Rate constant of two reactions are given below. Indentifying their order of reaction.
(i) `k = 6.3 xx 10^(-2) L mol^(-1) s^(-1)`
(ii) `k = 2.8 xx 10^(-4) s^(-1)`

Options

A. (i) second order, (ii) first order
B. (i) first order, (ii) second order
C. (i) zero order, (ii) first order
D. (i) second order, (ii) zero order

Answer

To determine the order of the reactions based on the given rate constants and their units, we can follow these steps:

### Step 1: Analyze the units of the rate constants

1. **For the first reaction**: The rate constant is given as \( k = 6.3 \times 10^{-2} \, \text{L mol}^{-1} \text{s}^{-1} \).
- The unit \( \text{L mol}^{-1} \text{s}^{-1} \) indicates that the reaction is dependent on the concentration of reactants in terms of molarity (mol/L).
- The general form for the rate law is:
\[ \text{Rate} = k [A]^n \]
- Here, the unit of rate is \( \text{mol L}^{-1} \text{s}^{-1} \).

### Step 2: Relate the units to the order of reaction

2. **For the first reaction**:
- The unit of the rate constant can be expressed as:
\[ k = \frac{\text{Rate}}{[\text{A}]^n} \]
- Rearranging gives:
\[ k = \frac{\text{mol L}^{-1} \text{s}^{-1}}{(\text{mol L}^{-1})^n} = \text{mol L}^{-1} \text{s}^{-1} \cdot \text{L}^{n} \cdot \text{mol}^{-n} \]
- This simplifies to:
\[ k = \text{L}^{1-n} \cdot \text{mol}^{-n} \cdot \text{s}^{-1} \]
- Setting the powers of L and mol equal to the units of \( k \):
- For \( n = 2 \): \( k \) has units of \( \text{L}^{-1} \text{mol}^{-1} \text{s}^{-1} \) (second order).
- For \( n = 1 \): \( k \) has units of \( \text{L}^{0} \text{mol}^{-1} \text{s}^{-1} \) (first order).
- Since \( k = 6.3 \times 10^{-2} \, \text{L mol}^{-1} \text{s}^{-1} \), it indicates a **second-order reaction** (since \( n = 2 \)).

### Step 3: Analyze the second reaction

3. **For the second reaction**: The rate constant is given as \( k = 2.8 \times 10^{-4} \, \text{s}^{-1} \).
- The unit \( \text{s}^{-1} \) indicates that the reaction rate does not depend on the concentration of reactants.
- This corresponds to a zero-order reaction, where the rate is constant and independent of the concentration.

### Conclusion

- The first reaction (with \( k = 6.3 \times 10^{-2} \, \text{L mol}^{-1} \text{s}^{-1} \)) is a **second-order reaction**.
- The second reaction (with \( k = 2.8 \times 10^{-4} \, \text{s}^{-1} \)) is a **first-order reaction**.