Question

At \(1000 \mathrm{~K}, K_{p}=1.85\) for the reaction $$ \mathrm{SO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{SO}_{3}(g) $$ (a) What is the value of \(K_{p}\) for the reaction \(\mathrm{SO}_{3}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) ?\) (b) What is the value of \(K_{p}\) for the reaction \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)\) ? (c) What is the value of \(K_{c}\) for the reaction in part (b)?

Short answer

(a) For the reverse reaction, \(K_p' = 0.54\). (b) For the doubled reaction, \(K_p'' = 3.4225\). (c) For the doubled reaction, \(K_c = 0.281\).

Step-by-step solution

(a) Find the Ka for reverse reaction

To find the equilibrium constant for the reaction \(\mathrm{SO}_{3}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g)\), we need to invert the given reaction. The \(K_p\) for the inverse reaction is the reciprocal of the given \(K_p\): $$ K_p' = \frac{1}{K_p} $$ For the given reaction, we have \(K_p = 1.85\), so: $$ K_p' = \frac{1}{1.85} = 0.54 $$ Therefore, the \(K_p\) for the reaction \(\mathrm{SO}_{3}(g) \rightleftharpoons \mathrm{SO}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g)\) is 0.54.

(b) Find the Ka for reaction with doubled coefficients

To find the \(K_p\) for the reaction \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)\), we must multiply the given reaction by 2. The relation between the initial \(K_p\) and the new \(K_p\) for a reaction multiplied by a factor \(n\) is: $$ K_p'' = (K_p)^n $$ In our case, we have \(K_p = 1.85\) and we are doubling the reaction (\(n=2\)), so: $$ K_p'' = (1.85)^2 = 3.4225 $$ Therefore, the \(K_p\) for the reaction \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)\) is 3.4225.

(c) Find the Kc for the reaction in part (b)

The relation between \(K_p\) and \(K_c\) for a reaction is: $$ K_p = K_c (RT)^{\Delta n} $$ where \(R\) is the ideal gas constant, \(T\) is the temperature in Kelvin, and \(\Delta n\) is the change in moles of gas for the reaction. For the reaction \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)\), we have \(\Delta n = 2 - (2 + 1) = -1\). The temperature is given as \(1000\,\mathrm{K}\), and the gas constant \(R=0.0821\, \mathrm{L\,atm\,K^{-1}mol^{-1}}\). So, we can use this information to find \(K_c\): $$ K_c = \frac{K_p}{(RT)^{\Delta n}} $$ with \(K_p = 3.4225\), \(R = 0.0821\,\mathrm{L\,atm\,K^{-1}mol^{-1}}\), \(T=1000\,\mathrm{K}\), and \(\Delta n=-1\), so: $$ K_c = \frac{3.4225}{(0.0821\cdot1000)^{-1}} = 3.4225\cdot0.0821 = 0.281 $$ Thus, the \(K_c\) for the reaction \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \rightleftharpoons 2 \mathrm{SO}_{3}(g)\) is 0.281.