Question

For the reaction \(\mathrm{C}(\mathrm{s})+\mathrm{CO}_{2}(\mathrm{~g}) \rightleftharpoons 2 \mathrm{CO}(\mathrm{g})\) the partial pressurc of \(\mathrm{CO}_{2}\) and \(\mathrm{CO}\) are 4 and 8 atm, respectively, then \(\mathrm{K}_{\mathrm{p}}\) for the reaction is (1) \(16 \mathrm{~atm}\) (2) \(2 \mathrm{~atm}\) (3) \(5 \mathrm{~atm}\) (4) \(4 \mathrm{~atm}\)

Short answer

16 atm

Step-by-step solution

- Write the equilibrium expression

For the reaction \(\mathrm{C}(\mathrm{s}) + \mathrm{CO}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{CO}(\mathrm{g})\) The equilibrium constant \(K_{p}\) can be written as \[ K_{p} = \frac{{(P_{\mathrm{CO}})^2}}{{P_{\mathrm{CO_2}}}} \] Here, \( P_{\mathrm{CO}} \) is the partial pressure of \( \mathrm{CO} \) and \( P_{\mathrm{CO_2}} \) is the partial pressure of \( \mathrm{CO_2} \).

- Insert the given values

Given that the partial pressure of \(\mathrm{CO_{2}}\)\( is 4 atm and the partial pressure of \(\mathrm{CO}\)\) is 8 atm. Plug these values into the equilibrium expression: \[ K_{p} = \frac{{(8)^2}}{{4}} \]

- Solve the equilibrium expression

Calculate the value by simplifying the expression: \[ K_{p} = \frac{{64}}{{4}} = 16 \]

Key concepts

partial pressure

In the context of gases, partial pressure is the pressure exerted by a single type of gas in a mixture of gases. It's important because it allows us to understand the individual behavior of each gas in a mixture.
For example, in our reaction \(\text{C(s) + CO}_2\text{(g)} \rightleftharpoons 2 \text{CO(g)}\), the partial pressure of \(\text{CO}_2\) is 4 atm and the partial pressure of \(\text{CO}\) is 8 atm.
The total pressure is the sum of these partial pressures. However, for the equilibrium constant calculation, we only need the individual partial pressures of the gases involved.

chemical equilibrium

Chemical equilibrium is the state in which both reactants and products are present in concentrations which have no further tendency to change with time. This happens when the forward reaction rate is equal to the reverse reaction rate.
In our reaction, chemical equilibrium is achieved when the rates at which \(\text{CO}_2\) reacts with carbon (C) to form \(\text{CO}\), and \(\text{CO}\) decomposes back to \(\text{CO}_2\) and C, are the same. Hence, the concentrations or partial pressures of \(\text{CO}_2\) and \(\text{CO}\) remain constant.

equilibrium expression

An equilibrium expression is a mathematical formula that relates the concentrations or partial pressures of reactants and products in a chemical reaction at equilibrium. For the given reaction \(\text{C(s) + CO}_2\text{(g)} \rightleftharpoons 2 \text{CO(g)}\), since \text{C(s)} is a solid, it's not included in the equilibrium expression.
The equilibrium constant \(K_p\) is given by: \[ K_{p} = \frac{{(P_{\text{CO}})^2}}{{P_{\text{CO}_2}}} \] Here, we raise the partial pressure of \(\text{CO}\) to the power of 2 because 2 moles of \(\text{CO}\) are formed for every mole of \(\text{CO}_2\) consumed.

Kp calculation

The equilibrium constant in terms of partial pressures, \(K_p\), helps us to understand the extent of a reaction. It can be calculated using the equilibrium expression.
Given the partial pressures: \(\text{CO}_2 = 4 \text{ atm}\) and \(\text{CO} = 8 \text{ atm}\), the equilibrium expression is: \[ K_{p} = \frac{{(8)^2}}{{4}} \] Simplifying this: \[ K_{p} = \frac{{64}}{{4}} = 16 \] Therefore, for our reaction at equilibrium, \(K_p\) is 16 atm, indicating the balance between products and reactants.