Question
State whether $\triangle S$ is positive, negative or zero for the reaction $2H_{(g)} → H_{2(g)}$. Explain.

Answer

(i) The given reaction, $2H_{(g)} → H_{2(g)}$ is the formation of $H_{2(g)}$ from free atoms.
(ii) Since two H atoms form one $H_2$ molecule, $\triangle n = 1 – 2= -1$ and disorder is decreased. Hence entropy change $\triangle S < 0$ (or negative).

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