For three reactions of first, second and third-order, the numerical value of the rate constant is the same. Which of the following is correct?
Given , [A] = the concentration of the reactant & `r_1 , r_2 and r_3` are the rates of first, second and third-order reaction respectively

Options

A. If `[A] = 1`, then `r_1 = r_2 = r_3`
B. If `[A] lt 1`, then , `r_1 gt r_2 gt r_3`
C. If `[A] gt 1`, then `r_3 gt r_2 gt r_1`
D. All of the above

Answer

To solve the problem, we need to analyze the rate equations for first, second, and third-order reactions, given that the numerical value of the rate constant (K) is the same for all three reactions.

### Step-by-Step Solution:

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

2. **Setting the Concentration of A**:
- Let's analyze the rates when the concentration of A is equal to 1 mol/L:
- For first-order:
\[ R_1 = K \cdot 1 = K \]
- For second-order:
\[ R_2 = K \cdot 1^2 = K \]
- For third-order:
\[ R_3 = K \cdot 1^3 = K \]
- Therefore, when [A] = 1, we have:
\[ R_1 = R_2 = R_3 = K \]

3. **Analyzing Different Concentrations**:
- **When [A] < 1**:
- In this case, since the powers of A are less than 1, the rates will be:
\[ R_1 > R_2 > R_3 \]
- This is because R1 depends linearly on [A], while R2 and R3 depend on higher powers of [A].

- **When [A] > 1**:
- Here, the rates will be:
\[ R_3 > R_2 > R_1 \]
- This is because R3 increases the fastest due to the cubic dependence on [A], followed by R2, and then R1.

4. **Conclusion**:
- From the analysis, we can conclude:
- When [A] = 1: \( R_1 = R_2 = R_3 \)
- When [A] < 1: \( R_1 > R_2 > R_3 \)
- When [A] > 1: \( R_3 > R_2 > R_1 \)
- Therefore, the correct statement is that all of the above conditions are true.

### Final Answer:
All of the above statements are correct.