If the rate of formation of C increases by a factor of 2.82 on doubling the concentration of A and increases by a factor of 9 on tripling the concentration of B,then what is the order of a chemical reaction `A to 2B to C`?

Options

A. `7//2`
B. `7//4`
C. `5//2`
D. `5//4`

Answer

To determine the order of the chemical reaction \( A \to 2B \to C \) based on the given information, we can follow these steps:

### Step 1: Write the Rate Law Expression
The rate of formation of C can be expressed as:
\[ \text{Rate} = k [A]^m [B]^n \]
where \( k \) is the rate constant, \( m \) is the order with respect to A, and \( n \) is the order with respect to B.

### Step 2: Analyze the Effect of Doubling the Concentration of A
When the concentration of A is doubled, the rate of formation of C increases by a factor of 2.82. This can be expressed mathematically as:
\[ \frac{\text{Rate}_2}{\text{Rate}_1} = \frac{k [2A]^m [B]^n}{k [A]^m [B]^n} = \frac{2^m [A]^m [B]^n}{[A]^m [B]^n} = 2^m \]
Setting this equal to the observed increase:
\[ 2^m = 2.82 \]
Taking the logarithm on both sides:
\[ m \log(2) = \log(2.82) \]
Calculating \( m \):
\[ m = \frac{\log(2.82)}{\log(2)} \approx 1.5 \]

### Step 3: Analyze the Effect of Tripling the Concentration of B
When the concentration of B is tripled, the rate of formation of C increases by a factor of 9. This can be expressed as:
\[ \frac{\text{Rate}_3}{\text{Rate}_1} = \frac{k [A]^m [3B]^n}{k [A]^m [B]^n} = \frac{[A]^m (3^n [B]^n)}{[A]^m [B]^n} = 3^n \]
Setting this equal to the observed increase:
\[ 3^n = 9 \]
Since \( 9 = 3^2 \), we can conclude:
\[ n = 2 \]

### Step 4: Calculate the Overall Order of the Reaction
The overall order of the reaction is the sum of the individual orders:
\[ \text{Overall Order} = m + n = 1.5 + 2 = 3.5 \]

### Final Answer
The order of the chemical reaction \( A \to 2B \to C \) is \( 3.5 \).

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