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Q. For the reaction given below the values of standard Gibbs free energy of formation at $298 \,K$ are given. What is the nature of the reaction?
$I_2 + H_2S \to 2HI + S$
$\Delta G^{\circ}_{f}\left(HI\right) = 1.8\,kJ\,mol^{-1}$, $\Delta G^{\circ }_{f}\left(H_{2}S\right) = 33.8\,kJ\,mol^{-1}$

Thermodynamics

Solution:

$I_{2} + H_{2}S \to 2HI + S$
$\Delta G^{\circ} = \Sigma G^{\circ}_{f\left(Products\right)} - \Sigma G^{\circ}_{f\left(Reactants\right)}$
$\Delta G^{\circ} = \left(2 \times 1.8 + 0\right) - \left(0 + 33.8\right) = -30.2 \,kJ$
Hence, reaction is spontaneous in forward direction.