Question

At \(1100 \mathrm{~K}, K_{\mathrm{p}}=0.25\) for the reaction $$ 2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g) $$ What is the value of \(K\) at this temperature?

Short answer

At 1100 K, the value of K for the given reaction is approximately \(K = 0.00002745\).

Step-by-step solution

Write the expression for the reaction quotient Qc, which is equal to K

Start by writing the expression for reaction quotient Qc in terms of concentrations for the given reaction: \( Q_c = \frac{[\mathrm{SO_3}]^2}{[\mathrm{SO_2}]^2 [\mathrm{O_2}]} \) At equilibrium, Qc = K, so \( K = \frac{[\mathrm{SO_3}]^2}{[\mathrm{SO_2}]^2 [\mathrm{O_2}]} \)

Write the expression for Kp and find its value

Kp can be written as: \( K_p = \frac{(\frac{P_{\mathrm{SO_3}}}{RT})^2}{(\frac{P_{\mathrm{SO_2}}}{RT})^2 (\frac{P_{\mathrm{O_2}}}{RT})} \) Given, Kp = 0.25. Now, we need to find the relationship between Kp and K.

Establish the relationship between Kp and K

To establish the relationship between Kp and K, we will divide both the expressions for K and Kp: \( \frac{K_p}{K} = \frac{(\frac{P_{\mathrm{SO_3}}^2}{(RT)^2})}{(\frac{P_{\mathrm{SO_2}}^2P_{\mathrm{O_2}}}{(RT)^3})} \) \( \frac{K_p}{K} = \frac{(\frac{P_{\mathrm{SO_3}}^2}{(RT)^3})}{(\frac{P_{\mathrm{SO_2}}^2P_{\mathrm{O_2}}}{(RT)^3})} \cdot \frac{RT}{RT} = \frac{P_{\mathrm{SO_2}}^2P_{\mathrm{O_2}}}{P_{\mathrm{SO_3}}^2} \cdot RT \)

Find the value of K

We know that Kp = 0.25, and from the reaction, the stoichiometric coefficients are equal so we will have delta n = 0, which means the ratio in the above expression will be 1. Thus, the relation becomes: \( \frac{0.25}{K} = RT\) Now, find the value of K at 1100 K. The temperature is given in Kelvin, so we will use the gas constant value R = 8.314 J/(mol·K): \( K = \frac{0.25}{(8.314 \,\mathrm{J/(mol\cdot K)})(1100 \,\mathrm{K})} \) Now, calculate the value of K: \( K = 0.00002745 \) So, at 1100 K, the value of K for the given reaction is approximately 0.00002745.