Question

The standard heat of formation in kcal $mo{l}^{-1}$ of $N{O}_2\left(g\right)$ and $N_2{O}_4\left(g\right)$ are 8.0 and 2.0 respectively. The heat of dimerization of $N{O}_2$ in kcal is:
$2N{O}_2\left(g\right)=N_2{O}_4\left(g\right)$

• A. $10.0$
• B. $-6.0$
• C. $-12.0$
• D. $-14.0$

Answer 1

The correct option is D $-14.0$

Solution:- (D) $-14.0$

Dimerization of ${N{O}_2}_{\left(g\right)}$-

$2{N{O}_2}_{\left(g\right)}\rightleftharpoons {N_2{O}_4}_{\left(g\right)}$

Given:-

${\Delta H_f}_{{N{O}_2}_{\left(g\right)}}=8kcal/mol$

${\Delta H_f}_{{N_2{O}_4}_{\left(g\right)}}=2kcal/mol$

$\Delta H={\sum \Delta H_f}_{\left(product\right)}-{\sum \Delta H_f}_{\left(reactant\right)}$

$\Rightarrow \Delta H=\left(2\right)-(2\times 8)$

$=-14kcal/mol$

Hence the heat of dimerization of ${N{O}_2}_{\left(g\right)}$ is $-14kcal/mol$.