The rate law for a reaction between `A` and `B` 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 becomes

Options

A. `2^(1//(m+n))`
B. `m + n`
C. `m-n`
D. `2^(n-m)`

Explanation

To solve the problem, we need to find the ratio of the new rate of the reaction to the earlier rate after changing the concentrations of reactants A and B.

### Step-by-Step Solution:

1. **Write the Rate Law Expression**:
The rate law for the reaction is given by:
\[ \text{Rate} = k[A]^n[B]^m \]
where \( k \) is the rate constant, \( [A] \) is the concentration of reactant A, and \( [B] \) is the concentration of reactant B.

2. **Define the Initial Rate**:
Let the initial rate of the reaction be \( R_1 \):
\[ R_1 = k[A]^n[B]^m \]

3. **Change the Concentrations**:
According to the problem, the concentration of A is doubled and the concentration of B is halved:
\[ [A] \rightarrow 2[A] \quad \text{and} \quad [B] \rightarrow \frac{[B]}{2} \]

4. **Write the New Rate Expression**:
Let the new rate of the reaction be \( R_2 \):
\[ R_2 = k(2[A])^n\left(\frac{[B]}{2}\right)^m \]

5. **Expand the New Rate Expression**:
Expanding \( R_2 \):
\[ R_2 = k \cdot (2^n[A]^n) \cdot \left(\frac{[B]^m}{2^m}\right) = k \cdot 2^n \cdot [A]^n \cdot \frac{[B]^m}{2^m} \]
This simplifies to:
\[ R_2 = k \cdot [A]^n \cdot [B]^m \cdot \frac{2^n}{2^m} = k[A]^n[B]^m \cdot 2^{n-m} \]

6. **Find the Ratio of New Rate to Initial Rate**:
Now, we can find the ratio of the new rate to the initial rate:
\[ \frac{R_2}{R_1} = \frac{k[A]^n[B]^m \cdot 2^{n-m}}{k[A]^n[B]^m} \]
The \( k[A]^n[B]^m \) terms cancel out:
\[ \frac{R_2}{R_1} = 2^{n-m} \]

7. **Final Result**:
Therefore, the ratio of the new rate to the earlier rate is:
\[ \frac{R_2}{R_1} = 2^{n-m} \]