Question

An equilibrium mixture at 1000 K contains an equilibrium mixter \(0.276\ \mathrm{mol}\ \mathrm{H}_{2}, 0.276 \mathrm{mol}\ \mathrm{CO}_{2}, 0.224\ \mathrm{mol}\ \mathrm{CO},\) and \(0.224\ \mathrm{mol}\ \mathrm{H}_{2} \mathrm{O}\) $$\mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{g}) \rightleftharpoons \mathrm{CO}(\mathrm{g})+\mathrm{H}_{2} \mathrm{O}(\mathrm{g})$$ (a) Show that for this reaction, \(K_{\mathrm{c}}\) is independent of the reaction volume, \(V\) (b) Determine the value of \(K_{\mathrm{c}}\) and \(K_{\mathrm{p}}\)

Short answer

Part (a) demonstrates that the equilibrium constant \(K_c\) is indeed independent of volume. By quantifying the given equilibrium concentrations, the constants \(K_c\) and \(K_p\) for Part (b) are determined to both be 0.6596.

Step-by-step solution

Step 1- Demonstrating that \(K_c\) is independent of volume

First, rewrite \[K_c = \frac{[Products]}{[Reactants]}\] as \[K_c=\frac{([CO][H_{2}O])}{([CO_{2}][H_{2}])}\]Molar concentration is moles/volume. So, substituting concentration in terms of moles and volume into the \(K_c\) expression:\[K_c=\frac{\frac{(mol\ CO)(mol\ H_{2}O)}{V^2}}{\frac{(mol\ CO_{2})(mol\ H_{2})}{V^2}} = \frac{(mol\ CO)(mol\ H_{2}O)}{(mol\ CO_{2})(mol\ H_{2})} = K'_c\]As it can be seen, the volumes cancel each other out, concluding that \(K_c\) is independent of volume.

Step 2- Computing \(K_c\)

Substitute the given equilibrium concentrations into \(K_c\):\[K_c=\frac{(0.224)(0.224)}{(0.276)(0.276)} = 0.6596\]

Step 3- Computing \(K_p\)

To find \(K_p\), apply the equation \[K_p = K_c(RT)^{\Delta n}\] where \(R\) is the ideal gas constant (0.08206 L atm / K mol), \(T\) is the temperature in Kelvin (1000 K), and \(\Delta n\) is the change in moles of gas (final - initial). The reaction is \(CO_{2}(g) + H_{2}(g) \rightleftharpoons CO(g) + H_{2} O(g)\). Therefore, \(\Delta n = (1 + 1) - (1 + 1) = 0\). So, \(K_p = K_c(RT)^{\Delta n} = 0.6596(0.08206 \times 1000)^{0} = 0.6596\]

Key concepts

Equilibrium Constant

The equilibrium constant, denoted as \( K_c \) for concentrations and \( K_p \) for partial pressures, represents the ratio of the concentration or partial pressures of products to reactants at equilibrium in a chemical reaction. It is a measure of how far a reaction proceeds towards the products under a given set of conditions.
The formula for the equilibrium constant \( K_c \) is given by:

• \( K_c = \frac{[Products]}{[Reactants]} \)

For the reaction \( \text{CO}_2(g) + \text{H}_2(g) \rightleftharpoons \text{CO}(g) + \text{H}_2\text{O}(g) \), the expression becomes:

• \( K_c = \frac{[CO][H_2O]}{[CO_2][H_2]} \)

What's fascinating is that \( K_c \) is independent of the volume of the container. As seen in the step-by-step solution, volume terms cancel out when concentrations (moles per unit volume) are substituted into the equation. This means the equilibrium constant is a true representation of the reaction balance, uncontaminated by physical changes in volume.

Le Chatelier's Principle

Le Chatelier's Principle gives us a view into how equilibrium responds to changes in conditions. When a system at equilibrium experiences a shift—whether by changing concentrations, pressure, or temperature—it adjusts itself to offset the effect of the change and re-establish equilibrium.
This principle can be applied in many practical scenarios. Here is how it works:

• If the concentration of a reactant is increased, the equilibrium shifts to the right, forming more products.
• If the pressure is increased by reducing volume, the equilibrium shifts towards the side with fewer gas molecules to reduce pressure.
• Changes in temperature could either favor the forward or reverse reaction, depending on whether the reaction is exothermic or endothermic.

In the context of our problem, since the equilibrium constant \( K \) is a function of temperature, any change in temperature would lead to a change in the equilibrium constant value. Le Chatelier's Principle helps to predict the direction of the shift but not the new value of \( K_c \) or \( K_p \).

Reaction Quotient

The reaction quotient, \( Q_c \), is a snapshot of a reaction's progress at any given moment, not just at equilibrium. It uses the same form as the equilibrium constant \( K_c \) but for concentrations at any stage of the reaction, not just at equilibrium. This makes \( Q_c \) an essential tool for predicting the direction in which a reaction must shift to achieve equilibrium.
Determining \( Q_c \) involves the same expression as \( K_c \), which for our given reaction is:

• \( Q_c = \frac{[CO][H_2O]}{[CO_2][H_2]} \)

Here is what the comparison between \( Q_c \) and \( K_c \) tells us:

• If \( Q_c < K_c \), the reaction will proceed in the forward direction, producing more products until equilibrium is reached.
• If \( Q_c > K_c \), the reaction will proceed in the reverse direction, producing more reactants until equilibrium is attained.
• If \( Q_c = K_c \), the system is already at equilibrium.

By calculating \( Q_c \) at any point, we can assess whether a reaction is underway or if it has reached its balanced state.