Question

In reaction $N_2\left(g\right)+3H_2\left(g\right)⟶2N{H}_3\left(g\right)$ the rate of appearance of $N{H}_3$ is $2.5\times {10}^{-4}mol$ $L^{-1}$ $se{c}^{-1}$. The rate of reaction & rate of disappearance of $H_2$ will be (in $mol$ $L^{-1}se{c}^{-1}$) is:

• A. $3.75\times {10}^{-4},1.25\times {10}^{-4}$
• B. $1.25\times {10}^{-4},2.5\times {10}^{-4}$
• C. $1.25\times {10}^{-4},3.75\times {10}^{-4}$
• D. $5.0\times {10}^{-4},3.75\times {10}^{-4}$

Answer 1

The correct option is C $1.25\times {10}^{-4},3.75\times {10}^{-4}$

Given reaction is :

$N_2\left(g\right)+3H_2\left(g\right)\to 2N{H}_3\left(g\right)$

Rate of appearance of $N{H}_3=2.5\times {10}^{-4}mol{L}^{-1}$

We will write rate of reaction which are as follows:

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

$ROR=\frac{2.5\times {10}^{-4}}{2}$

$ROR=1.25\times {10}^{-4}$

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

$-\frac{d\left[H_2\right]}{dt}=\frac{3}{2}\times 2.5\times {10}^{-4}$

$-\frac{d\left[H_2\right]}{dt}=3.75\times {10}^{-4}$

So,

$ROR=1.25\times {10}^{-4}$ & Rate of disappearance of $H_2$ is $-\frac{d\left[H_2\right]}{dt}=3.75\times {10}^{-4}$

Option C is correct.