In the elementary reaction `A + B to AB ` , if the concentration A and B is doubled , the rate of reaction will

Options

A. Be doubled
B. Be halved
C. Increase bu 6 times
D. Incrase by 4

Explanation

To solve the problem, we need to analyze the elementary reaction given:

**Reaction:** A + B → AB

### Step 1: Write the Rate Law Expression
For an elementary reaction, the rate law can be directly derived from the stoichiometry of the reaction. The rate of the reaction (R) is given by:

\[ R = k [A]^m [B]^n \]

Where:
- \( k \) is the rate constant,
- \( [A] \) and \( [B] \) are the concentrations of reactants A and B,
- \( m \) and \( n \) are the stoichiometric coefficients of A and B, respectively.

In this case, since the coefficients of A and B are both 1, we have:

\[ R = k [A]^1 [B]^1 \]
\[ R = k [A][B] \]

### Step 2: Determine the Initial Rate
Let’s denote the initial concentrations of A and B as \( [A] \) and \( [B] \). The initial rate (R1) of the reaction can be expressed as:

\[ R_1 = k [A][B] \]

### Step 3: Doubling the Concentrations
Now, if the concentrations of A and B are doubled, we have:

\[ [A]_{new} = 2[A] \]
\[ [B]_{new} = 2[B] \]

### Step 4: Calculate the New Rate
The new rate (R2) with the doubled concentrations will be:

\[ R_2 = k [A_{new}][B_{new}] \]
\[ R_2 = k (2[A])(2[B]) \]
\[ R_2 = k \cdot 4[A][B] \]
\[ R_2 = 4(k [A][B]) \]
\[ R_2 = 4R_1 \]

### Conclusion
Thus, when the concentrations of A and B are doubled, the rate of the reaction increases by a factor of 4.

**Final Answer:** The rate of reaction will increase by 4 times.

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