Question

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

$2\mathrm{NO}\left(\mathrm{g}\right)+2{\mathrm{H}}_2\left(\mathrm{g}\right)\to {\mathrm{N}}_2\left(\mathrm{g}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)$

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 beobserved?

Short answer

The rate of reaction is given by the expression

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

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

$2\mathrm{NO}\left(\mathrm{g}\right)+2{\mathrm{H}}_2\left(\mathrm{g}\right)\to {\mathrm{N}}_2\left(\mathrm{g}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)$

Rate expression is given as

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

Where k is the rate constant of the reaction

Rate constant units

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

${\mathrm{mol}}^{-\left(\mathrm{n}-1\right)}{\mathrm{L}}^{\mathrm{n}-1}{\mathrm{s}}^{-1}$

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

${\mathrm{mol}}^{-\left(4-1\right)}{\mathrm{L}}^{4-1}{\mathrm{s}}^{-1}={\mathrm{mol}}^{-3}{\mathrm{L}}^3{\mathrm{s}}^{-1}$

Rate of the reaction

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

$$\begin{gathered} \mathrm{Rate}=\mathrm{k}\left[2{\mathrm{H}}_2\right]\left[3\mathrm{NO}\right]{}^2 \\ \mathrm{Rate}=\mathrm{k}2\mathrm{x}\left[{\mathrm{H}}_2\right]\mathrm{x}9\left[\mathrm{NO}\right]{}^2 \\ \mathrm{Rate}=18\mathrm{k}\left[{\mathrm{H}}_2\right]\left[\mathrm{NO}\right]{}^2 \end{gathered}$$

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