Concentration dependence of rate is called differential rate equation. Integrated differential equations give relation between directly measured experimental data i.e., concentration at different times and rate constant. The integrated rate equations are different for the reactions of differennt reaction orders. the first-order reaction has a rate constant `1.15xx10^(-3)s^(-1)`.
Q. When the rate constant has same units as the rate of reaction, the order of the reaction is:

Options

A. Zero order
B. First order
C. Second order
D. Fractional order

Explanation

To determine the order of the reaction when the rate constant has the same units as the rate of reaction, we can follow these steps:

### Step 1: Understand the Units of Rate
The rate of a reaction is defined as the change in concentration of reactants or products over time. The unit of rate can be expressed as:
\[ \text{Rate} = -\frac{d[R]}{dt} \]
This means the unit of rate is typically in terms of concentration per unit time, which is:
\[ \text{Unit of Rate} = \text{mol L}^{-1} \text{s}^{-1} \]

### Step 2: Write the General Rate Equation
For a reaction of order \( n \), the rate can be expressed as:
\[ \text{Rate} = k [A]^n \]
where \( k \) is the rate constant and \([A]\) is the concentration of the reactant.

### Step 3: Determine the Units of the Rate Constant
From the rate equation, we can rearrange it to find the units of the rate constant \( k \):
\[ k = \frac{\text{Rate}}{[A]^n} \]
Substituting the units we have:
\[ k = \frac{\text{mol L}^{-1} \text{s}^{-1}}{(\text{mol L}^{-1})^n} = \frac{\text{mol L}^{-1} \text{s}^{-1}}{\text{mol}^n \text{L}^{-n}} \]
This simplifies to:
\[ k = \text{mol}^{1-n} \text{L}^{n-1} \text{s}^{-1} \]

### Step 4: Set the Units of Rate Constant Equal to the Units of Rate
According to the question, the rate constant \( k \) has the same units as the rate of reaction. Therefore, we set:
\[ \text{mol}^{1-n} \text{L}^{n-1} \text{s}^{-1} = \text{mol L}^{-1} \text{s}^{-1} \]

### Step 5: Equate the Exponents
From the equation above, we can equate the exponents of mol and L:
1. For mol:
\[ 1 - n = 1 \implies n = 0 \]
2. For L:
\[ n - 1 = -1 \implies n = 0 \]

### Step 6: Conclusion
Since we found \( n = 0 \), this indicates that the reaction is a **zero-order reaction**.

### Final Answer
The order of the reaction is **zero order**.

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