Question

For the reaction,$N_2+3H_2⟶2N{H}_3$ 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. $4\times {10}^{-4}mol{L}^{-1}{s}^{-1}$
• B. $6\times {10}^{-4}mol{L}^{-1}{s}^{-1}$
• C. $1\times {10}^{-4}mol{L}^{-1}{s}^{-1}$
• D. $3\times {10}^{-4}mol{L}^{-1}{s}^{-1}$

Answer 1

The correct option is D $3\times {10}^{-4}mol{L}^{-1}{s}^{-1}$

$\frac{1}{2}\frac{d\left[N{H}_3\right]}{dt}$ = $1\times {10}^{-4}$

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

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