The unit of the rate constant of nth order is

Options

A. `"mol"^(1-n)L^(n-1)s^(-1)`
B. `"mol"^(n-1)L^(1-n)s^(-1)`
C. `"mol"^(n-1)L^(n-1)s`
D. `"mol"^(n)L^(1-n)s^(-1)`

Explanation

To determine the unit of the rate constant (k) for a reaction of nth order, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding 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. **Identifying Units**:
The unit of rate is typically expressed in terms of concentration per time. In SI units, this is:
\[ \text{Rate} = \text{mole} \cdot \text{litre}^{-1} \cdot \text{second}^{-1} \]

3. **Concentration Units**:
The concentration \( [A] \) has units of:
\[ [A] = \text{mole} \cdot \text{litre}^{-1} \]

4. **Substituting Units into the Rate Law**:
Rearranging the rate equation to solve for the rate constant \( k \):
\[ k = \frac{\text{Rate}}{[A]^n} \]

5. **Substituting the Units**:
Now substituting the units into the equation:
\[ k = \frac{\text{mole} \cdot \text{litre}^{-1} \cdot \text{second}^{-1}}{(\text{mole} \cdot \text{litre}^{-1})^n} \]

6. **Simplifying the Expression**:
This can be simplified:
\[ k = \frac{\text{mole} \cdot \text{litre}^{-1} \cdot \text{second}^{-1}}{\text{mole}^n \cdot \text{litre}^{-n}} \]
\[ k = \text{mole}^{1-n} \cdot \text{litre}^{n-1} \cdot \text{second}^{-1} \]

7. **Final Unit of Rate Constant**:
Therefore, the unit of the rate constant \( k \) for a reaction of nth order is:
\[ k = \text{mole}^{1-n} \cdot \text{litre}^{n-1} \cdot \text{second}^{-1} \]