6.12  Enthalpies of formation of CO (g ) , CO_2 ( g ), N_ 2 O ( g ) \: \: and \: \: N_2 O_4 (g) are –110, – 393, 81 and 9.7 KJ mol ^{-1}  respectively. Find the value of \Delta _ r H for the reaction:

           N_2O_4(g) + 3CO(g) \rightarrow N_2O(g) + 3CO_2(g)

Answers (1)
M manish

Given,
Enthalpies of formation of CO (g ) , CO_2 ( g ), N_ 2 O ( g ) \: \: and \: \: N_2 O_4 (g) are –110, – 393, 81  and 9.7 KJ mol ^{-1}
N_2O_4(g) + 3CO(g) \rightarrow N_2O(g) + 3CO_2(g)
 We know that the \Delta _rH=\sum \Delta _fH(product)-\sum \Delta _fH(reactants)

For the above reaction,

\Delta _rH =[{\Delta _fH(N_{2}O)+3\Delta _fH(CO_{2})}]-[\Delta _fH(N_{2}O_{4})+3\Delta _fH(CO)]
Substituting the given values we get,
             \\= [81+3(-393)]-[9.7+3(-110)]\\ =-777.7 kJ/mol
Thus the value of \Delta _ r H of the reaction is -777.7 kJ/mol

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