# 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:           $N_2O_4(g) + 3CO(g) \rightarrow N_2O(g) + 3CO_2(g)$

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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