Summary of Equilibrium Reactions
Summary
The following rules will help in predicting the changes that take place in equilibrium reactions:
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If the concentration of a reactant (on the left) is increased, then some of it must be used to form the products (on the right) for equilibrium to be maintained. The equilibrium position will shift to the right. \(\text{K}_{\text{c}}\) is unchanged.
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If the concentration of a reactant (on the left) is decreased, then some of the products (on the right) must be used to form reactants for equilibrium to be maintained. The equilibrium position will shift to the left. \(\text{K}_{\text{c}}\) is unchanged.
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If the forward reaction is endothermic, then an increase in temperature will favour this reaction and the product yield and \(\text{K}_{\text{c}}\) will increase. A decrease in temperature will decrease product yield and likewise decrease \(\text{K}_{\text{c}}\).
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If the forward reaction is exothermic, then an decrease in temperature will favour this reaction and the product yield and \(\text{K}_{\text{c}}\) will increase. An increase in temperature will decrease product yield and likewise decrease \(\text{K}_{\text{c}}\).
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Increasing the pressure favours the side of the equilibrium with the least number of gas molecules. This is shown in the balanced chemical equation. This rule applies in reactions with one or more gaseous reactants or products. \(\text{K}_{\text{c}}\) is unchanged.
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Decreasing the pressure favours the side of the equilibrium with more gas molecules. This rule applies in reactions with one or more gaseous reactants or products. \(\text{K}_{\text{c}}\) is unchanged.
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A catalyst does not affect the equilibrium position of a reaction. It only influences the rate of the reaction, in other words, how quickly equilibrium is reached.
The following simulation will help you to understand these concepts.
Example: Chemical Equilibrium
Question
\(2\text{NO}(\text{g}) + \text{O}_{2}(\text{g})\) \(\leftrightharpoons\) \(2\text{NO}_{2}(\text{g})\) \(\quad \Delta{H} < 0\)
How will:
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the reverse reaction
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the equilibrium position be affected by:
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A decrease in temperature?
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The addition of a catalyst?
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The addition of more \(\text{NO}_{2}\) gas?
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The reverse reaction:
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The forward reaction is exothermic (\(\Delta{H} < 0\)) so the reverse reaction must be endothermic. A decrease in temperature will cause the equilibrium to shift to favour the exothermic reaction. Therefore the reverse reaction rate will decrease sharply, and then gradually increase until equilibrium is re-established.
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The addition of a catalyst will speed up both the forward and reverse reactions.
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\(\text{NO}_{2}\) is a product. Therefore the addition of more \(\text{NO}_{2}\) will increase the rate of the formation of reactants. The rate of the reverse reaction will therefore increase sharply, and then gradually decrease until equilibrium is re-established.
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The equilibrium position:
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A decrease in temperature will favour the exothermic reaction and the forward reaction is exothermic. Therefore the equilibrium position will shift to the right.
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The addition of a catalyst will have no effect on the equilibrium position as both the forward and reverse reactions rates would be increased equally.
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The addition of more \(\text{NO}_{2}\) will favour the formation of the reactants and so the equilibrium will shift to the left.
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Example: Graphs of Equilibrium
Question
Study the graph and answer the questions that follow:
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Does the equilibrium favour the reactants or the products?
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Determine the value of \(\text{K}_{\text{c}}\) if the coefficients of the balanced equation all equal 1.
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The forward reaction has \(\Delta{H} > 0\). What effect will an increase in temperature have on [A], [B] and [C]?
Step 1: Determine which compounds are the reactants and which are the products
The concentration of reactants decreases from the start of the reaction to equilibrium. Therefore A and B are reactants.
The concentration of products increases from the start of the reaction to equilibrium. Therefore C is a product.
\(\text{aA} + \text{bB}\) \(\leftrightharpoons\) \(\text{cC}\)
We are told that all coefficients in the balanced equation equal 1. Therefore the general equation is: \(\text{A} + \text{B}\) \(\leftrightharpoons\) \(\text{C}\)
Step 2: Do reactants or products have higher concentrations at equilibrium?
A and B (the reactants) have higher concentrations at equilibrium.
Step 3: Does the equilibrium favour the reactants or products?
The reactants have higher concentrations than the products, therefore the equilibrium must favour the reactants.
Step 4: Determine the equilibrium concentration values of A, B and C
[A] = \(\text{2.5}\) \(\text{mol.dm$^{-3}$}\)
[B] = \(\text{2.0}\) \(\text{mol.dm$^{-3}$}\)
[C] = \(\text{1.5}\) \(\text{mol.dm$^{-3}$}\)
Step 5: Calculate \(\text{K}_{\text{c}}\)
\[\text{K}_{\text{c}}=\frac{\text{[C]}}{\text{[A][B]}}=\frac{\text{1.5}}{(\text{2.5})(\text{2.0})} = \text{0.3}\]
Step 6: Determine which reaction is exothermic and which is endothermic
The forward reaction has \(\Delta{H} > 0\). This means that the forward reaction is endothermic. The reverse reaction must therefore be exothermic.
Step 7: Which reaction is favoured by an increase in temperature?
The endothermic reaction would be favoured by an increase in temperature (to lower the temperature). This is the forward reaction.
Step 8: How would [A], [B] and [C] change?
The forward reaction is favoured, therefore the equilibrium would shift to the right. This means that the reactant concentrations ([A] and [B]) would decrease and the product concentration ([C]) would increase.
This lesson is part of:
Chemical Equilibrium