For the reaction at 298K, `A_((g))+B_((g))rarrC_((g))`
`DeltaE=-5 cal and DeltaS =-10cal K^(-1)`

Options

A. `DeltaG=+2612 cal`
B. `DeltaG=-2612 cal`
C. `DeltaG=+261.2 cal`
D. `DeltaG=2379 cal`

Explanation

To solve the problem, we need to find the Gibbs free energy change (ΔG) for the reaction given the values of ΔE and ΔS at 298 K. We will also calculate ΔH using the relationship between ΔE, ΔH, and the change in the number of moles of gas (Δn).

### Step 1: Calculate Δn
For the reaction:
\[ A_{(g)} + B_{(g)} \rightarrow C_{(g)} \]

- Reactants: 2 moles (1 mole of A and 1 mole of B)
- Products: 1 mole (1 mole of C)

Thus, the change in the number of moles of gas (Δn) is:
\[ \Delta n = \text{moles of products} - \text{moles of reactants} = 1 - 2 = -1 \]

### Step 2: Calculate ΔH using ΔE
The relationship between ΔE and ΔH is given by:
\[ \Delta H = \Delta E + \Delta n \cdot R \cdot T \]
Where:
- R (gas constant) = 2 cal K\(^{-1}\) mol\(^{-1}\)
- T = 298 K

Substituting the values:
\[ \Delta H = -5 \text{ cal} + (-1) \cdot (2 \text{ cal K}^{-1} \text{ mol}^{-1}) \cdot (298 \text{ K}) \]
\[ \Delta H = -5 \text{ cal} - 596 \text{ cal} = -601 \text{ cal} \]

### Step 3: Calculate ΔG using ΔH and ΔS
The Gibbs free energy change (ΔG) is calculated using the formula:
\[ \Delta G = \Delta H - T \cdot \Delta S \]
Substituting the values:
\[ \Delta G = -601 \text{ cal} - (298 \text{ K} \cdot -10 \text{ cal K}^{-1}) \]
\[ \Delta G = -601 \text{ cal} + 2980 \text{ cal} = 2379 \text{ cal} \]

### Final Answer
Thus, the Gibbs free energy change (ΔG) for the reaction at 298 K is:
\[ \Delta G = 2379 \text{ cal} \]