

Q7

2023

QCAA

Paper 2

4 marks

Q7

When heated in a sealed container, solid mercury(II) oxide (HgO) decomposed to form metallic mercury (Hg) and oxygen gas (O $_2$ ).

$2\text{HgO(s)} \rightleftharpoons 2\text{Hg(l)} + \text{O}_2\text{(g)}$
Orange $\quad$ Silver $\quad$ Colourless

Q7a

1 mark

Identify whether the reaction occurs in an open or closed system.

Descriptor	Marks
identifies closed system	1

Q7b

3 marks

Explain why the colour of the system does not change once equilibrium is established.

At equilibrium there is no net change in the concentration of the reactants and the products.
The forward and reverse reactions are still occurring.
As the forward reaction is equal to the rate of the reverse reaction the colour of the system at equilibrium remains constant.

Marking Criteria

Descriptor	Marks
identifies the concentration of the reactants and products remain constant at equilibrium	1
explains that the forward and reverse reactions are occurring simultaneously	1
explains forward reaction is equal to reverse reaction and therefore there is no colour change	1

NESA Chemistry Static and Dynamic Equilibrium

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