If the standard free energy change for a reaction is 1.546 kJ mol−1 at 500∘C, then the value of standard equilibrium constant for the reaction is

If the standard free energy change for a reaction is $1.546 kJ {mol}^{- 1} at 500^{\circ} C$ , then the value of standard equilibrium constant for the reaction is

A
antilog (0.105)
B
antilog(-0.105)
C
antilog (0.241)
D
antilog (- 0.241)

Solution:

$\begin{array}{l} Δ G^{\circ} = - R T \ln K^{\circ} \\ 1.546 \times 10^{3} J {mol}^{- 1} = - (8.314 {JK}^{- 1} {mol}^{- 1}) (773 K) \ln K^{\circ} \\ i.e. \log K^{\circ} = - \frac{1.546 \times 10^{3}}{2.303 \times 8.314 \times 773} i.e. K^{\circ} = antilog (- 0.105) \end{array}$

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