Question

The rate for the following reaction is given by: \[ A + B \rightarrow C, \quad \text{Rate} = k[A][B]^2 \] How is the rate affected if we double the concentration of B?

Hint:

If rate ∝ [B]², doubling B increases rate by 4× (square dependence rule).

Answer 1

Step 1: Understanding the Rate Law.
The rate law for the reaction is given as:
\[ \text{Rate} = k[A][B]^2 \] where:
- \( k \) is the rate constant,
- \( [A] \) and \( [B] \) are the concentrations of reactants A and B, respectively.

Step 2: Effect of Doubling the Concentration of B.
If the concentration of \( B \) is doubled, the effect on the rate can be determined by looking at the power of \( [B] \) in the rate law.
Since \( [B] \) is raised to the power of 2, when the concentration of \( B \) is doubled, the rate will increase by a factor of \( 2^2 = 4 \).

Step 3: Conclusion.
When the concentration of \( B \) is doubled, the rate of the reaction increases by a factor of 4, because the rate law shows that the rate is proportional to the square of the concentration of \( B \).