Question

The reaction $2H_2\left(g\right)+2NO\left(g\right)\to 2H_{2}O\left(g\right)+N_2\left(g\right)$ is first order in $H_2$ and second order in NO. The rate constant of the reaction at a certain temperature is $0.42M^{-2}{s}^{-1}$. Calculate the rate when $\left[H_2\right]=0.015M$ and [NO] = $0.025$M.

Answer 1

The given reaction is :-

$2H_2+2NO⟶2H_{2}O+N_2$

$\left(g\right)$ $\left(g\right)$ $\left(g\right)$ $\left(g\right)$

Given that, the reaction is first order in $H_2$ & second order in $NO$. So,

Rate= $K\left[H_2{]}^1\right[NO{]}^2$ $-\left(i\right)$

Now, $K$ is given as $0.42M^{-2}{s}^{-1}$, $\left[H_2\right]=0.15\times {10}^{-1}M$

$\left[NO\right]=0.025M$

So using these values in $e{q}^n\left(i\right)$

Rate= $0.42\times 0.015\times (0.025{)}^{2}M{s}^{-1}$

=$0.0000039375M{s}^{-1}$