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CHEMISTRY, M5 2024 HSC 10 MC

The following system is at equilibrium.

\(\underset{\text { propan-2-ol }}{\ce{CH_3CHOHCH_3(g)}} \rightleftharpoons \underset{\text {propan-2-one}}{\ce{CH_3COCH_3(g)}}\)\(\ce{+ H_2(g)}\)

A catalyst is added to the system.

Which row of the table correctly identifies the change in the yield of propan-2-one and the reaction rates?

\begin{align*}
\begin{array}{l}
\ & \\
\\
\textbf{A.}\\
\\
\textbf{B.}\\
\\
\textbf{C.}\\
\\
\textbf{D.}\\
\\
\end{array}
\begin{array}{|l|l|}
\hline
\quad\quad \textit{Yield of } & \quad \quad \quad \quad \textit{Reaction Rates} \\
\quad\textit{propan-2-one} & \textit{} \\
\hline
\text{Remains the same} & \text{Both forward and reverse rates} \\
\text{} & \text{are unchanged.} \\
\hline
\text{Remains the same} & \text{Both forward and reverse rates} \\
\text{} & \text{increase equally} \\
\hline
\text{Decreases} & \text{Reverse rate increases more than} \\
\text{} & \text{the forward rate increases.} \\
\hline
\text{Increases} & \text{Forward rate increases more than}\\
\text{} & \text{the reverse rate increases.}\\
\hline
\end{array}
\end{align*}

Show Answers Only

\(B\)

Show Worked Solution
  • Adding a catalyst to an equilibrium system will decrease the activation energy of both the forward and reverse reactions equally.
  • Thus, the reaction rate of the forward and reverse reactions will both increase and cause no change to the yield of the reaction.

\(\Rightarrow B\)

Filed Under: Factors that Affect Equilibrium Tagged With: Band 3, smc-3670-90-Catalyst

CHEMISTRY, M5 EQ-Bank 30

A catalyst increases the value of the equilibrium constant, thus favouring the extent of the forward reaction, resulting in a greater yield of product.

Determine if this statement is correct or incorrect, giving reasons for your answer.   (2 marks)

--- 5 WORK AREA LINES (style=lined) ---

Show Answers Only

The statement is incorrect:

  • A catalyst does not affect relative amounts of reactants and products at equilibrium.
  • Only a change in the position of equilibrium will change the value of the equilibrium constant (i.e. the yield of product).
  • A catalyst increases the rates of forward and reverse reactions equally, getting the system to equilibrium faster.
Show Worked Solution

The statement is incorrect:

  • A catalyst does not affect relative amounts of reactants and products at equilibrium.
  • Only a change in the position of equilibrium will change the value of the equilibrium constant (i.e. the yield of product).
  • A catalyst increases the rates of forward and reverse reactions equally, getting the system to equilibrium faster.

Filed Under: Factors that Affect Equilibrium Tagged With: Band 4, smc-3670-90-Catalyst

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?

  1. Adding a catalyst
  2. Decreasing the volume
  3. Decreasing the pressure
  4. 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`

Filed Under: Factors that Affect Equilibrium Tagged With: Band 4, smc-3670-10-Temperature, smc-3670-29-Pressure, smc-3670-35-Volume, smc-3670-70-Max yield conditions, smc-3670-90-Catalyst

CHEMISTRY, M5 2019 HSC 7 MC

How does the addition of a catalyst affect a reversible reaction?

  1. It increases the activation energy of the forward reaction only.
  2. It decreases the activation energy of the forward reaction only.
  3. It increases the activation energy of both the forward and reverse reactions.
  4. It decreases the activation energy of both the forward and reverse reactions.
Show Answers Only

`D`

Show Worked Solution
  • A catalyst helps to make a chemical reaction happen more easily by offering a different pathway for the reaction to occur.
  • This reduction in the required activation energy applies to both the forward and reverse reactions.

`=>D`

Filed Under: Factors that Affect Equilibrium Tagged With: Band 4, smc-3670-90-Catalyst

CHEMISTRY, M5 2022 HSC 23

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

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

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

--- 1 WORK AREA LINES (style=lined) ---

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

--- 5 WORK AREA LINES (style=lined) ---

  1. 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)

--- 8 WORK AREA LINES (style=lined) ---

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.

Filed Under: Factors that Affect Equilibrium Tagged With: Band 3, Band 4, smc-3670-10-Temperature, smc-3670-30-Concentration, smc-3670-40-Collision theory, smc-3670-90-Catalyst

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