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A catalyst increases the value of the equilibrium constant, thus favouring the extent of the forward reaction, resulting in a greater yield of product.
Evaluate this statement, giving reasons why it is correct or incorrect. (2 marks)
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→ 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.
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.
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?
`B`
→ By decreasing the volume, the equilibrium will shift to the left so that less gas molecules are present (Le Chatelier’s principle).
`=>B`
How does the addition of a catalyst affect a reversible reaction?
`D`
→ 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`
Consider the following system which is at equilibrium in a rigid, sealed container.
\( \ce{4NH3(g) + 5O2(g) \rightleftharpoons 4NO(g) + 6H2O(g)} \ \ \ \ \ \ \Delta H = -950\ \text{kJ mol}^{-1} \)
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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.
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.