Question

For the reaction: \[ 2A + B \rightarrow 3C \] The rate law is given as: \[ \text{Rate} = k[A]^2[B] \] If the concentration of A is doubled and the concentration of B is halved, how does the rate of the reaction change?

• A. It remains the same.
• B. It doubles.
• C. It increases by a factor of 4.
• D. It decreases by a factor of 2.

Hint:

In chemical kinetics, the rate law describes how the rate of a reaction depends on the concentrations of reactants. Pay attention to the powers of the concentration terms in the rate law when altering reactant concentrations.

Answer 1

The Correct Option is C

The rate law is given by: \[ \text{Rate} = k[A]^2[B] \] Let's compare the initial rate with the rate after changing the concentrations. - Initial concentration of A = \( [A] \), initial concentration of B = \( [B] \). - New concentration of A = \( 2[A] \) (doubled), new concentration of B = \( \frac{1}{2}[B] \) (halved). Substitute these into the rate law: - Initial rate: \( \text{Rate}_\text{initial} = k[A]^2[B] \) - New rate: \( \text{Rate}_\text{new} = k(2[A])^2\left(\frac{1}{2}[B]\right) \) Simplifying: \[ \text{Rate}_\text{new} = k \times 4[A]^2 \times \frac{1}{2}[B] = 2k[A]^2[B] \] Thus, the new rate is twice the initial rate, meaning the rate increases by a factor of 4.