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 step by step, we need to analyze the rate law and how the changes in concentrations affect the rate of the reaction.

### Step 1: Write the original rate law
The rate of the reaction is given by:
\[ \text{Rate} = k[A]^n[B]^m \]
Let’s denote this as the original rate \( R_1 \).

### Step 2: Define the new concentrations
According to the problem, we are doubling the concentration of \( A \) and halving the concentration of \( B \):
- New concentration of \( A \) = \( 2[A] \)
- New concentration of \( B \) = \( \frac{1}{2}[B] \)

### Step 3: Write the new rate law with the new concentrations
The new rate \( R_2 \) with the modified concentrations will be:
\[ R_2 = k(2[A])^n\left(\frac{1}{2}[B]\right)^m \]

### Step 4: Simplify the new rate expression
Now, we can simplify \( R_2 \):
\[ R_2 = k \cdot (2^n[A]^n) \cdot \left(\frac{1}{2^m}[B]^m\right) \]
This can be rewritten as:
\[ R_2 = k \cdot 2^n \cdot \frac{1}{2^m} \cdot [A]^n \cdot [B]^m \]
\[ R_2 = k \cdot 2^{n-m} \cdot [A]^n \cdot [B]^m \]

### Step 5: Calculate the ratio of new rate to old rate
Now, we find the ratio of the new rate \( R_2 \) to the old rate \( R_1 \):
\[ \frac{R_2}{R_1} = \frac{k \cdot 2^{n-m} \cdot [A]^n \cdot [B]^m}{k \cdot [A]^n \cdot [B]^m} \]
The \( k \), \( [A]^n \), and \( [B]^m \) terms cancel out:
\[ \frac{R_2}{R_1} = 2^{n-m} \]

### Step 6: Conclusion
Thus, the ratio of the new rate to the old rate is:
\[ \frac{R_2}{R_1} = 2^{n-m} \]