The rate law for a reaction between the substances A and B is given by rate `= K[A]^(n) [B]^(m)`. On doubling the concentration of A and having the concentration of B, the ratio of the new rate to the earlier rate of the reactio will be:

Options

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

Explanation

To solve the problem, we need to determine the ratio of the new rate to the earlier rate when the concentration of A is doubled and the concentration of B remains unchanged.

### Step-by-Step Solution:

1. **Write the Rate Law Expression**:
The rate law for the reaction is given as:
\[ \text{Rate} = k[A]^n[B]^m \]

2. **Define the Initial Rate**:
Let the initial concentrations of A and B be \([A]\) and \([B]\) respectively. The initial rate (\(R_1\)) can be expressed as:
\[ R_1 = k[A]^n[B]^m \]

3. **Change the Concentration of A**:
When the concentration of A is doubled, the new concentration of A becomes \(2[A]\). The concentration of B remains unchanged at \([B]\).

4. **Define the New Rate**:
The new rate (\(R_2\)) with the modified concentration of A is:
\[ R_2 = k[2A]^n[B]^m \]
This can be rewritten as:
\[ R_2 = k(2[A])^n[B]^m = k \cdot 2^n[A]^n[B]^m \]

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

6. **Final Result**:
Therefore, the ratio of the new rate to the earlier rate when the concentration of A is doubled is:
\[ \frac{R_2}{R_1} = 2^n \]