Conisder the following elementary reaction,
`2A + B + C rarr` Products. All reactants are present in the gaseous state and reactant `C` is taken in excess.
How will the rate change if the concentration of A is doubled and that of B is tripled ?

Options

A. The rate increases `12` times of original value.
B. The rate increases `8` times of original value.
C. The rate reduces `12` times of original value.
D. The rate reduces `8` times of original value.

Explanation

To solve the problem, we need to analyze the given elementary reaction and how the changes in concentrations of reactants A and B affect the rate of the reaction.

### Step-by-Step Solution:

1. **Identify the Reaction and Rate Expression**:
The given reaction is:
\[ 2A + B + C \rightarrow \text{Products} \]
Since it is an elementary reaction, the rate of reaction can be expressed as:
\[ \text{Rate} = k [A]^2 [B]^1 \]
where \( k \) is the rate constant, \([A]\) is the concentration of A, and \([B]\) is the concentration of B. The concentration of C is in excess and does not affect the rate.

2. **Initial Rate Calculation**:
Let the initial concentrations of A and B be \([A]\) and \([B]\), respectively. The initial rate can be expressed as:
\[ \text{Rate}_0 = k [A]^2 [B] \]

3. **Change in Concentrations**:
According to the problem, the concentration of A is doubled and the concentration of B is tripled:
- New concentration of A: \( [A]' = 2[A] \)
- New concentration of B: \( [B]' = 3[B] \)

4. **New Rate Calculation**:
Substitute the new concentrations into the rate expression:
\[ \text{Rate}' = k [A']^2 [B'] = k (2[A])^2 (3[B]) \]
Simplifying this gives:
\[ \text{Rate}' = k (4[A]^2)(3[B]) = 12 k [A]^2 [B] \]

5. **Comparison of Rates**:
Now, we can compare the new rate to the initial rate:
\[ \text{Rate}' = 12 \times \text{Rate}_0 \]
This indicates that the rate of the reaction increases by a factor of 12.

### Final Answer:
The rate of the reaction increases 12 times the original value.

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