The rate law for a reaction between the substances A and is given by Rate = k[A]n [B]m: On doubling the concentration of A and halving the concentration of B, the ratio of the new rate to the earlier rate of the reaction will be as:
Answer & explanation
Correct answer: option 4
The correct answer is option 4. \(2^{(n – m)}\).
The rate law for a reaction between the substances A and B is given by:
\( \text{Rate} = k[A]^n [B]^m \)
On doubling the concentration of A and halving the concentration of B, the ratio of the new rate to the earlier rate of the reaction will be as follows:
Doubling the concentration of A: \( [A] \rightarrow 2[A] \)
Halving the concentration of B: \( [B] \rightarrow \frac{1}{2}[B] \)
Substituting these changes into the rate expression:
\(\text{New Rate} = k(2[A])^n (\frac{1}{2}[B])^m\)
\(\text{New Rate} = k2^n [A]^n \left(\frac{1}{2}\right)^m [B]^m\)
\(\text{New Rate} = k 2^n 2^{-m} [A]^n [B]^m\)
Therefore, the ratio of the new rate to the earlier rate is:
\(\frac{\text{New Rate}}{\text{Earlier Rate}} = \frac{k 2^n 2^{-m} [A]^n [B]^m}{k[A]^n [B]^m} = 2^{n - m}\)
Hence, the correct option would be (4) \( 2^{(n - m)} \), as it represents the ratio of the new rate to the earlier rate.