If `DeltaG` for the reaction is `A^(+) + B^(-) rarr A^(2+) + B^(2-) ` is x, the `DeltaG` for the reaction is
`(1)/(2)A^(+) + (1)/(2)B^(-) rarr (1)/(2) A^(2+) + (1)/(2) B^(2-)` is

Options

A. `(x)/(2)`
B. 2x
C. `x^(2)`
D. `sqrtx`

Answer

To solve the problem, we need to find the Gibbs free energy change (ΔG) for the reaction:

\[ \frac{1}{2} A^{+} + \frac{1}{2} B^{-} \rightarrow \frac{1}{2} A^{2+} + \frac{1}{2} B^{2-} \]

given that ΔG for the reaction:

\[ A^{+} + B^{-} \rightarrow A^{2+} + B^{2-} \]

is \( x \).

### Step-by-Step Solution:

1. **Understanding the Relationship**:
The given reaction has ΔG = \( x \). This means that for every mole of \( A^{+} \) and \( B^{-} \) reacting to form \( A^{2+} \) and \( B^{2-} \), the change in Gibbs free energy is \( x \).

2. **Scaling the Reaction**:
The reaction we want to analyze is half of the original reaction. We can express this mathematically by multiplying the entire reaction by \( \frac{1}{2} \):
\[ \frac{1}{2} A^{+} + \frac{1}{2} B^{-} \rightarrow \frac{1}{2} A^{2+} + \frac{1}{2} B^{2-} \]

3. **Effect on ΔG**:
Since ΔG is an extensive property, it scales with the amount of substance involved in the reaction. Therefore, if we multiply the coefficients of the reaction by \( \frac{1}{2} \), we must also multiply ΔG by \( \frac{1}{2} \):
\[ \Delta G = \frac{1}{2} \times x = \frac{x}{2} \]

4. **Conclusion**:
Thus, the ΔG for the reaction \( \frac{1}{2} A^{+} + \frac{1}{2} B^{-} \rightarrow \frac{1}{2} A^{2+} + \frac{1}{2} B^{2-} \) is \( \frac{x}{2} \).

### Final Answer:
\[ \Delta G = \frac{x}{2} \]