The rate constant of reaction `2 A + B rarr C` is `2.57 xx 10^(-5) L "mole"^(-1) sec^(-1)` after `10 sec`, `2.65 xx 10^(-5) L "mole"^(-1)sec^(-1)` after `20sec` and `2.55 xx 10^(-5) L "mole"^(-1) sec^(-1)` after `30 sec`. The order of the reaction is :

Options

A. `0`
B. `1`
C. `2`
D. `3`

Explanation

To determine the order of the reaction \(2 A + B \rightarrow C\) based on the given rate constants, we can follow these steps:

### Step 1: Identify the Units of the Rate Constant
The rate constant \(k\) is given with units of \(L \, mole^{-1} \, sec^{-1}\).

### Step 2: Relate the Units to Reaction Order
The units of the rate constant \(k\) can help us determine the order of the reaction. The general form for the rate of a reaction is given by:
\[ R = k \cdot [A]^n \]
where \(R\) is the rate of the reaction, \([A]\) is the concentration of reactant \(A\), and \(n\) is the order of the reaction.

### Step 3: Write the Units for the Rate
The units for the rate \(R\) of the reaction are typically expressed as:
\[ R = \frac{mole}{L \cdot sec} \]

### Step 4: Set Up the Equation for Units
From the equation \(R = k \cdot [A]^n\), we can express the units of \(k\) as:
\[ k = \frac{R}{[A]^n} \]
Substituting the units we have:
\[ \text{Units of } k = \frac{\frac{mole}{L \cdot sec}}{(mole/L)^n} \]

### Step 5: Simplify the Units
This simplifies to:
\[ \text{Units of } k = \frac{mole}{L \cdot sec} \cdot \frac{L^n}{mole^n} \]
\[ = \frac{L^{n-1} \cdot mole^{1-n}}{sec} \]

### Step 6: Equate the Units
Now we equate this to the given units of \(k\) which are \(L \, mole^{-1} \, sec^{-1}\):
\[ L^{n-1} \cdot mole^{1-n} = L^1 \cdot mole^{-1} \]

### Step 7: Set Up the Equations
From this, we can set up two equations based on the powers of \(L\) and \(mole\):
1. For \(L\): \(n - 1 = 1\) → \(n = 2\)
2. For \(mole\): \(1 - n = -1\) → \(n = 2\)

### Step 8: Conclusion
Both equations give us the same result, \(n = 2\). Therefore, the order of the reaction is:
\[ \text{Order of the reaction} = 2 \]