Question

For the following reactions, predict whether the pressure of the reactants or products increases or remains the same when the volume of the reaction vessel is increased. (a) \(\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{2} \mathrm{O}(g)\) (b) \(\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g)\) (c) \(\mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(g)\)

Short answer

Answer: For the given reactions, when the volume of the reaction vessel is increased: (a) The pressure of the products (gaseous H2O) increases. (b) The pressure of the reactants (N2 and H2) increases. (c) The pressure of the reactants (C2H4 and H2O) increases.

Step-by-step solution

Identify the number of moles of gaseous reactants and products in each reaction

First, let's enumerate the moles of gaseous reactants and products for each reaction: (a) \(\text{H}_2\text{O}(l) \rightleftharpoons \text{H}_2\text{O}(g)\) - 0 moles of gaseous reactants and 1 mole of gaseous product (b) \(\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)\) - 4 moles of gaseous reactants and 2 moles of gaseous products (c) \(\text{C}_2\text{H}_4(g) + \text{H}_2\text{O}(g) \rightleftharpoons \text{C}_2\text{H}_5\text{OH}(g)\) - 2 moles of gaseous reactants and 1 mole of gaseous product

Determine the response of each reaction to an increase in volume

Now let's examine each reaction and determine the response to an increase in volume based on Le Châtelier's principle: (a) When the volume is increased, the pressure decreases. For this reaction, there are no gaseous reactants and 1 mole of a gaseous product. Therefore, the equilibrium will shift to the side with more gas molecules, which in this case is the product side. So, the pressure on the product side will increase. (b) When the volume is increased in this reaction, the pressure also decreases. Since there are more moles of gaseous reactants than products (4 vs. 2), the equilibrium will shift to the side with more gas molecules, which is the reactant side. Therefore, the pressure on the reactant side will increase. (c) In this reaction, when the volume is increased, the pressure decreases again. There are more moles of gaseous reactants than products (2 vs. 1). The equilibrium will shift to the side with more gas molecules, which is the reactant side. As a result, the pressure on the reactant side will increase.

Conclusion

In conclusion, for each reaction, we used Le Châtelier's principle to determine how the equilibrium would shift in response to an increase in volume: (a) The pressure of the products (gaseous H2O) increases. (b) The pressure of the reactants (N2 and H2) increases. (c) The pressure of the reactants (C2H4 and H2O) increases.