Question

Nitrogen oxide reacts with hydrogen at elevated temperatures according to the following chemical equation:

$2\text{NO}\mathit{\hspace{0.17em}}\left(g\right)\mathit{+}\mathit{2}{\text{H}}_{\mathit{2}}\left(g\right)\mathit{\to }{\text{N}}_{\mathit{2}}\mathit{\left(}\mathit{g}\mathit{\right)}\mathit{+}\mathit{2}{\text{H}}_{\mathit{2}}O$

It is observed that, when the concentration of H₂ is cut in half, the rate of the reaction is also cut in half. When the concentration of NO is multiplied by 10, the rate of the reaction increases by a factor of 100

. (a) Write the rate expression for this reaction, and give the units of the rate constant k.

(b) If [NO] were multiplied by 3 and [H₂] by 2, what change in the rate would be observed?

Short answer

The rate of reaction is given by expression

$\text{Rate=18k}\left[{\text{H}}_2\right]{\left[NO\right]}^2$

Step-by-step solution

Rate expression for a reaction

The rate of chemical reaction can be explained as expression which describes relation between rate and product of concendration of reactents raised to power.

The given chemical reaction is

$2\text{NO}\mathit{\hspace{0.17em}}\mathit{\left(}g\mathit{\right)}\mathit{+}\mathit{2}{\text{H}}_{\mathit{2}}\left(g\right)\mathit{\to }{\text{N}}_{\mathit{2}}\mathit{\left(}g\mathit{\right)}\mathit{+}\mathit{2}{\text{H}}_{\mathit{2}}O.........\left(1\right)$

Rate expression is given as

$\text{Rate=k}\left[{{\text{H}}_2}^1\right]{\left[\mathrm{NO}\right]}^2$

Rate constant units:

Units of rate constant for overall reaction of the order n is given as

${\text{mol}}^{-(n-1)}{\text{L}}^{\left(n-1\right)}{\text{S}}^{-1}$

The overall order of the reaction is 4 so it can be written as

$mo{l}^{-\left(4-1\right)}{L}^{4-1}S{-1}^{}$

Rate of the reaction:

b) The concendration of NO and H₂ are multiplied by 3 and 2 ,then the rate of the reaction is written as

$$\begin{gathered} \text{Rate}=\text{k}\left[2{\text{H}}_2\right]{\left[3\text{NO}\right]}^2 \\ \text{Rate}={\text{k}}_2\times \left[{\text{H}}_{\text{2}}\text{×9}\right]{\left[NO\right]}^2 \\ \text{Rate}=\hspace{0.17em}18\text{k}\left[{\text{H}}_2\right]{\left[\text{NO}\right]}^{\text{2}} \end{gathered}$$

Therefore the rate of the reaction increases by 18 times, when the concentration of NO and H₂ is multiplied by 3 and 2 respectively.