In the reaction `A+B hArr AB`, if the concentration of A and B is increased by a factor of `2`, it will cause the equilibrium concentration of AB to change to

Options

A. Two times to original value
B. Three times to original value
C. Same
D. Zero

Explanation

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

\[ A + B \rightleftharpoons AB \]

### Step-by-Step Solution:

1. **Define Initial Conditions**:
Let's assume we start with 1 mole of A and 1 mole of B, and initially, there are 0 moles of AB.
- At \( t = 0 \):
- \([A] = 1\)
- \([B] = 1\)
- \([AB] = 0\)

2. **Define Change at Equilibrium**:
Let \( x \) be the amount of A and B that reacts to form AB at equilibrium. Therefore, at equilibrium:
- \([A] = 1 - x\)
- \([B] = 1 - x\)
- \([AB] = x\)

3. **Establish New Conditions**:
If the concentrations of A and B are increased by a factor of 2, we have:
- New concentration of A = \( 2 \times 1 = 2 \)
- New concentration of B = \( 2 \times 1 = 2 \)

The new reaction can be represented as:
\[ 2A + 2B \rightleftharpoons 2AB \]

4. **Define New Equilibrium Changes**:
Let \( y \) be the amount of A and B that reacts to form AB in this new scenario:
- At \( t = 0 \) for the new conditions:
- \([A] = 2\)
- \([B] = 2\)
- \([AB] = 0\)

At equilibrium:
- \([A] = 2 - y\)
- \([B] = 2 - y\)
- \([AB] = 2y\)

5. **Relate New Equilibrium to Original**:
From the original reaction, we established that at equilibrium, \( x \) moles of AB were formed. In the new scenario, since the concentrations of A and B are doubled, the amount of AB formed will also double. Therefore:
\[ 2y = 2x \]
This indicates that the equilibrium concentration of AB has increased by a factor of 2.

6. **Conclusion**:
Thus, the equilibrium concentration of AB changes to 2 times its original value.

### Final Answer:
The equilibrium concentration of AB will change to **2 times its original value**.

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