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Consider the following reaction approaching equilibrium at 27∘^\circ∘C and 1 atm pressure
A+B\mathrm{A+B}A+B ⇌kr=102kf=103\mathrel{\mathop{\kern0pt\rightleftharpoons} \limits_{{k_r} = {{10}^2}}^{{k_f} = {{10}^3}}} kr=102⇌kf=103 C+D\mathrm{C+D}C+D
The standard Gibb's energy change (ΔrGθ)\mathrm{(\Delta_r G^\theta)}(ΔrGθ) at 27∘^\circ∘C is (−-−) ___________ kJ mol−1^{-1}−1 (Nearest integer).
(Given : R=8.3 J K−1 mol−1\mathrm{R=8.3~J~K^{-1}~mol^{-1}}R=8.3 J K−1 mol−1 and ln10=2.3\mathrm{\ln 10=2.3}ln10=2.3)
∵ Δ G0=−RT ln Keq\because \Delta \mathrm{G}^{0}=-\mathrm{RT} \ln \mathrm{K}_{\mathrm{eq}}∵ Δ G0=−RT ln Keq
and Keq=KfKb\mathrm{K}_{\mathrm{eq}}=\frac{\mathrm{K}_{\mathrm{f}}}{\mathrm{K}_{\mathrm{b}}}Keq=KbKf
∴ Keq=103102=10\therefore \mathrm{K}_{\mathrm{eq}}=\frac{10^{3}}{10^{2}}=10∴ Keq=102103=10
∴ Δ G=−RT ln 10\therefore \Delta \mathrm{G}=-\mathrm{RT} \ln 10∴ Δ G=−RT ln 10
⇒−(8.3 × 300 × 2.3) =−5.7 kJ mole−1 ≈ 6 kJ\Rightarrow-(8.3 \times 300 \times 2.3) \\ =-5.7 \mathrm{~kJ} \mathrm{~mole}^{-1} \approx 6 \mathrm{~kJ}⇒−(8.3 × 300 × 2.3) =−5.7 kJ mole−1 ≈ 6 kJ mole−1\mathrm{mole}^{-1}mole−1 (nearest integer)
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