Question

For a reaction, the rate law is given by \( \text{rate} = k[A]^2[B] \). If the concentration of \( A \) is doubled and the concentration of \( B \) is halved, how will the rate of the reaction change?

• A. The rate will be doubled.
• B. The rate will be halved.
• C. The rate will be quadrupled.
• D. The rate will remain unchanged.

Hint:

The rate of a reaction depends on the concentrations of the reactants raised to the powers specified by the rate law. Always use the correct exponents and factor in how concentration changes affect the rate.

Answer 1

The Correct Option is C

The rate law for the given reaction is: \[ \text{rate} = k[A]^2[B] \] Where: - \( k \) is the rate constant, - \( [A] \) and \( [B] \) are the concentrations of reactants \( A \) and \( B \). We are told that the concentration of \( A \) is doubled and the concentration of \( B \) is halved. Let the initial concentrations of \( A \) and \( B \) be \( [A]_0 \) and \( [B]_0 \), respectively. Then, the new concentrations are: \[ [A] = 2[A]_0 \quad \text{and} \quad [B] = \frac{1}{2}[B]_0 \] Now, substitute these into the rate law: \[ \text{new rate} = k(2[A]_0)^2\left(\frac{1}{2}[B]_0\right) \] \[ \text{new rate} = k \times 4[A]_0^2 \times \frac{1}{2}[B]_0 \] \[ \text{new rate} = 2k[A]_0^2[B]_0 \] Comparing the new rate with the original rate: \[ \text{rate} = k[A]_0^2[B]_0 \] Thus, the new rate is twice the original rate. Therefore, the rate will be quadrupled.