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?
Answer & explanation
Correct answer: option 4
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.