In Haber's process of manufacturing of ammonia :
$N_{2} (g) + 3 H_{2} (g) ⟶ 2 {NH}_{3} (g); H_{25}^{\circ} C = - 92.2 kJ$
Molecule $N_{2} (g) H_{2} (g) {NH}_{3} (g)$
$\begin{array}{llll} C_{P} {JK}^{- 1} 29.1 28.8 35.1 \end{array}$
If $C_{p}$ is independent of temperature, then reaction at 100° C as compared to that of 25° C will be :

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Less endothermic

Less exothermic

More endothermic

More exothermic

answer is C.

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Detailed Solution

In the reaction we have

$\begin{array}{l} {(Δ C_{P})}_{R x^{n}} = 2 \times 35.1 - 3 \times 28.8 - 29.1 = - 45.3 \\ H_{T_{2}} - H_{T_{1}} = Δ C_{P} (T_{2} - T_{1}) \\ H_{100^{\circ} C} = H_{25^{\circ} C} + 75 \times Δ C_{P} \\ |H_{100^{\circ} C}| > |H_{25^{\circ} C}| \\ H_{100^{\circ} C} < 0 \end{array}$