Rate constant of a reaction (k) is `175 "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)`

Explanation

To determine the order of the reaction from the given rate constant \( k \) with units \( 175 \, \text{litre}^2 \, \text{mol}^{-2} \, \text{s}^{-1} \), we can follow these steps:

### Step 1: Understand the units of the rate constant
The rate of a reaction can be expressed as:
\[ \text{Rate} = k [A]^n \]
where \( [A] \) is the concentration of the reactant and \( n \) is the order of the reaction.

### Step 2: Identify the units of the rate
The unit of rate is typically expressed in terms of concentration per unit time. For example:
\[ \text{Rate} = \text{mol} \, \text{L}^{-1} \, \text{s}^{-1} \]

### Step 3: Write the expression for the units of \( k \)
From the rate law, we can express the units of \( k \) as:
\[ k = \frac{\text{Rate}}{[A]^n} \]
Substituting the units, we get:
\[ k = \frac{\text{mol} \, \text{L}^{-1} \, \text{s}^{-1}}{(\text{mol} \, \text{L}^{-1})^n} \]

### Step 4: Simplify the units of \( k \)
This can be simplified to:
\[ k = \frac{\text{mol} \, \text{L}^{-1} \, \text{s}^{-1}}{\text{mol}^n \, \text{L}^{-n}} = \text{mol}^{1-n} \, \text{L}^{-1-n} \, \text{s}^{-1} \]

### Step 5: Set the units of \( k \) equal to the given units
Now, we can equate this to the given units of \( k \):
\[ \text{mol}^{1-n} \, \text{L}^{-1-n} \, \text{s}^{-1} = \text{mol}^{-2} \, \text{L}^2 \, \text{s}^{-1} \]

### Step 6: Compare the powers of the units
From this equation, we can set up two equations by comparing the powers of \( \text{mol} \) and \( \text{L} \):
1. For \( \text{mol} \):
\[ 1 - n = -2 \implies n = 3 \]
2. For \( \text{L} \):
\[ -1 - n = 2 \implies n = -3 \quad (\text{This should be consistent with the previous result}) \]

### Conclusion
Thus, the order of the reaction is \( n = 3 \), indicating that it is a third-order reaction.