First slide

2nd Law of Thermodynaimcs

Question

$∆_{f}^{0} S$ of H₂O_(l), NH_3(g) and NO_(g) respectively are –237, –108 and +104 K.Cal.mole^–1. Then $∆_{f}^{0} S$ for 4NH_3(g) + 5O₂→ 4NO_(g) + 6H₂O (in K.Cal) will be

Easy

Solution

4NH_3(g) + 5O₂→ 4NO_(g) + 6H₂O , $∆_{f}^{0} S = ?$

$\begin{array}{l} {\Delta _r}{S^ \odot } = \sum {S_{products}} - \sum {S_{reac\tan ts}}\\ \,\,\,\,\,\,\,\,\, = [4{S_{NO}} + 6{S_{{H_2}O}}] - [4{S_{N{H_3}}} + 5{S_{{O_2}}}]\\ \,\,\,\,\,\,\,\, = 4(104) + 6( - 237) - [4( - 108) + 5(0)]\\ \,\,\,\,\,\,\, = \, - 574Kcal \end{array}$

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