Question

How much and in what direction will each of the following effect the rate of the reaction:

CO(g) + \({\bf{NO}}{}_{\bf{2}}\) (g)⟶ \({\bf{CO}}{}_{\bf{2}}\) (g) + NO(g) if the rate law for the reaction is rate =\({\bf{k(NO}}{}_{\bf{2}}{{\bf{)}}^{\bf{2}}}{\bf{a}}\)?

1. Decreasing the pressure of \({\bf{NO}}{}_{\bf{2}}\) from 0.50 atm to 0.250 atm.
2. Increasing the concentration of CO from 0.01 M to 0.03 M.

Short answer

1. Decreasing the pressure of \({\bf{NO}}{}_{\bf{2}}\) from 0.50 atm to 0.250 atm is factor 4
2. Increasing the concentration of CO from 0.01 M to 0.03 M. is not affect the rate of reaction

Rate law predict the relationship between rate of reaction and concentration of reaction.

Step-by-step solution

Rate of reaction

\(\frac{{{\bf{rate(R}}{}_{\bf{1}}{\bf{)}}}}{{{\bf{rate(R}}{}_2{\bf{)}}}}{\bf{ = }}\frac{{{\bf{k(0}}{\bf{.25NO}}{}_{\bf{2}}{{\bf{)}}^{\bf{2}}}}}{{{\bf{k(0}}{\bf{.50NO}}{}_{\bf{2}}{{\bf{)}}^{\bf{2}}}}}{\bf{ = }}\frac{{{\bf{0}}{\bf{.0625}}}}{{{\bf{0}}{\bf{.25}}}}{\bf{ = }}\frac{{\bf{1}}}{{\bf{4}}}\)

Here rate (\({\bf{R}}{}_{\bf{1}}\)) is four time large from rate (\({\bf{R}}{}_2\)). Pressure reduces the rate by factor of 4.

Rate of reaction

Carbon monoxide does not affect to rate law. So, it does not change the rate of reaction.