The rate for a reaction is found to be: Rate = k[NO₂^–][I^–][H⁺]²

How would the rate of reaction change when the concentration of NO₂^–, I^–, and H⁺ are tripled?

81 times

The correct answer is option 4. 81 times.

To determine how the rate of reaction changes when the concentrations of \(NO_2^–\), \(I^–\), and \(H^+\) are tripled, we need to understand the effect of concentration changes on the rate expression.

Given the rate expression:

\[ \text{Rate} = k[\text{NO}_2^-][\text{I}^-][\text{H}^+]^2 \]

If the concentrations of \(NO_2^–\), \(I^–\), and \(H^+\) are all tripled, the new rate would be:

\[ \text{New Rate} = k(3[\text{NO}_2^-])(3[\text{I}^-])(3[\text{H}^+])^2 \]

\[ \text{New Rate} = k \times 3 \times 3 \times 3 \times 3 \times [\text{NO}_2^-][\text{I}^-][\text{H}^+]^2 \]

\[ \text{New Rate} = 81 \times \text{Rate} \]

This means that the rate of reaction would increase by a factor of 81 times when the concentrations of \(NO_2^–\), \(I^–\), and \(H^+\) are all tripled.

Therefore, the correct answer is 81 times.