Question

For the process $$ \mathrm{CO}(g)+\mathrm{H}_{2} \mathrm{O}(g) \rightleftharpoons \mathrm{CO}_{2}(g)+\mathrm{H}_{2}(g) $$ it is found that the equilibrium concentrations at a particular temperature are \(\left[\mathrm{H}_{2}\right]=1.4 M,\left[\mathrm{CO}_{2}\right]=1.3 M,[\mathrm{CO}]=0.71 M,\) and \(\left[\mathrm{H}_{2} \mathrm{O}\right]=0.66 M .\) Calculate the equilibrium constant \(K\) for the reaction under these conditions.

Short answer

The equilibrium constant K for this reaction under these conditions is approximately \(3.88\).

Step-by-step solution

Write down the given values.

We have the following equilibrium concentrations given in the exercise: - \(\left[\mathrm{H}_{2}\right] = 1.4\,\text{M}\) - \(\left[\mathrm{CO}_{2}\right] = 1.3\,\text{M}\) - \(\left[\mathrm{CO}\right] = 0.71\,\text{M}\) - \(\left[\mathrm{H}_{2} \mathrm{O}\right] = 0.66\,\text{M}\)

Write down the equilibrium constant equation.

Using the equation for K mentioned in the analysis, we can write down the equilibrium constant equation for this reaction: \[K = \frac{\left[\mathrm{CO}_{2}\right]\left[\mathrm{H}_{2}\right]}{\left[\mathrm{CO}\right]\left[\mathrm{H}_{2}\mathrm{O}\right]}\]

Substitute the given values into the equation.

Now, substitute the given equilibrium concentrations into the equation to compute the value of K: \[K = \frac{(1.3\,\text{M})(1.4\,\text{M})}{(0.71\,\text{M})(0.66\,\text{M})}\]

Calculate the value of K.

Perform the multiplication and division to compute the value of the equilibrium constant: \[K \approx \frac{1.82}{0.4686}\] \[K \approx 3.88\] Therefore, the equilibrium constant K for this reaction under these conditions is approximately \(3.88\).

Key concepts

Chemical Equilibrium

Chemical equilibrium refers to the state in which the concentrations of reactants and products do not change with time in a closed system. It indicates a balance between the forward and reverse reactions, where both occur at the same rate. At this point, the macroscopic properties remain constant even though microscopic reactions still occur.
The concept of equilibrium is pivotal in understanding how reactions behave under specific conditions. It provides a snapshot of how concentrations of substances adjust themselves to reach stability. In the context of the given reaction:

• When \( \mathrm{CO}\,(g)+\mathrm{H}_{2}\mathrm{O}\,(g) \rightleftharpoons \mathrm{CO}_{2}\,(g)+\mathrm{H}_{2}\,(g) \) reaches equilibrium, the rate at which \( \mathrm{CO} \) and \( \mathrm{H}_{2}\mathrm{O} \) convert into \( \mathrm{CO}_{2} \) and \( \mathrm{H}_{2} \) equals the rate of the reverse reaction.
• The concentrations of the reactants and products remain constant, not necessarily equal.
• This balance is characterized by the equilibrium constant \( K \), which is a ratio of product concentrations to reactant concentrations, each raised to the power of their coefficients in the balanced chemical equation.

Reaction Kinetics

Reaction kinetics is the study of the speed or rate of a chemical reaction and how that rate is affected by parameters such as concentration, temperature, and catalysts. This field focuses on understanding the different factors that influence how quickly a reaction approaches equilibrium.
For any given reaction, such as the one between carbon monoxide and water to form carbon dioxide and hydrogen, kinetics helps predict how changes in conditions can shift the equilibrium position or alter the speed of achieving equilibrium.

• Increased concentration of reactants or products generally enhances the rate of reaction, pushing the system more rapidly towards equilibrium.
• Temperature plays a crucial role; an increase typically raises reaction rates as particles move faster and collision frequency and energy increase.
• Catalysts could lower activation energy, allowing equilibrium to be reached more quickly without shifting its position.
• Understanding these factors allows chemists to design conditions to optimize reactions for industrial applications or laboratory research.

Concentration Calculations

Concentration calculations involve determining the amount of a substance present in a given volume, expressed typically in terms of molarity (M), moles per liter. Accurate concentration calculations are crucial for determining the equilibrium constant, \( K \), as it requires precise values of the reactant and product concentrations at equilibrium.
To perform these calculations effectively:

• Gather the equilibrium concentrations for all reactants and products. In this example, these values were: \([\mathrm{H}_{2}] = 1.4\, \text{M}, [\mathrm{CO}_{2}] = 1.3\, \text{M}, [\mathrm{CO}] = 0.71\, \text{M}, [\mathrm{H}_{2}\mathrm{O}] = 0.66\, \text{M}\).
• Using these values, plug them into the equilibrium constant expression: \[\ K = \frac{[\mathrm{CO}_{2}][\mathrm{H}_{2}]}{[\mathrm{CO}][\mathrm{H}_{2}\mathrm{O}]}\].
• Simplify and solve the expression to find the value of \( K \), ensuring calculations are performed with care to maintain accuracy in substations and divisions.

Concentration calculations help predict reaction tendencies and determine whether a system is at equilibrium or needs adjustment.