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Q.

 ∆f0S of H2O(l), NH3(g) and NO(g) respectively are –237, –108 and +104 K.Cal.mole–1. Then ∆f0S for 4NH3(g) + 5O2→ 4NO(g) + 6H2O (in K.Cal) will be

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a

– 574

b

–1148

c

+1148

d

+ 574

answer is A.

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Detailed Solution

4NH3(g) + 5O2→ 4NO(g) + 6H2O , ∆f0S=?

 

\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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