The rate constant of `n^(th)` order has units :

Options

A. `"litre"^(1-n)mol^(1-n)sec^(-1)`
B. `"Mol"^(1-n)litre^(1-n)sec`
C. `"Mol"^(1-n^(2))litre^(n^(2))sec^(-1)`
D. `"Mole"^(1-n)litre^(n-1)sec^(-1)`

Answer

To determine the units of the rate constant (k) for an nth order reaction, we start from the rate law expression:

1. **Write the rate law equation**:
The rate of a reaction (r) can be expressed as:
\[ r = k [A]^n \]
where:
- \( r \) is the rate of the reaction,
- \( k \) is the rate constant,
- \([A]\) is the concentration of the reactant.

2. **Identify the units of the rate**:
The rate \( r \) is typically expressed in terms of concentration change over time. In SI units, this is:
\[ r = \text{mol L}^{-1} \text{s}^{-1} \]

3. **Identify the units of concentration**:
Concentration \([A]\) is expressed in moles per liter (mol/L), or:
\[ [A] = \text{mol L}^{-1} \]

4. **Substituting the units into the rate law**:
Rearranging the rate law gives us:
\[ k = \frac{r}{[A]^n} \]
Substituting the units:
\[ k = \frac{\text{mol L}^{-1} \text{s}^{-1}}{(\text{mol L}^{-1})^n} \]

5. **Simplifying the units**:
This simplifies to:
\[ k = \frac{\text{mol L}^{-1} \text{s}^{-1}}{\text{mol}^n \text{L}^{-n}} = \text{mol}^{1-n} \text{L}^{n-1} \text{s}^{-1} \]

6. **Final expression for the units of k**:
Thus, the units of the rate constant \( k \) for an nth order reaction are:
\[ k = \text{mol}^{1-n} \text{L}^{n-1} \text{s}^{-1} \]