smc-3670-10-Temperature

CHEMISTRY, M5 2025 HSC 14 MC

The equation for the decomposition of hydrogen iodide is shown.

$\ce{2HI(g)\rightleftharpoons I2(g) + H2(g)} \quad \quad \Delta H=+52 \ \text{kJ mol}^{-1}$

The equilibrium formed during this reaction was investigated in two experiments carried out at different temperatures. The initial and equilibrium concentrations for both experiments are shown in the table, with only the $K_{e q}$ for Experiment 1 shown.

Which row in the table correctly compares features of the two experiments?

\begin{align*}
\begin{array}{l}
\rule{0pt}{2.5ex} \ \rule[-1ex]{0pt}{0pt}& \\
\ \rule[-1ex]{0pt}{0pt}& \\
\rule{0pt}{2.5ex}\textbf{A.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{B.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{C.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{D.}\rule[-1ex]{0pt}{0pt}\\
\end{array}
\begin{array}{|c|c|}
\hline
\rule{0pt}{2.5ex}\quad \quad \quad K_{e q}\quad \quad\quad& \ \ \text{Temperature of}\ \ \\
\ \rule[-1ex]{0pt}{0pt}& \textit{experiment} \\
\hline
\rule{0pt}{2.5ex}\text{Lower in 1}\rule[-1ex]{0pt}{0pt}&\text{Lower in 2}\\
\hline
\rule{0pt}{2.5ex}\text{Lower in 1}\rule[-1ex]{0pt}{0pt}& \text{Higher in 2}\\
\hline
\rule{0pt}{2.5ex}\text{Higher in 1}\rule[-1ex]{0pt}{0pt}& \text{Lower in 2} \\
\hline
\rule{0pt}{2.5ex}\text{Higher in 1}\rule[-1ex]{0pt}{0pt}& \text{Higher in 2} \\
\hline
\end{array}
\end{align*}

Show Answers Only

$B$

Show Worked Solution

The equilibrium constant for experiment two:
$K_{eq} = \dfrac{\ce{[H2][I2]}}{\ce{[HI]^2}} = \dfrac{0.02 \times 0.02}{0.04^2} = 0.25$
Therefore the $K_{eq}$ for experiment 1 is lower than the $K_{eq}$ for experiment 2.
For an endothermic reaction (thinking of heat as a reactant), increasing the temperature of the reaction will shift the equilibrium to the products and increase $K_{eq}$ which is what occured in experiment 2.
Hence, the temperature of the experiment is higher in 2.

$\Rightarrow B$

CHEMISTRY, M5 EQ-Bank 13

A student hypothesises that increasing the temperature of an exothermic reaction slows down the reaction rate because less products are produced. Is this student correct? Give reasons. (4 marks)

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The student is incorrect.
Exothermic reactions release heat to the surrounding environment, as modelled in this general reaction equation:
$\ce{R_{(1)} + R_{(2)} \rightleftharpoons P_{(1)} + P_{(2)} + Heat}$
Increasing the temperature will result in fewer products being produced as the position of equilibrium shifts to the left (as per Le Chatelier’s principle).
However, contrary to the student’s hypothesis, fewer products being produced does not cause the reaction rate to slow down.
In fact, the rate of reaction will increase as the kinetic energy of particles will increase, leading to more successful particle collisions that exceed $\ce{E_{a}}$.
The mistake the student has made is viewing fewer products as causing a decrease in reaction rate. The actual occurrence has been a shift in equilibrium to the left and an increase in the (forward) reaction rate to reach equilibrium.

Show Worked Solution

The student is incorrect.
Exothermic reactions release heat to the surrounding environment, as modelled in this general reaction equation:
$\ce{R_{(1)} + R_{(2)} \rightleftharpoons P_{(1)} + P_{(2)} + Heat}$
Increasing the temperature will result in fewer products being produced as the position of equilibrium shifts to the left (as per Le Chatelier’s principle).
However, contrary to the student’s hypothesis, fewer products being produced does not cause the reaction rate to slow down.
In fact, the rate of reaction will increase as the kinetic energy of particles will increase, leading to more successful particle collisions that exceed $\ce{E_{a}}$.
The mistake the student has made is viewing fewer products as causing a decrease in reaction rate. The actual occurrence has been a shift in equilibrium to the left and an increase in the (forward) reaction rate to reach equilibrium.

CHEMISTRY, M5 2023 HSC 18 MC

Carbon dioxide reacts with hydrogen gas to form carbon monoxide and water vapour in a sealed flask, according to the following equation.

$ \ce{CO2(g) + H2(g)} \rightleftharpoons \ce{CO(g) + H2O(g)} $

A temperature change was imposed on the equilibrium system at time $t$ and the rates of both the forward and reverse reactions were monitored.

Which row of the table correctly identifies the nature of both temperature change at time $ t $ and the $ \Delta H $ of the forward reaction?

\begin{align*}
\begin{array}{l}
\rule{0pt}{2.5ex}\textit{}& \textit{} \\
\textit{}\rule[-1ex]{0pt}{0pt}& \textit{} \\
\rule{0pt}{2.5ex}\textbf{A.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{B.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{C.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{D.}\rule[-1ex]{0pt}{0pt}\\
\end{array}
\begin{array}{|c|c|}
\hline
\rule{0pt}{2.5ex}\textit{Temperature change}& \textit{$\Delta H$ of the forward} \\
\textit{at time $t$}\rule[-1ex]{0pt}{0pt}& \textit{reaction} \\
\hline
\rule{0pt}{2.5ex}\text{Decrease}\rule[-1ex]{0pt}{0pt}&\text{+}\\
\hline
\rule{0pt}{2.5ex}\text{Decrease}\rule[-1ex]{0pt}{0pt}& \text{–}\\
\hline
\rule{0pt}{2.5ex}\text{Increase}\rule[-1ex]{0pt}{0pt}& \text{+} \\
\hline
\rule{0pt}{2.5ex}\text{Increase}\rule[-1ex]{0pt}{0pt}& \text{–} \\
\hline
\end{array}
\end{align*}

Show Answers Only

$A$

Show Worked Solution

The rate decrease is associated with a decrease in temperature (collision theory).
Since the rate slowed, the forward reaction is endothermic (requires heat to react) and thus has a positive (+) $ \Delta H $

$\Rightarrow A$

♦♦ Mean mark 39%.

CHEMISTRY, M5 2023 HSC 33

Gases $ \ce{A_2} $ and $ \ce{B_2} $ are placed in a closed container of variable volume, as shown.

The reaction between these substances is as follows.

$ \ce{A2(g) + 2B_2(g) \rightleftharpoons 2AB_2(g) \quad \Delta \textit{H} = -10 \text{kJ mol}^{-1}} $

The following graph shows changes in the amounts (in mol) of these three substances over time in this container.

Explain what is happening in this system between 6 minutes and 8 minutes. (2 marks)

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Explain TWO different factors that could result in the disturbance at 8 minutes. (4 marks)

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a. Between 6 and 8 minutes:

The system is in equilibrium.
The horizontal lines of each reactant in the graph indicate that the amount of reactants and products remain constant and hence the forward and reverse reactions are proceeding at the same rate.

b. After 8 minutes $\ce{AB2}$ is consumed, and $\ce{A2}$ and $\ce{B2}$ are produced.

Factor 1:

An increase in temperature that decreases the equilibrium constant, $\text{K}$.
In this case, the reaction quotient $\text{Q}$ will be greater than $\text{K}$. This will result in $\ce{AB2}$ being consumed and $\ce{A2}$ and $\ce{B2}$ being produced until $\text{Q}$ approaches $\text{K}$ and the system reaches equilibrium again.

Factor 2:

Increase in volume of the container.
This will increase the reaction quotient $\text{Q}$ while $\text{K}$ stays the same. Again, this will cause $\ce{AB2}$ to be consumed and $\ce{A2}$ and $\ce{B2}$ to be produced until $\text{Q}$ approaches $\text{K}$ and the system reaches equilibrium again.

Show Worked Solution

a. Between 6 and 8 minutes:

The system is in equilibrium.
The horizontal lines of each reactant in the graph indicate that the amount of reactants and products remain constant and hence the forward and reverse reactions are proceeding at the same rate.

b. After 8 minutes $\ce{AB2}$ is consumed, and $\ce{A2}$ and $\ce{B2}$ are produced.

Factor 1:

An increase in temperature that decreases the equilibrium constant, $\text{K}$.
In this case, the reaction quotient $\text{Q}$ will be greater than $\text{K}$. This will result in $\ce{AB2}$ being consumed and $\ce{A2}$ and $\ce{B2}$ being produced until $\text{Q}$ approaches $\text{K}$ and the system reaches equilibrium again.

Factor 2:

Increase in volume of the container.
This will increase the reaction quotient $\text{Q}$ while $\text{K}$ stays the same. Again, this will cause $\ce{AB2}$ to be consumed and $\ce{A2}$ and $\ce{B2}$ to be produced until $\text{Q}$ approaches $\text{K}$ and the system reaches equilibrium again.

CHEMISTRY, M5 EQ-Bank 12 MC

Nitrogen dioxide (a brown gas) and dinitrogen tetroxide (a colourless gas) are both forms of oxides of nitrogen. They are in equilibrium according to the equation

$ \ce{2NO2(g) \rightleftharpoons N2O4(g)}$.

An equilibrium mixture of the two gases at room temperature is light brown but at higher temperatures the colour becomes a much deeper brown.

What conclusion can be drawn from this observation?

The reverse reaction in the equation is endothermic.
The forward reaction in the equation is endothermic.
The brown colour is due to the strong nitrogen–oxygen bonds in $\ce{NO2}$.
The equilibrium concentration of $ \ce{N2O4}$ is not dependent on temperature.

Show Answers Only

`A`

Show Worked Solution

Le Chatelier’s principle states that when an equilibrium system is subject to a change in conditions, it will shift such as to partially counteract the imposed change.
When the temperature of the system is increased, it will shift to favour the endothermic reaction, counteracting the increase in temperature.
In this case, the deeper brown colour shows the conversion to nitrogen dioxide is favoured following an increase in temperature. The reverse reaction is therefore endothermic.

`=>A`

CHEMISTRY, M5 EQ-Bank 12

An industrial plant makes ammonia from nitrogen gas and hydrogen gas. The reaction is exothermic.

The graph shows the adjustments made to increase the yield of ammonia.

Account for the changes in conditions that have shaped the graph during the time the system was observed. Include a relevant chemical equation in your answer. (5 marks)

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$\ce{N2(g) + 3H2(g) \rightleftharpoons 2NH3(g) \ \ \ \ ΔH = -92 kJ mol}^{-1} $

From $ t_0$ to $t_1: $ the system is in equilibrium.

At $t_1:$

Nitrogen is introduced to the system and its concentration increases sharply.
Le Chatelier’s principle states that when a system in equilibrium is disturbed, the equilibrium will shift in the direction that minimises the change. In this case, the equilibrium will shift to the right to use up more nitrogen. A greater yield of ammonia will result until equilibrium is re-established.

At $t_2:$

The concentration of both reactants and products increases. This effect could be caused by a decrease in volume of the reaction vessel which will result in an increase in pressure on the system.
The above equation shows that 4 moles of gas (on the left-hand side) react to form 2 moles of gas (on the right-hand side). Le Chatelier’s principle dictates that this increase in pressure will cause the system to again shift right, to the side with fewer moles of gas, to counteract the change.
This right shift will further increase the yield of ammonia until equilibrium is re-established.

At $t_3:$

There is a change to the system that shifts the reaction back to the left. The gradual change in concentrations indicate that this could be due to a change in temperature.
Since this reaction is exothermic, the reverse reaction (left shift) absorbs heat. An increase in temperature would cause this shift, lowering the yield of ammonia until equilibrium is again restored.

Show Worked Solution

$\ce{N2(g) + 3H2(g) \rightleftharpoons 2NH3(g) \ \ \ \ ΔH = -92 kJ mol}^{-1} $

From $ t_0$ to $t_1: $ the system is in equilibrium.

At $t_1:$

Nitrogen is introduced to the system and its concentration increases sharply.
Le Chatelier’s principle states that when a system in equilibrium is disturbed, the equilibrium will shift in the direction that minimises the change. In this case, the equilibrium will shift to the right to use up more nitrogen. A greater yield of ammonia will result until equilibrium is re-established.

At $t_2:$

The concentration of both reactants and products increases. This effect could be caused by a decrease in volume of the reaction vessel which will result in an increase in pressure on the system.
The above equation shows that 4 moles of gas (on the left-hand side) react to form 2 moles of gas (on the right-hand side). Le Chatelier’s principle dictates that this increase in pressure will cause the system to again shift right, to the side with fewer moles of gas, to counteract the change.
This right shift will further increase the yield of ammonia until equilibrium is re-established.

At $t_3:$

There is a change to the system that shifts the reaction back to the left. The gradual change in concentrations indicate that this could be due to a change in temperature.
Since this reaction is exothermic, the reverse reaction (left shift) absorbs heat. An increase in temperature would cause this shift, lowering the yield of ammonia until equilibrium is again restored.

CHEMISTRY, M5 2018 HSC 25

The graph shows the number of molecules of $\ce{N2}$ and $\ce{H2}$ that possess a certain kinetic energy at two different temperatures.

With reference to the graph, explain why changing the temperature and adding a catalyst would change the rate of production of ammonia. (4 marks)

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Gas particles need to attain activation energy $ \text{E}_\text{A} $ in order to react when they collide.
$ \text{E}_\text{A} $ is a level of kinetic energy that is sufficient for reactions to occur.
Some $\ce{N2}$ and $\ce{H2}$ particles have a kinetic energy above $ \text{E}_\text{A} $ when the temperature is lower at $\text{T}_1$ (see graph above).
However, the number of $\ce{N2}$ and $\ce{H2}$ particles with enough energy to react and produce ammonia is significantly greater at the higher temperature $\text{T}_2$, where the graph has shifted to the right.
A catalyst lowers the $ \text{E}_\text{A} $. On the graph, this would shift the dotted $ \text{E}_\text{A} $ line to the left.
In the presence of a catalyst, both graphs show there will be a greater number of particles that can react in a given time and therefore the rate of ammonia production will increase in both cases.

Show Worked Solution

Gas particles need to attain activation energy $ \text{E}_\text{A} $ in order to react when they collide.
$ \text{E}_\text{A} $ is a level of kinetic energy that is sufficient for reactions to occur.
Some $\ce{N2}$ and $\ce{H2}$ particles have a kinetic energy above $ \text{E}_\text{A} $ when the temperature is lower at $\text{T}_1$ (see graph above).
However, the number of $\ce{N2}$ and $\ce{H2}$ particles with enough energy to react and produce ammonia is significantly greater at the higher temperature $\text{T}_2$, where the graph has shifted to the right.
A catalyst lowers the $ \text{E}_\text{A} $. On the graph, this would shift the dotted $ \text{E}_\text{A} $ line to the left.
In the presence of a catalyst, both graphs show there will be a greater number of particles that can react in a given time and therefore the rate of ammonia production will increase in both cases.

Mean mark 53%.

CHEMISTRY, M5 2016 HSC 28

A mixture of carbon monoxide, chlorine and phosgene $\ce{(COCl_2)}$ gases was placed in a closed container. The concentrations of the gases were monitored over time.

At what time does the system first reach equilibrium? Justify your answer. (2 marks)

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At four minutes, the temperature of the container was increased.
Explain, with reference to the graph, whether the decomposition of $\ce{COCl_2}$ into $\ce{CO}$ and $\ce{Cl_2}$ is exothermic or endothermic. (3 marks)

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a. At two minutes (where the concentrations stop changing)

b. $\ce{CO(g) + Cl2(g) \rightleftharpoons COCl2(g)}$

From the graph, when the temperature is increased, the $\ce{COCl2}$ concentration decreases while $\ce{CO}$ and $\ce{Cl2}$ concentrations increases.
The reaction is countering the increase in temperature by shifting to the left and absorbing heat (Le Chatelier’s principle).
Therefore the decomposition of $\ce{COCl2(g)}$ is endothermic.

Show Worked Solution

a. At two minutes (where the concentrations stop changing)

b. $\ce{CO(g) + Cl2(g) \rightleftharpoons COCl2(g)}$

From the graph, when the temperature is increased, the $\ce{COCl2}$ concentration decreases while $\ce{CO}$ and $\ce{Cl2}$ concentrations increases.
The reaction is countering the increase in temperature by shifting to the left and absorbing heat (Le Chatelier’s principle).
Therefore the decomposition of $\ce{COCl2(g)}$ is endothermic.

CHEMISTRY, M5 2016 HSC 14 MC

Consider the following endothermic reaction taking place in a closed vessel.

$\ce{N_2O_4($g$) \rightleftharpoons 2NO_2($g$)}$

Which of the following actions would cause more $\ce{N_2O_4}$ to be produced?

Adding a catalyst
Decreasing the volume
Decreasing the pressure
Increasing the temperature

Show Answers Only

`B`

Show Worked Solution

By decreasing the volume, the equilibrium will shift to the left so that less gas molecules are present (Le Chatelier’s principle).

`=>B`

CHEMISTRY, M5 2015 HSC 16 MC

The equation describes an equilibrium reaction occurring in a closed system.

$\ce{X(g) + Y(g) \rightleftharpoons 4Z(g)}\hspace{1.5em}\Delta{H} = +58 \:\text{kJ} $

Under which set of conditions would the highest yield of `\text{Z}(g)` be obtained?

\begin{align*}
\begin{array}{l}
\rule{0pt}{1.5ex} \ \rule[-0.5ex]{0pt}{0pt}& \\
\rule{0pt}{2.5ex}\textbf{A.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{B.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{C.}\rule[-1ex]{0pt}{0pt}\\
\rule{0pt}{2.5ex}\textbf{D.}\rule[-1ex]{0pt}{0pt}\\
\end{array}
\begin{array}{|c|c|}
\hline
\rule{0pt}{1.5ex}\textit{Temperature}\ \text{(°C)} \rule[-0.5ex]{0pt}{0pt}& \textit{Pressure}\ \text{(kPa)}\\
\hline
\rule{0pt}{2.5ex}50\rule[-1ex]{0pt}{0pt}&100\\
\hline
\rule{0pt}{2.5ex}50\rule[-1ex]{0pt}{0pt}& 200\\
\hline
\rule{0pt}{2.5ex}300\rule[-1ex]{0pt}{0pt}& 100 \\
\hline
\rule{0pt}{2.5ex}300\rule[-1ex]{0pt}{0pt}& 200 \\
\hline
\end{array}
\end{align*}

Show Answers Only

`C`

Show Worked Solution

Forward reaction is endothermic `(DeltaH=+58\ text{kJ})`
High temperature will shift reaction to the right.
Right hand side has more gas molecules (4 vs 2) and therefore the forward reaction will benefit from lower pressure.
Highest yield of `\text{Z}(g)` when temperature is higher and pressure lower.

`=>C`

CHEMISTRY, M5 2019 HSC 12 MC

Methanol can be produced from the reaction of carbon monoxide and hydrogen, according to the following equation:

$ \ce{CO(g) + 2H2(g) \rightleftharpoons CH3OH(g)}\ \ \ \ \ \ \text{Δ}H_r ^{\ \ominus} =-90\ \text{kJ mol}^{-1} $

Which set of conditions will produce the maximum yield of methanol?

Low pressure and low temperature
Low pressure and high temperature
High pressure and low temperature
High pressure and high temperature

Show Answers Only

`C`

Show Worked Solution

To maximise the methanol yield, the equation must shift towards the right hand side.
Pressure: If the pressure is increased, the system will shift to reduce this increase. In this reaction, a shift to the right hand side will occur because there are fewer gas molecules (1 on right vs 3 on the left).
Temperature: If the temperature is decreased, the equilibrium will shift to the exothermic side which absorbs heat (right hand side). This compensation will result in more methanol.

`=>C`

CHEMISTRY, M5 2020 HSC 26

Nitric oxide gas (`text{NO}`) can be produced from the direct combination of nitrogen gas and oxygen gas in a reversible reaction.

Write the balanced chemical equation for this reaction. (1 mark)

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The energy profile diagram for this reaction is shown.

Explain, using collision theory, how an increase in temperature would affect the value of `K_{eq}` for this system. Refer to the diagram in your answer. (4 marks)

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a. $\ce{N2(g) + O2(g) \rightleftharpoons 2NO(g) }$

b. Increased temperature’s effect on `K_{eq}`:

From the graph, the forward reaction is endothermic.
The activation energy of the forward endothermic reaction is greater than the activation energy of the reverse exothermic reaction.
An increase in temperature would cause the rates of both the forward and reverse reaction due to the higher average kinetic energy, resulting in a larger likelihood of a successful collisions.
However, the rate of the forward reaction would increase to a higher extent than the reverse reaction, since it is an endothermic reaction.
Using `K_(eq) = ([text{NO}]^2)/([text{N}_2][text{O}_2])`, as the equilibrium shifts right, the equilibrium constant would increase.

Show Worked Solution

a. $\ce{N2(g) + O2(g) \rightleftharpoons 2NO(g) }$

b. Increased temperature’s effect on `K_{eq}`:

From the graph, the forward reaction is endothermic.
The activation energy of the forward endothermic reaction is greater than the activation energy of the reverse exothermic reaction.
An increase in temperature would cause the rates of both the forward and reverse reaction due to the higher average kinetic energy, resulting in a larger likelihood of a successful collisions.
However, the rate of the forward reaction would increase to a higher extent than the reverse reaction, since it is an endothermic reaction.
Using `K_(eq) = ([text{NO}]^2)/([text{N}_2][text{O}_2])`, as the equilibrium shifts right, the equilibrium constant would increase.

CHEMISTRY, M5 2020 HSC 16 MC

Compounds `text{X}`, `text{Y}` and `text{Z}` are in equilibrium. The diagram shows the effects of temperature and pressure on the equilibrium yield of compound `text{Z}`.

Which equation would be consistent with this data?

`text{X}(g)+3 text{Y}(g) ⇌ 2 text{Z}(g) qquad Delta text{H} > 0`
`text{X}(g)+3 text{Y}(g) ⇌ 2 text{Z}(g) qquad Delta text{H} < 0`
`2 text{X}(g) ⇌ 2 text{Y}(g) + text{Z}(g) qquad Delta text{H} > 0`
`2 text{X}(g) ⇌ 2 text{Y}(g)+ text{Z}(g) qquad Delta text{H} < 0`

Show Answers Only

`C`

Show Worked Solution

The yield of `Z` increases as temperature increases, thus, endothermic reaction `Delta text{H} > 0`.
The yield of `Z` increases as pressure decreases, thus more gaseous moles on the product side.

`=> C`

♦ Mean mark 49%.

CHEMISTRY, M5 2022 HSC 23

Consider the following system which is at equilibrium in a rigid, sealed container.

$ \ce{4NH3(g) + 5O2(g) \rightleftharpoons 4NO(g) + 6H2O(g)} \ \ \ \ \ \ \Delta H = -950\ \text{kJ mol}^{-1} $

Identify what would happen to the amount of $ \ce{NO(g)} $ if the temperature was increased. (1 mark)

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Explain why a catalyst does not affect the equilibrium position of this system. (2 marks)

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Using collision theory, explain what would happen to the concentration of $ \ce{NO(g)} $ if $ \ce{H2O(g)} $ was removed from the system. (3 marks)

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Show Answers Only

a. The amount of $ \ce{NO2}$ decreases.

b. Catalyst affect on equilibrium:

A catalyst lowers the activation energy and equally increases the rate of both the forward and reverse reactions.
As a result, there is no net change in equilibrium, and thus the equilibrium position remains unchanged.

c. If $ \ce{H2O}$ is removed from the system:

This would cause the reverse reaction to decrease because there would be a lower likelihood of successful collisions between $ \ce{NO}$ and $ \ce{H2O}$ molecules.
As a result, the forward reaction rate is greater than the reverse reaction rate, thus, the equilibrium would shift to the right, causing $ \ce{[NO]}$ and $ \ce{[H2O]}$ to increase.
As the equilibrium is shifting to the right, the forward reaction rate decreases, whilst the reverse reaction rate increases, until they reach equilibrium.
At equilibrium, the concentration of all substances remain constant.

Show Worked Solution

a. The amount of $ \ce{NO2}$ decreases.

b. Catalyst affect on equilibrium:

A catalyst lowers the activation energy and equally increases the rate of both the forward and reverse reactions.
As a result, there is no net change in equilibrium, and thus the equilibrium position remains unchanged.

c. If $ \ce{H2O}$ is removed from the system:

This would cause the reverse reaction to decrease because there would be a lower likelihood of successful collisions between $ \ce{NO}$ and $ \ce{H2O}$ molecules.
As a result, the forward reaction rate is greater than the reverse reaction rate, thus, the equilibrium would shift to the right, causing $ \ce{[NO]}$ and $ \ce{[H2O]}$ to increase.
As the equilibrium is shifting to the right, the forward reaction rate decreases, whilst the reverse reaction rate increases, until they reach equilibrium.
At equilibrium, the concentration of all substances remain constant.

CHEMISTRY, M5 2021 HSC 22

Consider the following equilibrium system.

The solution is orange.

Justify TWO ways to shift the equilibrium to the left to change the colour of the solution. (3 marks)

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Answers could include two of the following methods.

Add `text{Cr}_2 text{O}_7 \ ^(2-)` ions:

This would increase the concentration of `text{Cr}_2 text{O}_7\ ^(2-)` ions.
According to Le Chatelier’s Principle, this would cause the equilibrium to shift left to counteract the change and decrease the concentration of `text{Cr}_2 text{O}_7\ ^(2-)` ions.

Decrease the concentration of `text{H}^+` ions by adding a base:

This would decrease the concentration of `text{H}^+` ions through an acid-base reaction.
According to Le Chatelier’s Principle, this would cause the equilibrium to shift left to counteract the change and increase the concentration of `text{H}^+` ions.

Increase the temperature:

According to Le Chatelier’s Principle, this would cause the equilibrium to shift left towards the endothermic side to counteract the change and decrease the temperature.

Show Worked Solution

Answers could include two of the following methods.

Add `text{Cr}_2 text{O}_7 \ ^(2-)` ions:

This would increase the concentration of `text{Cr}_2 text{O}_7\ ^(2-)` ions.
According to Le Chatelier’s Principle, this would cause the equilibrium to shift left to counteract the change and decrease the concentration of `text{Cr}_2 text{O}_7\ ^(2-)` ions.

Decrease the concentration of `text{H}^+` ions by adding a base:

This would decrease the concentration of `text{H}^+` ions through an acid-base reaction.
According to Le Chatelier’s Principle, this would cause the equilibrium to shift left to counteract the change and increase the concentration of `text{H}^+` ions.

Increase the temperature:

According to Le Chatelier’s Principle, this would cause the equilibrium to shift left towards the endothermic side to counteract the change and decrease the temperature.