First slide

2nd Law of Thermodynaimcs

Question

For the reaction $2 H (g) ⟶ H_{2} (g), the sign of ΔH and ΔS$ respectively are

Easy

A

+,-
B

+, +
C

-,+
D

-,-

Solution

$\begin{array}{l} ΔH = - ve (H - H bond formed) \\ ΔS = - ve ({Δn}_{g} = - 1) \end{array}$
2H $\to$ H₂the entropy of products is less than the reactants .so the $∆ S = - ve$ from Gibbs equation
$∆ G =∆ H - T .∆ S$
$∆ G = ∆ H + T . ∆ S (∆ S = - v e)$
for spontaneous reaction $∆ G = - v e$
so the above reaction $∆ H is - ve$

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