Question

For the reaction:
$4N{H}_3\left(g\right)+3O_2\left(g\right)\to 2N_2\left(g\right)+6H_{2}O\left(l\right)$
at $298K$. Given that heat of formation of $N{H}_3(g$) and $H_{2}O\left(l\right)$ are $-46kJmo{l}^{-1}$ and $-286kJmo{l}^{-1}$ respectively. Calculate heat of reaction at $298K$.

• A. $-2232kJmo{l}^{-1}$
• B. $-1532kJmo{l}^{-1}$
• C. $-1852kJmo{l}^{-1}$
• D. $-2832kJmo{l}^{-1}$

Answer 1

The correct option is B $-1532kJmo{l}^{-1}$

We know,
${\Delta }_{r}H={\Delta }_f{H}_{products}^o-{\Delta }_f{H}_{reactants}^o$
So,
${\Delta }_{r}H=2\times {\Delta }_f{H}_{N_2\left(g\right)}^o+6\times {\Delta }_f{H}_{H_{2}O\left(l\right)}^o-[4\times {\Delta }_f{H}_{N{H}_3\left(g\right)}^o+3\times {\Delta }_f{H}_{O_2\left(g\right)}^o]$

${\Delta }_f{H}_{O_2}^{o}and{\Delta }_f{H}_{N_2}^o$ are taken to be 0 as by convention, standard enthalpy for formation of an element in reference state, i.e., its most stable state of aggregation is taken as zero.

putting the values,
${\Delta }_{r}H=2\times \left(0\right)+6\times (-286)-[4\times (-46)+3\times (0\left)\right]$
We get,
${\Delta }_{r}H=-1532kJmo{l}^{-1}$