The rate constant of a reaction with nth order will have units:

Options

A. `litre^(1-n) mol^(n-1) sec^-1`
B. `mol^(1-n) litre^(1-n) sec^-1`
C. `mol^(1-n) litre^(n-1) sec^-1`
D. None of these

Answer

To find the units of the rate constant \( k \) for a reaction of order \( n \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Rate Law**:
The rate of a reaction can be expressed as:
\[ \text{Rate} = k [A]^n \]
where \( k \) is the rate constant, \( [A] \) is the concentration of the reactant, and \( n \) is the order of the reaction.

2. **Rearranging the Rate Law**:
To find the units of \( k \), we can rearrange the equation:
\[ k = \frac{\text{Rate}}{[A]^n} \]

3. **Define the Rate**:
The rate of a reaction is defined as the change in concentration over time. Therefore, the units of rate can be expressed as:
\[ \text{Rate} = \frac{\text{moles}}{\text{liter} \cdot \text{second}} = \text{mol L}^{-1} \text{s}^{-1} \]

4. **Substituting the Units**:
Now substitute the units of the rate into the equation for \( k \):
\[ k = \frac{\text{mol L}^{-1} \text{s}^{-1}}{([A])^n} \]

5. **Expressing Concentration Units**:
The concentration \( [A] \) has units of \( \text{mol L}^{-1} \). Therefore, \( [A]^n \) will have units:
\[ [A]^n = (\text{mol L}^{-1})^n = \text{mol}^n \text{L}^{-n} \]

6. **Final Units of \( k \)**:
Now substituting this back into the equation for \( k \):
\[ k = \frac{\text{mol L}^{-1} \text{s}^{-1}}{\text{mol}^n \text{L}^{-n}} = \text{mol}^{1-n} \text{L}^{n-1} \text{s}^{-1} \]

Thus, the units of the rate constant \( k \) for a reaction of order \( n \) are:
\[ \text{mol}^{1-n} \text{L}^{n-1} \text{s}^{-1} \]