Question

For the reaction:-
$N_{2\left(g\right)}+3H_{2\left(g\right)}\to 2N{H}_{3\left(g\right)}$
Under certain conditions of temperature and partial pressure of the reactants, the rate of formation of $N{H}_3$ is $10^{-3}kgh{r}^{-1}.$ The rate of consumption of $H_2$ under same condition is:

• A. $1.5\times 10^{-3}kgh{r}^{-1}$
• B. $6.67\times 10^{-4}kgh{r}^{-1}$
• C. $0.0001764kgh{r}^{-1}$
• D. None of these

Answer 1

Answer: (C)

$N_{2(g)}+3H_{2(g)}\to 2N{H}_{3(g)}$

Rate of rxn $=\frac{-d}{dt}\left(N_2\right)=\frac{-1}{3}\frac{d}{dt}\left(H_2\right)$

$=+\frac{1}{2}\frac{d}{dt}\left(N{H}_3\right)$

Rate of formation of $N{H}_3=+\frac{d}{dt}\left(N_2\right)=\frac{10^{-3}\times 10^3}{17}$

$=0.0588molh{r}^{-1}$

$\frac{-1}{3}\frac{d}{dt}\left(H_2\right)=+\frac{1}{2}\frac{d}{dt}\left(N{H}_3\right)$

$\frac{-d}{dt}\left(H_2\right)=\frac{3}{2}\times \frac{d}{dt}\left(N{H}_3\right)=\frac{3}{2}\times 0.0588$

$=0.0882mollt{r}^{-1}h{r}^{-1}$

$H_2$ 1 mole $\to 2g$ or $0.002kg$

0.0882 moles $\to 0.002\times 0.0882=0.0001764kgh{r}^{-1}$