The order of reaction is an experimentally determined quanity. It may be zero, poistive, negative, or fractional. The kinetic equation of `nth` order reaction is
`k xx t = (1)/((n-1))[(1)/((a-x)^(n-1)) - (1)/(a^(n-1))]` …(i)
Half life of `nth` order reaction depends on the initial concentration according to the following relation:
`t_(1//2) prop (1)/(a^(n-1))` ...(ii)
The unit of the rate constant varies with the order but general relation for the unit of `nth` order reaction is
Units of `k = [(1)/(Conc)]^(n-1) xx "Time"^(-1)` ...(iii)
The differential rate law for `nth` order reaction may be given as:
`(dX)/(dt) = k[A]^(n)` ...(iv)
where `A` denotes the reactant.
The unit of rate and rate constant are same for

Options

A. Zero order reaction
B. First order reaction
C. Second order reaction
D. Half order reaction

Explanation

To determine for which order of reaction the unit of rate and the unit of rate constant are the same, we can analyze the units for different orders of reactions.

### Step 1: Understand the Units of Rate
The rate of a reaction is defined as the change in concentration of a reactant or product over time. The unit of rate can be expressed as:
\[ \text{Rate} = \frac{d[A]}{dt} \]
Where \([A]\) is the concentration of the reactant. The units for concentration are typically moles per liter (mol/L), and time is in seconds (s). Therefore, the unit of rate is:
\[ \text{Units of Rate} = \text{mol L}^{-1} \text{s}^{-1} \]

### Step 2: Determine the Units of Rate Constant (k) for Different Orders
1. **Zero-Order Reaction:**
The rate law is given by:
\[ \text{Rate} = k \]
Here, \(k\) has the same unit as the rate:
\[ \text{Units of } k = \text{mol L}^{-1} \text{s}^{-1} \]

2. **First-Order Reaction:**
The rate law is:
\[ \text{Rate} = k[A] \]
Rearranging gives:
\[ k = \frac{\text{Rate}}{[A]} \]
Thus, the units of \(k\) are:
\[ \text{Units of } k = \frac{\text{mol L}^{-1} \text{s}^{-1}}{\text{mol L}^{-1}} = \text{s}^{-1} \]

3. **Second-Order Reaction:**
The rate law is:
\[ \text{Rate} = k[A]^2 \]
Rearranging gives:
\[ k = \frac{\text{Rate}}{[A]^2} \]
Thus, the units of \(k\) are:
\[ \text{Units of } k = \frac{\text{mol L}^{-1} \text{s}^{-1}}{(\text{mol L}^{-1})^2} = \text{L mol}^{-1} \text{s}^{-1} \]

4. **Half-Order Reaction:**
The rate law is:
\[ \text{Rate} = k[A]^{1/2} \]
Rearranging gives:
\[ k = \frac{\text{Rate}}{[A]^{1/2}} \]
Thus, the units of \(k\) are:
\[ \text{Units of } k = \frac{\text{mol L}^{-1} \text{s}^{-1}}{(\text{mol L}^{-1})^{1/2}} = \text{L}^{1/2} \text{mol}^{1/2} \text{s}^{-1} \]

### Step 3: Conclusion
From the analysis:
- For zero-order reactions, the unit of rate and the unit of rate constant \(k\) are both \(\text{mol L}^{-1} \text{s}^{-1}\).
- For first-order reactions, the unit of \(k\) is \(\text{s}^{-1}\).
- For second-order reactions, the unit of \(k\) is \(\text{L mol}^{-1} \text{s}^{-1}\).
- For half-order reactions, the unit of \(k\) is \(\text{L}^{1/2} \text{mol}^{1/2} \text{s}^{-1}\).

Thus, the order of reaction for which the unit of rate and the unit of rate constant are the same is **zero-order reaction**.

### Final Answer
**Zero-order reaction (Option A)**