Equilibrium constant for the reaction:
`H_(2)(g) +I_(2) (g) hArr 2HI(g) is K_(c) = 50 at 25^(@)C`
The standard Gibbs free enegry change for the reaction will be:

Options

A. `-6.964 kJ`
B. `-9.694 kJ`
C. `-4.964 kJ `
D. `-6.496 kJ`

Explanation

To find the standard Gibbs free energy change (\( \Delta G^\circ \)) for the reaction \( H_2(g) + I_2(g) \rightleftharpoons 2HI(g) \) with an equilibrium constant (\( K_c \)) of 50 at 25°C, we can use the following formula:

\[ \Delta G^\circ = -RT \ln K \]

Where:
- \( R \) is the universal gas constant, which is approximately \( 8.314 \, \text{J/(mol K)} \)
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