Question

Calculate the standard free energy change for the reaction,
$H_2\left(g\right)+I_2\left(g\right)⟶2HI\left(g\right)$; .........$\Delta H^{\ominus }=+51.9kJ$

Given that the standard entropies $\left(\Delta S^{\ominus }\right)$ of $H_2$, $I_2$ and $HI$ are $130.6$, $116.7$ and $206.3J{K}^{-1}mo{l}^{-1}$, respectively.

Predict whether the reaction is feasible at the standard state or not?

• A. Reaction is not feasible.
• B. Reaction is feasible.
• C. Can't predict
• D. None of these

Answer 1

The correct option is A Reaction is not feasible.

The standard entropy change for the reaction is ${\Delta }^{\ominus }S=\sum S^{\ominus }\left(Product\right)-\sum S^{\ominus }\left(Reactant\right)$ $=2S_{HI}^{\ominus }-(S_{H_2}^{\ominus }+S_{I_2}^{\ominus })$ $=2\times 206.3-130.6-116.7$ $=165.3J{K}^{-1}mo{l}^{-1}$

The standard Gibbs free energy change for the reaction is:

$\Delta G^{\ominus }=\Delta H^{\ominus }-T\Delta S^{\ominus }$ $=51.9kJ-298\times 165.3\times {10}^{-3}kJ$$=51.9-49.25$ $=2.6406kJmo{l}^{-1}$ $=2640.6Jmo{l}^{-1}$

$\Delta G=+ve$.

Since, the standard free energy change for the reaction is positive, the reaction is not feasible at the standard state.