Question

At 700 K, the equilibrium constant Kp, for the reaction $2S{O}_3\left(g\right)\rightleftharpoons 2S{O}_2\left(g\right)+O_{2}Is1.8\times {10}^{-3}KPa$. What is the numerical value in moles per liter of Kc for the reaction at the same temperature?

Answer 1

$N{O}_2\left(g\right)\rightleftharpoons NO\left(g\right)+\frac{1}{2}{O}_2$

One mole of $N{O}_2\left(g\right)$ is placed in vessel and allowed to come to equilibrium at a total pressure of 1 bar. An analysis of the contents of the vessel gives the following results in the picture.

$K_p\left(700K\right)=\frac{\left[NO\right]{\left[O_2\right]}^{1/2}}{\left[N{O}_2\right]}=\frac{\left[P_{NO}\right]{\left[P_{O_2}\right]}^{1/2}}{\left[P_{{NO}_2}\right]}=\frac{\left[0.872\right]{\left[0.872\right]}^{1/2}}{\left[1\right]}$

$K_p\left(800K\right)=\frac{\left[NO\right]{\left[O_2\right]}^{1/2}}{\left[N{O}_2\right]}=\frac{\left[P_{NO}\right]{\left[P_{O_2}\right]}^{1/2}}{\left[P_{{NO}_2}\right]}=\frac{\left[2.5\right]\left[2.5\right]}{\left[1\right]}=3.952$

$△{G°}_{\left(700K\right)}=-RTln{K}_p\left(700K\right)=-0.397kJ/㏖$

$△{G°}_{\left(800K\right)}=-RTln{K}_p\left(800K\right)=-9.140kJ/㏖$

$△H=(\frac{△{G°}_{T_1}}{T_1}-\frac{△G°}{T_2}){(\frac{1}{T_2}-\frac{1}{T_1})}^{-1}$

$=(-\frac{0.397}{700}-(-\frac{9.140}{800})){(\frac{1}{300}-\frac{1}{700})}^{-1}$

$=-15.167358$

$\frac{△H}{13}=-1.1667$

$\therefore$ Answer$=1$