smc-3670-29-Pressure

CHEMISTRY, M5 2025 HSC 29

Consider the following reaction.

$\ce{2NO2(g) \rightleftharpoons N2O4(g) \quad \quad \Delta H= -57.2 \ \text{kJ mol}^{-1}}$

A sealed reaction vessel of fixed volume contains a mixture of $\ce{NO2}$ and $\ce{N2O4}$ gases at equilibrium.

Explain the impact of the addition of argon, an inert gas, on the temperature of the system. (4 marks)

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The equilibrium expression for the reaction is:
$Q = \dfrac{\ce{[N2O4]}}{\ce{[NO2]^2}} = K_{eq}$
When argon is added at constant volume, the total pressure increases, but the concentrations of $\ce{NO2}$ and $\ce{N2O4}$ remain unchanged because the amount of each gas and the volume remain constant. Since the concentrations in the equilibrium expression do not change, the value of $Q$ remains equal to $K_{eq}$, so the equilibrium position does not shift.
Because the equilibrium does not shift, no extra forward or reverse reaction occurs, meaning there is no absorption or release of heat. Even though the reaction is exothermic, the system does not produce or consume heat if equilibrium is unchanged.

Show Worked Solution

The equilibrium expression for the reaction is:
$Q = \dfrac{\ce{[N2O4]}}{\ce{[NO2]^2}} = K_{eq}$
When argon is added at constant volume, the total pressure increases, but the concentrations of $\ce{NO2}$ and $\ce{N2O4}$ remain unchanged because the amount of each gas and the volume remain constant. Since the concentrations in the equilibrium expression do not change, the value of $Q$ remains equal to $K_{eq}$, so the equilibrium position does not shift.
Because the equilibrium does not shift, no extra forward or reverse reaction occurs, meaning there is no absorption or release of heat. Even though the reaction is exothermic, the system does not produce or consume heat if equilibrium is unchanged.

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 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

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`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*}

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`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 2017 HSC 16 MC

The following equilibrium is established in a closed system.

`text{CO}_(2)(g)+ text{H}_(2) text{O} (l) ⇌ text{H}_(2) text{CO}_(3)(aq) qquadqquad Delta H=-19.4 \ text{kJ mol}^(-1)`

How can the gas pressure in the system be decreased?

Add more `text{CO}_2 (g)`
Add hydroxide ions to the solution
Decrease the volume of the container
Increase the temperature of the system

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`B`

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Gas pressure can be decreased if the equilibrium shifts to the right.
If hydroxide ions are added to the solution by adding NaOH, this will neutralise the carbonic acid and cause an equilibrium shift to the right.

`=>B`

♦♦ Mean mark 32%.

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

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`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 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`

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`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 14 MC

Nitrogen dioxide can react with itself to produce dinitrogen tetroxide.

$ \ce{2NO2(g) \rightleftharpoons N2O4(g)\ \ \ \ \ \ $K_{eq}$ = 0.010} $

In an experiment, 100.0 cm³ of $ \ce{NO2}$ is placed in a syringe. The plunger is then pushed in until the volume is 50.0 cm³, while maintaining a constant temperature. The system is allowed to return to equilibrium.

Which statement is true for the system at equilibrium?

The value of `K_{eq}` has increased.
The ratio `([\text{NO}_2])/([\text{N}_2\text{O}_4])` has decreased.
The concentration of `\text{N}_2\text{O}_4` has decreased.
The concentrations of `\text{NO}_2` and `\text{N}_2\text{O}_4` have doubled.

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`B`

Show Worked Solution

As the volume is decreased, the pressure increases. According to Le Chatelier’s Principle, the equilibrium will shift toward the right with fewer moles of gas in order to counteract the change in equilibrium (ie decrease pressure).
As a result, $ \ce{[N2O4]} $ will increase and $ \ce{[NO2]} $ will decrease.
Therefore, `([\text{NO}_2])/([\text{N}_2\text{O}_4])` will decrease.

`=> B`

♦♦ Mean mark 35%.