Rate constant of a reaction (k) is `4 xx 10^(-2) "litre"^(-1) molsec^(-1)`. What is the order of reaction ?

Options

A. 0
B. 2
C. 1
D. 3

Answer

To determine the order of the reaction based on the given rate constant (k), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Units of the Rate Constant (k)**:
The given rate constant is \( k = 4 \times 10^{-2} \, \text{litre}^{-1} \, \text{mol} \, \text{sec}^{-1} \).

2. **Write the General Rate Law**:
The rate law for a reaction can be expressed as:
\[ \text{Rate} = k \cdot [A]^x \]
where \( [A] \) is the concentration of reactant A and \( x \) is the order of the reaction.

3. **Identify the Units of Rate**:
The unit of rate is typically expressed as:
\[ \text{Rate} = \text{mol} \, \text{litre}^{-1} \, \text{sec}^{-1} \]

4. **Relate the Units of k to the Concentration**:
The units of k can be expressed in terms of concentration and time. From the rate law:
\[ \text{Rate} = k \cdot [A]^x \]
Rearranging gives:
\[ k = \frac{\text{Rate}}{[A]^x} \]
Therefore, the units of k can be expressed as:
\[ k = \frac{\text{mol} \, \text{litre}^{-1} \, \text{sec}^{-1}}{(\text{mol} \, \text{litre}^{-1})^x} = \frac{\text{mol} \, \text{litre}^{-1} \, \text{sec}^{-1}}{\text{mol}^x \, \text{litre}^{-x}} \]

5. **Simplify the Units of k**:
This simplifies to:
\[ k = \text{mol}^{1-x} \, \text{litre}^{x-1} \, \text{sec}^{-1} \]

6. **Set the Units Equal**:
Now, we can set the units of k equal to the given units:
\[ \text{mol}^{1-x} \, \text{litre}^{x-1} \, \text{sec}^{-1} = \text{mol}^{1} \, \text{litre}^{-1} \, \text{sec}^{-1} \]

7. **Compare the Exponents**:
From the comparison of the units:
- For \(\text{mol}\): \(1 - x = 1\) → \(x = 0\)
- For \(\text{litre}\): \(x - 1 = -1\) → \(x = 0\)
- For \(\text{sec}\): This part is consistent as it is \(-1\) on both sides.

8. **Conclusion**:
Since \(x = 0\), the order of the reaction is **zero order**.

### Final Answer:
The order of the reaction is **0**.