Rate constant of a reaction (k) is `2 xx 10^(-3) "litre"^(2) "mol"^(-2)sec^(-1)`. What is the order of reaction ?">

Options

A. `1^(st)`
B. `2^(nd)`
C. `4^(th)`
D. `3^(rd)`

Explanation

To determine the order of the reaction based on the given rate constant \( k \), we will analyze the units of \( k \) and relate them to the general form of the rate law.

### Step-by-Step Solution:

1. **Identify the given rate constant and its units**:
The rate constant \( k \) is given as:
\[ k = 2 \times 10^{-3} \, \text{litre}^2 \, \text{mol}^{-2} \, \text{sec}^{-1} \]

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

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

4. **Express the units of the rate constant \( k \)**:
The units of \( k \) can be expressed in terms of the units of concentration raised to the power of \( n \):
\[ k = \frac{\text{Rate}}{[A]^n} = \frac{\text{mol} \, \text{litre}^{-1} \, \text{sec}^{-1}}{(\text{mol} \, \text{litre}^{-1})^n} \]

5. **Simplify the units of \( k \)**:
This gives us:
\[ k = \frac{\text{mol} \, \text{litre}^{-1} \, \text{sec}^{-1}}{\text{mol}^n \, \text{litre}^{-n}} = \text{mol}^{1-n} \, \text{litre}^{n-1} \, \text{sec}^{-1} \]

6. **Set the units of \( k \) equal to the given units**:
We can now equate the units of \( k \):
\[ \text{mol}^{1-n} \, \text{litre}^{n-1} \, \text{sec}^{-1} = \text{mol}^{-2} \, \text{litre}^2 \, \text{sec}^{-1} \]

7. **Compare the powers of each unit**:
- For \(\text{mol}\):
\[ 1 - n = -2 \implies n = 3 \]
- For \(\text{litre}\):
\[ n - 1 = 2 \implies n = 3 \]
- For \(\text{sec}\):
\[ \text{Both sides already match as sec}^{-1} \]

8. **Conclusion**:
Since \( n = 3 \), the order of the reaction is **third order**.

### Final Answer:
The order of the reaction is 3 (third order).

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