Question

Enthalpies of formation of CO(g), $C{O}_2\left(g\right)$ $N_{2}O\left(g\right)$ and $N_2{O}_4\left(g\right)$ are $-110,-393.81$ and $9.7kJ$ $mo{l}^{-1}$ respectively. Find the value of ${△}_{r}H$ for the reaction :

$N_2{O}_4\left(g\right)+3CO\left(g\right)\to N_{2}O\left(g\right)+3C{O}_2\left(g\right)$

Answer 1

The balanced chemical reaction is $N_2{O}_4\left(g\right)+3CO\left(g\right)\to N_{2}O\left(g\right)+3C{O}_2\left(g\right)$
The enthalpy change for the reaction is,

${\Delta }_{r}H={\Delta }_{f}H\left[N_{2}O\right(g\left)\right]+3{\Delta }_{f}H\left[C{O}_2\right(g\left)\right]-\left[{\Delta }_{f}H{N}_2{O}_{4}9g\right)+3{\Delta }_{f}H\left[CO\right(g\left)\right]]$

${\Delta }_{r}H=81+3(-393)-[9.7+3(-110\left)\right]$

${\Delta }_{r}H=-777.7kJ/mol$

Hence, the enthalpy change for the reaction is $-777.7kJ/mol$.