At 300 K,for the reaction PCl_{5(g)} rightle | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The relationship between standard Gibbs free energy change $\Delta G^0$ and equilibrium constant $K_p$ is given by the formula: $\Delta G^0 = -RT

Vedclass · Accepted Answer

$38.72 \ kJ \ mol^{-1}$

Vedclass · Answer

(A) The relationship between standard Gibbs free energy change $\Delta G^0$ and equilibrium constant $K_p$ is given by the formula: $\Delta G^0 = -RT \ln K_p$. Given: $R = 8.314 \ J \ K^{-1} \ mol^{-1}$,$T = 300 \ K$,$K_p = 1.8 	imes 10^{-7}$. Substituting the values: $\Delta G^0 = -(8.314 \ J \ K^{-1} \ mol^{-1}) 	imes (300 \ K) 	imes \ln(1.8 	imes 10^{-7})$. $\Delta G^0 = -2494.2 	imes (\ln(1.8) + \ln(10^{-7}))$. $\Delta G^0 = -2494.2 	imes (0.5878 - 16.118)$. $\Delta G^0 = -2494.2 	imes (-15.5302) \approx 38735 \ J \ mol^{-1}$. Converting to $kJ \ mol^{-1}$: $\Delta G^0 \approx 38.74 \ kJ \ mol^{-1}$. Thus,the closest option is $38.72 \ kJ \ mol^{-1}$.

At $300 \ K$,for the reaction $PCl_{5(g)} \rightleftharpoons PCl_{3(g)} + Cl_{2(g)}$,the equilibrium constant $K_p = 1.8 \times 10^{-7}$. Calculate its standard Gibbs free energy change $\Delta G^0$.

Similar Questions

$2O_{3(g)} \rightleftharpoons 3O_{2(g)}$
At $300 \ K$,ozone is $50\%$ dissociated. The standard free energy change at this temperature and $1 \ atm$ pressure is $(-) \dots \ J \ mol^{-1}$ (Nearest integer).
[Given: $\ln 1.35 = 0.3$ and $R = 8.3 \ J \ K^{-1} \ mol^{-1}$ ]

For the reaction $CH_{4(g)} + H_{2(g)} \longrightarrow C_2H_{6(g)}$,$K_p = 3.356 \times 10^{17}$,calculate $\Delta G^{\circ}$ for the reaction at $298 \ K$.

Which of the following correctly represents the relationship between $\Delta G$ and $\Delta G^{\circ}$? $[P = \text{products}, R = \text{reactants}]$

At $60^{\circ} C$,dinitrogen tetroxide is $50$ per cent dissociated. Calculate the standard free energy change at this temperature and at one atmosphere.

For a reaction,$\Delta G^{\circ} = -115 \, kJ$. What is the value of $\log \, K_p$ at $298 \, K$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module