The equilibrium constant of the following given reaction is $K_p = 6.022 \times 10^{-5}$ at $298 \ K$ temperature. $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$. Calculate the value of $\Delta_r G^o$.
(N/A) The relationship between standard Gibbs free energy change $\Delta_r G^o$ and equilibrium constant $K_p$ is given by the equation: $\Delta_r G^o = -RT \ln K_p$.
Given: $R = 8.314 \ J \ K^{-1} \ mol^{-1}$,$T = 298 \ K$,$K_p = 6.022 \times 10^{-5}$.
Substituting the values: $\Delta_r G^o = -(8.314 \ J \ K^{-1} \ mol^{-1}) \times (298 \ K) \times \ln(6.022 \times 10^{-5})$.
$\Delta_r G^o = -2477.572 \times (\ln(6.022) + \ln(10^{-5}))$.
$\Delta_r G^o = -2477.572 \times (1.795 - 11.513)$.
$\Delta_r G^o = -2477.572 \times (-9.718) \approx 24077 \ J \ mol^{-1} = 24.08 \ kJ \ mol^{-1}$.
Given: $R = 8.314 \ J \ K^{-1} \ mol^{-1}$,$T = 298 \ K$,$K_p = 6.022 \times 10^{-5}$.
Substituting the values: $\Delta_r G^o = -(8.314 \ J \ K^{-1} \ mol^{-1}) \times (298 \ K) \times \ln(6.022 \times 10^{-5})$.
$\Delta_r G^o = -2477.572 \times (\ln(6.022) + \ln(10^{-5}))$.
$\Delta_r G^o = -2477.572 \times (1.795 - 11.513)$.
$\Delta_r G^o = -2477.572 \times (-9.718) \approx 24077 \ J \ mol^{-1} = 24.08 \ kJ \ mol^{-1}$.
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