C 2 H 6 g - C 2 H 4 g - H 2 g - K p -0-05 atmThe value of - G - of the reaction at 627- C would be- Given - -ln 0-05 -3 - R -8-3 J K -1 mol -1A- 11-21 kJ mol -1B- 22-41 kJ mol -1C- 33-61 kJ mol -1D- 27-98 kJ mol -1

The $Δ G^{0}$ value for the reaction, $2 M g_{(s)} + O_{2 (g)} \to 2 M g O_{(s)}$ at $25^{\circ} C$ is,
(Given $Δ S_{s y s}^{0}) = - 217 J K^{- 1} m o l^{- 1}, Δ S_{s u r r}^{0} = 4 k J K^{- 1} m o l^{- 1})$

Q. The equilibrium constant for a reaction is 10.

Δ G^{\circ}

will be

(R = 8 J K^{- 1} m o l^{- 1}, T = 300 K)

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C2H6(g)⇌C2H4(g)+H2(g) ;Kp=0.05 atm. The value of ΔG∘ of the reaction at 627 ∘C would be: (Given:ln(0.05)≈−3; R=8.3 J K−1 mol−1)

The correct option is B 22.41 kJ mol−1ΔG∘=−RTlnKp =−(8.3 J K−1 mol−1)(900 K)ln(0.05) =22410 J mol−1=22.41 kJ mol−1

$C_{2} H_{6} (g) ⇌ C_{2} H_{4} (g) + H_{2} (g); K_{p} = 0.05 atm$ .
The value of $Δ G^{\circ}$ of the reaction at $627^{\circ} C$ would be:
$(Given : ln (0.05) \approx - 3; R = 8.3 {J K}^{- 1} {mol}^{- 1})$

The correct option is B $22.41 {kJ mol}^{- 1}$
$Δ G^{\circ} = - R T ln K_{p} = - (8.3 {J K}^{- 1} {mol}^{- 1}) (900 K) ln (0.05) = 22410 {J mol}^{- 1} = 22.41 {kJ mol}^{- 1}$