A gaseous reactant A forms two different product, in parallel reaction , B and C as follows:
`A rarr B, DeltaH^(@) =-3KJ, DeltaS^(@) = 20JK^(-1)" , " "A rarr C, DeltaH^(@) =-3.6KJ, DeltaS^(@) = 10 JK^(-1)`
Discuss the relative stability of B and C on the basis of Gibb's free energy change at `27^(@)C.`

Explanation

To determine the relative stability of products B and C formed from the gaseous reactant A based on Gibbs free energy change at 27°C, we will follow these steps:

### Step 1: Convert Temperature to Kelvin
First, we need to convert the temperature from Celsius to Kelvin. The formula for conversion is:

\[ T(K) = T(°C) + 273.15 \]

For 27°C:

\[ T = 27 + 273.15 = 300.15 \approx 300 \text{ K} \]

### Step 2: Calculate Gibbs Free Energy Change (ΔG) for A to B
Using the formula for Gibbs free energy change:

\[ \Delta G = \Delta H - T \Delta S \]

For the reaction A to B:
- ΔH = -3 kJ = -3000 J (since 1 kJ = 1000 J)
- ΔS = 20 J/K

Substituting the values:

\[ \Delta G_{A \to B} = -3000 J - (300 \text{ K} \times 20 \text{ J/K}) \]

Calculating:

\[ \Delta G_{A \to B} = -3000 J - 6000 J = -9000 J \]

### Step 3: Calculate Gibbs Free Energy Change (ΔG) for A to C
Now, we calculate ΔG for the reaction A to C using the same formula.

For the reaction A to C:
- ΔH = -3.6 kJ = -3600 J
- ΔS = 10 J/K

Substituting the values:

\[ \Delta G_{A \to C} = -3600 J - (300 \text{ K} \times 10 \text{ J/K}) \]

Calculating:

\[ \Delta G_{A \to C} = -3600 J - 3000 J = -6600 J \]

### Step 4: Compare the Gibbs Free Energy Changes
Now we compare the ΔG values for both reactions:

- ΔG for A to B = -9000 J
- ΔG for A to C = -6600 J

### Conclusion
Since ΔG for A to B is more negative than ΔG for A to C, product B is more stable than product C. A more negative ΔG indicates a more spontaneous reaction and greater stability of the product.