For a reaction A \(\rightarrow\) B, the rate law is given by:
\[
\text{Rate} = k[\text{A}]^2
\]
If the concentration of A is increased by a factor of 3, by what factor does the rate of the reaction increase?
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The rate of reaction is proportional to the concentration of reactants raised to the power of their respective order. For a reaction where the rate law is \( \text{Rate} = k[\text{A}]^2 \), a 3-fold increase in concentration results in a 9-fold increase in the rate.
The rate law for the reaction is given by:
\[
\text{Rate} = k[\text{A}]^2
\]
Where:
- \( k \) is the rate constant,
- \( [\text{A}] \) is the concentration of reactant A.
If the concentration of A is increased by a factor of 3, then:
\[
[\text{A}] \rightarrow 3[\text{A}]
\]
Substitute this new concentration into the rate law:
\[
\text{Rate}_{\text{new}} = k(3[\text{A}])^2 = k \times 9[\text{A}]^2
\]
Thus, the rate of the reaction increases by a factor of 9.
