Question

For the chemical reaction
$N_2$(g) + $3H_2$(g) $\rightleftharpoons$ $2N{H}_3$(g)
The correct option is:

• A. $-\frac{1}{3}$ $\frac{d\left[H_2\right]}{dt}$ = $-\frac{1}{2}$ $\frac{d\left[N{H}_3\right]}{dt}$
• B. $\frac{-d\left[N_2\right]}{dt}$ = $2\frac{d\left[N{H}_3\right]}{dt}$
• C. $\frac{-d\left[N_2\right]}{dt}$ = $\frac{1}{2}$ $\frac{d\left[N{H}_3\right]}{dt}$
• D. $3\frac{d\left[H_2\right]}{dt}$ = $2\frac{d\left[N{H}_3\right]}{dt}$

Answer 1

The correct option is C $\frac{-d\left[N_2\right]}{dt}$ = $\frac{1}{2}$ $\frac{d\left[N{H}_3\right]}{dt}$

$N_2$(g) + $3H_2$(g) $\rightleftharpoons$ $2N{H}_3$(g)
Rate of raection is given as

-$\frac{d\left[N_2\right]}{dt}$ = - $\frac{1}{3}$ $\frac{d\left[H_2\right]}{dt}$ = $\frac{1}{2}$ $\frac{d\left[N{H}_3\right]}{dt}$