Question

If $\frac{d\left[N{H}_3\right]}{dt}=34gh{r}^{-1}$ for the reaction $N_2+3H_2\rightleftharpoons 2N{H}_3$ then $\frac{d\left[H_2\right]}{dt}$ is ______ $gh{r}^{-1}$

• A. $-17$
• B. $-51$
• C. $-81$
• D. $32$

Answer 1

The correct option is B $-51$

For the reaction $N_2+3H_2\rightleftharpoons 2N{H}_3$

$-\frac{1}{3}\frac{d\left[H_2\right]}{dt}=\frac{1}{2}\frac{d\left[N{H}_3\right]}{dt}$

$\frac{d\left[H_2\right]}{dt}=-\frac{3}{2}\frac{d\left[N{H}_3\right]}{dt}$

But $\frac{d\left[N{H}_3\right]}{dt}=34gh{r}^{-1}$

then $\frac{d\left[H_2\right]}{dt}=-\frac{3}{2}\frac{d\left[N{H}_3\right]}{dt}=-\frac{3}{2}\times 34gh{r}^{-1}$ $=-51$ $gh{r}^{-1}$