Question #353c0

Mar 17, 2017

${K}_{p} = 2.25$

Explanation:

${\Delta}_{r} G = {\Delta}_{r} {G}^{0} + R T \ln Q$

The reaction is occurring at standard conditions then ${\Delta}_{r} {G}^{0} = 0$ and $Q = {K}_{e q}$

${\Delta}_{r} G = R T \ln Q$

$176.31 = 0.0821 \times 298 \times \ln Q$

$\ln Q = 7.2063$

$Q = 1347.895$

$Q = {K}_{e q} = 1347.895$

${K}_{p} = {K}_{c} {\left(R T\right)}^{\Delta n g}$

${K}_{p} = 1347.895 \times {\left(0.0821 \times 298\right)}^{- 2}$

${K}_{p} = 1347.895 \times {\left(24.4658\right)}^{- 2}$

${K}_{p} = \frac{1347.895}{24.4658} ^ \left(2\right)$

${K}_{p} = \frac{1347.895}{598.5753}$

${K}_{p} = 2.2518$

${K}_{p} = 2.25$