Question

For the reaction:

$N_2+3H_2\to 2N{H}_3$,

If $\frac{d\left[N{H}_3\right]}{dt}=2\times {10}^{-4}mol{L}^{-1}{s}^{-1}$, the value of $\frac{-d\left[H_2\right]}{dt}$ would be:

• A. $1\times {10}^{-4}mol{L}^{-1}{s}^{-1}$
• B. $3\times {10}^{-4}mol{L}^{-1}{s}^{-1}$
• C. $4\times {10}^{-4}mol{L}^{-1}{s}^{-1}$
• D. $6\times {10}^{-4}mol{L}^{-1}{s}^{-1}$

Answer 1

Answer: (B)

$3\times {10}^{-4}mol{L}^{-1}{s}^{-1}$

$N_2+{3H}_2⟶{2NH}_3$

Applying rate law.

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

Now, given that $\frac{d\left[{NH}_3\right]}{dt}=2\times {10}^{-4}mol{L}^{-1}{S}^{-1}$

Using above relation,

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

$\frac{-d\left[H_2\right]}{dt}=\frac{3}{2}\times 2\times {10}^{-4}mol{L}^{-1}{S}^{-1}$

$\therefore$ $\frac{-d\left[H_2\right]}{dt}=3\times {10}^{-4}mol{L}^{-1}{S}^{-1}$

Hence, option B is correct.