Question

The rate law for a reaction between the substances A and B is given by: \[ {Rate} = k[A]^m[B]^n \] 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:

• A. \( (m + n) \)
• B. \( (n - m) \)
• C. \( 2^{(n - m)} \)
• D. \( \frac{1}{2^{(m + n)}} \)

Hint:

When concentrations change, express new rate in terms of old rate using exponent laws.

Answer 1

The Correct Option is C

Step 1: Understanding the rate law. The rate law for the reaction between A and B is: \[ {Rate} = k[A]^m[B]^n \] where:
- \( k \) is the rate constant
- \( m \) is the order of the reaction with respect to A
- \( n \) is the order of the reaction with respect to B
Step 2: Effect of concentration change on the rate. Initially, the rate is: \[ {Rate}_{{initial}} = k[A]^m[B]^n \] When the concentration of A is doubled and the concentration of B is halved, the new rate is: \[ {Rate}_{{new}} = k[2A]^m\left[\frac{B}{2}\right]^n \] Simplifying: \[ {Rate}_{{new}} = k(2^m[A]^m)\left(\frac{1}{2^n}[B]^n\right) = 2^{m-n} \times k[A]^m[B]^n \] Step 3: Finding the ratio. The ratio of the new rate to the initial rate is: \[ \frac{{Rate}_{{new}}}{{Rate}_{{initial}}} = \frac{2^{m-n} \times k[A]^m[B]^n}{k[A]^m[B]^n} = 2^{n - m} \] Thus, the correct ratio is \( 2^{(n - m)} \).