Question

Predict the effect of increasing the container volume on the amounts of each reactant and product in the following reactions: (a) \(\mathrm{CH}_{3} \mathrm{OH}(l) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g)\) (b) \(\mathrm{CH}_{4}(g)+\mathrm{NH}_{3}(g) \rightleftharpoons \mathrm{HCN}(g)+3 \mathrm{H}_{2}(g)\)

Short answer

Increasing volume favors the gaseous products in (a) and the side with more moles of gas in (b).

Step-by-step solution

- Understand Le Chatelier's Principle

Le Chatelier's Principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium moves to counteract the change.

- Analyze Reaction (a)

Consider the reaction \(\text{CH}_3\text{OH}(l) \rightleftharpoons \text{CH}_3\text{OH}(g)\). When the volume of the container is increased, the pressure decreases. According to Le Chatelier's Principle, the system will shift to increase the pressure, which means it will shift towards the side with more moles of gas. Here, the gaseous side \( \text{CH}_3\text{OH}(g) \) will be favored. Consequently, the amount of \( \text{CH}_3\text{OH}(l) \) will decrease while the amount of \( \text{CH}_3\text{OH}(g) \) will increase.

- Analyze Reaction (b)

Consider the reaction \(\text{CH}_4(g) + \text{NH}_3(g) \rightleftharpoons \text{HCN}(g) + 3 \text{H}_2(g)\). Increasing the container volume will decrease the pressure. The system will shift toward the side with more moles of gas to increase the pressure. Here, the right side has a total of 4 moles of gas compared to 2 moles on the left side. Therefore, the equilibrium will shift towards the right, increasing the amounts of \( \text{HCN}(g) \) and \( \text{H}_2}(g) \) and decreasing the amounts of \( \text{CH}_4(g)\) and \( \text{NH}_3(g)\).

Key concepts

equilibrium shift

When a chemical reaction is at equilibrium, it means the rate of the forward reaction equals the rate of the reverse reaction. This balance can be tipped in favor of the forward or reverse reaction by changing the conditions. This is known as an equilibrium shift. Le Chatelier's Principle is a helpful guide to predict the direction of this shift when conditions such as pressure, temperature, or concentration are altered. For example, in reaction (a) \(\text{CH}_3\text{OH}(l) \rightleftharpoons \text{CH}_3\text{OH}(g)\), if the volume of the container is increased, the equilibrium will shift towards the side with more moles of gas, which is the gaseous \(\text{CH}_3\text{OH}(g)\) side.

pressure change

Pressure changes can significantly influence the position of equilibrium in a reaction involving gases. According to Le Chatelier's Principle, if the pressure of a system at equilibrium is decreased, the system will shift to increase the pressure again. This usually means moving towards the side with more moles of gas. Conversely, if the pressure is increased, the system will shift towards the side with fewer moles of gas. In reaction (b) \(\text{CH}_4(g) + \text{NH}_3(g) \rightleftharpoons \text{HCN}(g) + 3 \text{H}_2(g)\), increasing the volume of the container (which decreases the pressure) causes the equilibrium to shift to the right, where there are more moles of gas.

reaction dynamics

Reaction dynamics involve understanding how different variables affect the speed and direction of a reaction. Le Chatelier's Principle is key here, because it helps predict how a system will respond to changes and return to equilibrium. By knowing whether a system will shift left or right, you can predict the concentrations of reactants and products. For instance, in our example reaction (b), knowing that increasing the container's volume decreases pressure allows you to predict an increase in the amounts of \(\text{HCN}(g)\) and \(\text{H}_2(g)\).

moles of gas

In a gaseous reaction, the number of moles of gas on each side of the equation plays a crucial role in determining how the equilibrium shifts in response to pressure changes. In reaction (a) \(\text{CH}_3\text{OH}(l) \rightleftharpoons \text{CH}_3\text{OH}(g)\), the equilibrium will shift towards the side with more gas moles (gaseous \(\text{CH}_3\text{OH}(g)\)) when the container volume is increased. Similarly, in reaction (b) \(\text{CH}_4(g) + \text{NH}_3(g) \rightleftharpoons \text{HCN}(g) + 3 \text{H}_2(g)\), the right side has 4 moles of gas compared to 2 moles on the left side, prompting the shift towards the right when the volume is increased.