The rate constant is numerically the same for three reaction of first , second and third order respectively. Which one is true for the of three reaction, if concentration of reactant is greater than 1M ?

Options

A. `r_1 = r_2 = r_3`
B. `r_1 gt r_2 gt r_3`
C. `r_1 lt r_2 lt r_3`
D. All of these

Answer

To solve the problem, we need to analyze the rates of three different reactions of first, second, and third order, given that the rate constant (k) is numerically the same for all three reactions. We will also consider the condition that the concentration of the reactant (A) is greater than 1 M.

### Step-by-Step Solution:

1. **Identify the Rate Expressions**:
- For a **first-order reaction**, the rate (R1) is given by:
\[ R_1 = k[A]^1 = k[A] \]
- For a **second-order reaction**, the rate (R2) is given by:
\[ R_2 = k[A]^2 \]
- For a **third-order reaction**, the rate (R3) is given by:
\[ R_3 = k[A]^3 \]

2. **Assume Concentration Greater than 1 M**:
- Let’s assume the concentration of the reactant A is greater than 1 M, say \( [A] = 2 \, \text{M} \) (the exact value is not crucial as long as it is greater than 1 M).

3. **Calculate the Rates**:
- For **first-order**:
\[ R_1 = k \times 2 = 2k \]
- For **second-order**:
\[ R_2 = k \times (2)^2 = k \times 4 = 4k \]
- For **third-order**:
\[ R_3 = k \times (2)^3 = k \times 8 = 8k \]

4. **Compare the Rates**:
- From the calculations:
- \( R_1 = 2k \)
- \( R_2 = 4k \)
- \( R_3 = 8k \)
- Clearly, \( R_3 > R_2 > R_1 \).

5. **Conclusion**:
- Therefore, when the concentration of the reactant is greater than 1 M, the rates of the reactions will follow the order:
\[ R_3 > R_2 > R_1 \]
- This means that the third-order reaction will have the highest rate, followed by the second-order reaction, and the first-order reaction will have the lowest rate.

### Final Answer:
The correct order of the rates when the concentration of the reactant is greater than 1 M is:
\[ R_3 > R_2 > R_1 \]