Question
The oxidation number of $Cr$ in ${K_2}C{r_2}{O_7}$ is
$2 + 2x - 2 \times 7 = 0$; $2x - 14 + 2 = 0$
$2x = 12$; $x = \frac{{12}}{2} = + 6$.
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$\left. \begin{gathered}
{\text{4N}}{{\text{H}}_{3(g)}}{\text{ + 5}}{{\text{O}}_{2(g)}} \to {\text{ 4N}}{{\text{O}}_{(g)}}{\text{ + 6}}{{\text{H}}_{\text{2}}}{{\text{O}}_{{\text{(l)}}}} \hfill \\
{\text{2N}}{{\text{O}}_{(g)}}{\text{ + }}{{\text{O}}_{2(g)}} \to {\text{2N}}{{\text{O}}_{2(g)}} \hfill \\
{\text{3N}}{{\text{O}}_{2(g)}}{\text{ + }}{{\text{H}}_{{\text{ 2}}}}{{\text{O}}_{(l)}} \to {\text{2H N}}{{\text{O}}_{3(aq)}}{\text{ + N}}{{\text{O}}_{(g)}} \hfill \\
\end{gathered} \right|\begin{array}{*{20}{c}}
{\Delta H(kJ)} \\
{ - 904} \\
{ - 112} \\
{ - 140}
\end{array}$
[Assume theat the pre-exponential factor is same in both the cases.
Given $R =8.31 \,J\, K ^{-1} \,mol ^{-1}$ ]