Question

Consider the gas phase reaction: \[ CO + \frac{1}{2} O_2 \rightleftharpoons CO_2 \] At equilibrium for a particular temperature, the partial pressures of \( CO \), \( O_2 \), and \( CO_2 \) are found to be \( 10^{-6} \, {atm} \), \( 10^{-6} \, {atm} \), and \( 16 \, {atm} \), respectively. The equilibrium constant for the reaction is ......... \( \times 10^{10} \) (rounded off to one decimal place).

Hint:

The equilibrium constant is calculated using the partial pressures of the reactants and products at equilibrium. For reactions involving gases, remember to use the stoichiometric coefficients of the reactants and products as exponents in the equilibrium expression.

Answer 1

The equilibrium constant \( K_p \) for a gas-phase reaction is expressed in terms of the partial pressures of the reactants and products at equilibrium. For the given reaction: \[ CO + \frac{1}{2} O_2 \rightleftharpoons CO_2 \] The equilibrium constant \( K_p \) is given by: \[ K_p = \frac{P_{CO_2}}{P_{CO} P_{O_2}^{1/2}} \] Substituting the given partial pressures of CO, \( O_2 \), and \( CO_2 \): \[ K_p = \frac{16}{(10^{-6})(10^{-6})^{1/2}} = \frac{16}{10^{-6} \times 10^{-3}} = 16 \times 10^9 \] Therefore, the equilibrium constant \( K_p = 1.6 \times 10^{10} \). This tells us the ratio of the concentration of the product \( CO_2 \) to the products \( CO \) and \( O_2 \) at equilibrium.