Doubling the concentration of a reactant increases the rate of a reaction four times. With this knowledge, answer the following questions:

1. What is the order of the reaction with respect to that reactant?
2. Tripling the concentration of a different reactant increases the rate of a reaction three times. What is the order of the reaction with respect to that reactant?

The rate law for a chemical reaction is an expression that provides a relationship between the rate of the reaction and the concentration of the reactants participating in it.

\({\bf{rate = k(reactant}}{{\bf{)}}^{\bf{n}}}\)

Doubling the concentration of a reactant increases the rate of a reaction four times.

\({\bf{4(rate) = k(2reactant}}{{\bf{)}}^{\bf{n}}}\)

Hence, the order of the reaction can be calculated as

\(\begin{aligned}{c}\frac{{{\bf{4(Rate)}}}}{{{\bf{Rate}}}}{\bf{ = }}\frac{{\,{\bf{k(2(reactant)}}{{\bf{)}}^{\bf{n}}}}}{{{\bf{k(reactant}}{{\bf{)}}^{\bf{n}}}}}\\{\bf{n = 2}}\end{aligned}\)

The concentration of the reactant doubled and the rate quadrupled, hence the order with respect to the reactant is 2.

Tripling the concentration of a different reactant increases the rate of a reaction three times. It can be represented as

\({\bf{3(Rate) = }}\,{\bf{k(3(reactant)}}{{\bf{)}}^{\bf{n}}}\)

Hence, the order of the reaction can be calculated as

\(\begin{aligned}{}\frac{{{\bf{3(rate)}}}}{{{\bf{rate}}}}{\bf{ = }}\frac{{{\bf{k(3reactant}}{{\bf{)}}^{\bf{n}}}}}{{{\bf{k(reactant}}{{\bf{)}}^{\bf{n}}}}}\\{\bf{n = 1}}\end{aligned}\)

The concentration of the reactant and the rate both tripled, hence order with respect to this reactant is 1.