Rate constant of a reaction (k) is `2 xx 10^(-2) "litre"^(2) "mol"^(-2)sec^(-1)`. What is the order of reaction ?

Options

A. `1^(st)`
B. `2^(nd)`
C. `3^(rd)`
D. `4^(th)`

Answer

To determine the order of the reaction based on the given rate constant \( k \) with units \( 2 \times 10^{-2} \, \text{litre}^2 \, \text{mol}^{-2} \, \text{sec}^{-1} \), we can follow these steps:

### Step 1: Write the rate law expression
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 reactant A, and \( n \) is the order of the reaction.

### Step 2: Identify the units of rate
The unit of rate is given as:
\[ \text{Rate} = \text{mol} \, \text{litre}^{-1} \, \text{sec}^{-1} \]

### Step 3: Write the units of the rate constant \( k \)
The units of \( k \) are provided as:
\[ k = 2 \times 10^{-2} \, \text{litre}^2 \, \text{mol}^{-2} \, \text{sec}^{-1} \]

### Step 4: Relate the units of rate and \( k \)
From the rate law expression, we can express the units of \( k \) as:
\[ \text{Units of } k = \frac{\text{Units of Rate}}{[\text{Concentration}]^n} = \frac{\text{mol} \, \text{litre}^{-1} \, \text{sec}^{-1}}{(\text{mol} \, \text{litre}^{-1})^n} \]
This simplifies to:
\[ \text{Units of } k = \frac{\text{mol} \, \text{litre}^{-1} \, \text{sec}^{-1}}{\text{mol}^n \, \text{litre}^{-n}} = \text{mol}^{1-n} \, \text{litre}^{n-1} \, \text{sec}^{-1} \]

### Step 5: Set up the equation for units
Now we equate the units of \( k \):
\[ \text{mol}^{1-n} \, \text{litre}^{n-1} \, \text{sec}^{-1} = \text{mol}^{-2} \, \text{litre}^2 \, \text{sec}^{-1} \]

### Step 6: Compare the powers of units
From the equation, we can compare the powers of mol and litre:
1. For mol:
\[ 1 - n = -2 \implies n = 3 \]
2. For litre:
\[ n - 1 = 2 \implies n = 3 \]

### Conclusion
Both comparisons give us \( n = 3 \). Therefore, the order of the reaction is **third order**.