For a reaction `2NO(g) +Cl_(2)(g) rarr 2NOCl(g)`, when concentration of `Cl_(2)` is doubled, the rate becomes four times. What is the order of reaction with respect to `Cl_(2)` ?

Options

A. 1
B. 2
C. 3
D. 4

Explanation

To determine the order of the reaction with respect to \( Cl_2 \) for the reaction:

\[ 2NO(g) + Cl_2(g) \rightarrow 2NOCl(g) \]

we can follow these steps:

### Step 1: Write the rate law expression
The rate of a reaction can be expressed in terms of the concentrations of the reactants raised to their respective orders:

\[ \text{Rate} = k [NO]^x [Cl_2]^y \]

where \( k \) is the rate constant, \( x \) is the order with respect to \( NO \), and \( y \) is the order with respect to \( Cl_2 \).

### Step 2: Analyze the change in concentration
According to the problem, when the concentration of \( Cl_2 \) is doubled, the rate becomes four times. We can express this mathematically:

Let the initial concentration of \( Cl_2 \) be \( [Cl_2] \) and the initial rate be \( R_1 \):

\[ R_1 = k [NO]^x [Cl_2]^y \]

When the concentration of \( Cl_2 \) is doubled, the new concentration becomes \( 2[Cl_2] \) and the new rate \( R_2 \) is:

\[ R_2 = k [NO]^x (2[Cl_2])^y \]

### Step 3: Set up the equation based on the rate change
According to the problem, \( R_2 = 4R_1 \):

\[ 4R_1 = k [NO]^x (2[Cl_2])^y \]

Substituting \( R_1 \):

\[ 4(k [NO]^x [Cl_2]^y) = k [NO]^x (2[Cl_2])^y \]

### Step 4: Simplify the equation
We can cancel \( k [NO]^x \) from both sides (assuming \( k \) and \( [NO] \) are not zero):

\[ 4 [Cl_2]^y = (2[Cl_2])^y \]

### Step 5: Expand the right side
The right side can be expanded as follows:

\[ (2[Cl_2])^y = 2^y [Cl_2]^y \]

So, we have:

\[ 4 [Cl_2]^y = 2^y [Cl_2]^y \]

### Step 6: Divide both sides by \( [Cl_2]^y \)
Assuming \( [Cl_2] \) is not zero, we can divide both sides by \( [Cl_2]^y \):

\[ 4 = 2^y \]

### Step 7: Solve for \( y \)
Now, we can solve for \( y \):

\[ 2^2 = 2^y \implies y = 2 \]

### Conclusion
The order of the reaction with respect to \( Cl_2 \) is \( 2 \).

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