Question

Calculate \(K\) for the formation of methyl alcohol at \(100^{\circ} \mathrm{C}\) : $$ \mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(g) $$ given that at equilibrium, the partial pressures of the gases $$ \text { are } P_{\mathrm{CO}}=0.814 \mathrm{~atm}, \quad P_{\mathrm{H}_{2}}=0.274 \mathrm{~atm}, \text { and } P_{\mathrm{CH}_{3} \mathrm{OH}}= $$ \(0.0512 \mathrm{~atm} .\)

Short answer

Question: Calculate the equilibrium constant, K, for the formation of methyl alcohol at \(100^{\circ}\mathrm{C}\) given the equilibrium partial pressures of the gases involved: \(P_{\mathrm{CO}} = 0.814\,\mathrm{atm}\), \(P_{\mathrm{H}_{2}}=0.274\, \mathrm{atm}\), and \(P_{\mathrm{CH}_{3}\mathrm{OH}}=0.0512\,\mathrm{atm}\). Answer: The equilibrium constant, K, for the formation of methyl alcohol at \(100^{\circ}\mathrm{C}\) is approximately 2.569.

Step-by-step solution

Identify the reactants and products

The given chemical reaction can be written as: $$ \mathrm{CO}(g)+2 \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{CH}_{3}\mathrm{OH}(g) $$ Reactants: \(\mathrm{CO}(g)\) and \(\mathrm{H}_2(g)\) Products: \(\mathrm{CH}_3\mathrm{OH}(g)\)

Write the expression for K

Considering the stoichiometric coefficients in the reaction, we can write the expression for K, the equilibrium constant, as follows: $$ K_p = \dfrac{P_{\mathrm{CH}_3\mathrm{OH}}}{P_{\mathrm{CO}} \times {P_{\mathrm{H}_2}}^2} $$

Substitute the given partial pressures into the expression for K

We are given the following equilibrium partial pressures: $$ P_{\mathrm{CO}} = 0.814\,\mathrm{atm},\, P_{\mathrm{H}_{2}}=0.274\, \mathrm{atm}, \text{ and } P_{\mathrm{CH}_{3}\mathrm{OH}}=0.0512\,\mathrm{atm} $$ Substitute these values into the expression for K: $$ K_p = \dfrac{0.0512}{(0.814) \times (0.274)^2} $$

Calculate K

Calculate the equilibrium constant by evaluating the expression obtained in Step 3: $$ K_p = \dfrac{0.0512}{(0.814) \times (0.274)^2} \approx 2.569 $$ The equilibrium constant, K, for the formation of methyl alcohol at \(100^{\circ}\mathrm{C}\) is approximately 2.569.