Question

Consider the reaction $$\mathrm{CO}(g)+\mathrm{H}_{2}(g) \leftrightharpoons \mathrm{CH}_{3} \mathrm{OH}(g)$$ Suppose the system is at equilibrium, and then an additional mole of \(\mathrm{N}_{2} \mathrm{O}(g)\) is injected into the system at constant temperature. Once the reaction reestablishes equilibrium, has the amount of \(\mathrm{N}_{2} \mathrm{O}\) increased or decreased from its original equilibrium amount? Explain. What happens to the value of the equilibrium constant with this change?

Short answer

After adding N2O to the system, the equilibrium composition and the amount of N2O remain the same, since the total pressure change does not affect the given reaction's equilibrium. The value of the equilibrium constant, K, remains unchanged as well since the temperature is constant and the reaction's equilibrium is unaffected.

Step-by-step solution

Identify the reaction and given information

The given reaction is: \[\mathrm{CO}(g) + \mathrm{H}_2(g) \leftrightharpoons \mathrm{CH}_3\mathrm{OH}(g)\] We know that the system is initially at equilibrium, and then an additional mole of N2O is injected into the system at constant temperature.

Consider Le Châtelier's principle

Le Châtelier's principle states that if a change is made to a system at equilibrium, the system will adjust to oppose the change and restore the equilibrium. In this case, since N2O is not part of the reaction, adding N2O will not directly affect the reaction. However, the addition of N2O will increase the total pressure by increasing the total number of moles of gas.

Analyze the pressure change

The addition of N2O increases the total pressure inside the container. According to Le Châtelier's principle, the reaction will attempt to counteract this change by favoring the reaction direction that reduces pressure. In this case, the reaction will favor the direction where there are fewer moles of gas produced. Counting the moles of gas in the reaction, we see that 1 mole of CO and 1 mole of H2 produce 1 mole of CH3OH. There is no net change in the total moles of gas in the reaction. As a result, the reaction will not shift to counteract the pressure change, and the equilibrium will remain the same.

Determine the effect on the amount of N2O and the equilibrium constant

Since the total pressure change does not affect the given reaction's equilibrium, the amount of N2O will remain the same after equilibrium is reestablished. It neither increases nor decreases from its original amount. Furthermore, since the temperature is constant and the equilibrium of the reaction remains unchanged, the value of the equilibrium constant, K, will also remain the same. In summary, after adding N2O to the system, the equilibrium composition remains the same, and the amount of N2O stays the same. The value of the equilibrium constant remains unchanged as well.

Key concepts

Chemical Equilibrium

In a chemical reaction, equilibrium is a state where the concentrations of reactants and products no longer change, indicating a balance has been reached. This occurs when the rate of the forward reaction equals the rate of the reverse reaction. Chemical equilibrium is dynamic, meaning that even though the macroscopic properties like concentration remain constant, the individual molecules continue to react. In our example reaction, \[\mathrm{CO}(g) + \mathrm{H}_2(g) \leftrightharpoons \mathrm{CH}_3\mathrm{OH}(g)\]each component reaches a point where their concentrations do not change over time as long as temperature and pressure remain constant.

Equilibrium Constant

The equilibrium constant, denoted as \(K\), is a number that expresses the ratio of the concentrations of products to the concentrations of reactants at equilibrium, each raised to the power of their stoichiometric coefficients. For the reaction \[\mathrm{CO}(g) + \mathrm{H}_2(g) \leftrightharpoons \mathrm{CH}_3\mathrm{OH}(g)\]the equilibrium constant \( K \) can be expressed as:\[ K = \frac{[\mathrm{CH}_3\mathrm{OH}]}{[\mathrm{CO}][\mathrm{H}_2]} \]This ratio gives insight into the equilibrium position of the reaction. A constant equilibrium implies that, for the given reaction at a specific temperature, the ratio of products to reactants remains unchanged even if pressure is altered, provided no actual reactants or products are added.

Pressure Change

When pressure changes in a reaction system at equilibrium, Le Châtelier's principle predicts how the equilibrium will respond. While this principle suggests that a change in pressure will cause the system to adjust to counteract that change, it doesn’t always lead to a shift in equilibrium for every reaction. In the example of adding \(\mathrm{N}_2\mathrm{O}\) to \[\mathrm{CO}(g) + \mathrm{H}_2(g) \leftrightharpoons \mathrm{CH}_3\mathrm{OH}(g)\],there's no change in the net moles of gases during the reaction, as both sides of the equation have an equal total number of moles. Hence, despite the increase in total pressure due to \(\mathrm{N}_2\mathrm{O}\), the reaction's equilibrium and the amounts of each substance remain unchanged.

Mole Concept

The mole is a fundamental unit in chemistry that measures the amount of a substance. One mole equals \(6.022 \times 10^{23}\) entities, typically atoms or molecules. Understanding the mole concept is crucial when considering changes in reaction conditions. In our scenario, adding \(\mathrm{N}_2\mathrm{O}\) does not affect the mole balance of the \(\mathrm{CO}\), \(\mathrm{H}_2\), and \(\mathrm{CH}_3\mathrm{OH}\) reaction because \(\mathrm{N}_2\mathrm{O}\) is not part of the chemical equation. Analyzing reactions in terms of moles helps us predict the outcomes when conditions like pressure and volume change, especially when no stoichiometric changes happen as a result.