91Ó°ÊÓ

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.

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.

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.