The rate of reaction, `2NO + Cl_2 to 2NOCI` is doubled when concentration of `Cl_2` is doubled and it becomes 8 times when concentrations of both NO and `Cl_2` are doubled. Deduce the order of this reaction.

Explanation

To determine the order of the reaction \(2NO + Cl_2 \rightarrow 2NOCI\), we will analyze the given information step by step.

### Step 1: Write the Rate Law Expression
The rate law for the reaction can be expressed as:
\[ \text{Rate} = k [NO]^x [Cl_2]^y \]
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 First Condition
We are given that when the concentration of \(Cl_2\) is doubled, the rate of reaction is also doubled. This can be expressed mathematically as:
\[ \text{Rate}' = k [NO]^x [2Cl_2]^y = 2 \cdot k [NO]^x [Cl_2]^y \]
Substituting \(2Cl_2\) into the rate expression gives:
\[ k [NO]^x [2Cl_2]^y = k [NO]^x [Cl_2]^y \cdot 2^y \]
Since the rate doubles, we can equate:
\[ 2^y = 2 \]
This implies that:
\[ y = 1 \]

### Step 3: Analyze the Second Condition
Next, we are told that when the concentrations of both \(NO\) and \(Cl_2\) are doubled, the rate becomes 8 times the original rate. This can be expressed as:
\[ \text{Rate}'' = k [2NO]^x [2Cl_2]^y = 8 \cdot k [NO]^x [Cl_2]^y \]
Substituting \(2NO\) and \(2Cl_2\) into the rate expression gives:
\[ k [2NO]^x [2Cl_2]^y = k [NO]^x [Cl_2]^y \cdot 2^x \cdot 2^y \]
This simplifies to:
\[ k [NO]^x [Cl_2]^y \cdot 2^{x+y} \]
Setting this equal to \(8 \cdot k [NO]^x [Cl_2]^y\) gives:
\[ 2^{x+y} = 8 \]
Since \(8 = 2^3\), we have:
\[ x + y = 3 \]

### Step 4: Solve for the Order of the Reaction
We already found that \(y = 1\). Substituting this into the equation \(x + y = 3\):
\[ x + 1 = 3 \implies x = 2 \]

### Step 5: Determine the Overall Order
The overall order of the reaction is the sum of the orders with respect to each reactant:
\[ \text{Overall Order} = x + y = 2 + 1 = 3 \]

### Final Answer
The order of the reaction is **3**.

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