Molar heat capacities at constant pressure for A, B and C are 3, 1.5 and 2 J/K mol. The enthalpy of reaction and entropy of reaction, A +2 B `rarr` 3C are 20 kJ/mol and 20 J/K mol at 300 K. Calculate `DeltaG` (in kJ / mol) for the reaction,
`(1)/(2)A+Brarr(3)/(2)C`

Answer

To calculate the Gibbs free energy change (ΔG) for the reaction \( \frac{1}{2}A + B \rightleftharpoons \frac{3}{2}C \), we will follow these steps:

### Step 1: Write the given reaction and its parameters
The original reaction is:
\[ A + 2B \rightleftharpoons 3C \]
We are given:
- Enthalpy change (ΔH) = 20 kJ/mol
- Entropy change (ΔS) = 20 J/K·mol
- Temperature (T) = 300 K

### Step 2: Adjust the reaction to match the desired reaction
We need to find ΔG for the reaction:
\[ \frac{1}{2}A + B \rightleftharpoons \frac{3}{2}C \]

To do this, we can divide the original reaction by 2:
\[ \frac{1}{2}A + B \rightleftharpoons \frac{3}{2}C \]

### Step 3: Adjust the values of ΔH and ΔS
When we divide the reaction by 2, we also need to divide ΔH and ΔS by 2:
- New ΔH = \( \frac{20 \text{ kJ/mol}}{2} = 10 \text{ kJ/mol} \)
- New ΔS = \( \frac{20 \text{ J/K·mol}}{2} = 10 \text{ J/K·mol} \)

### Step 4: Convert ΔS to kJ
Since we want ΔG in kJ, we need to convert ΔS from J to kJ:
\[ \Delta S = 10 \text{ J/K·mol} = \frac{10}{1000} \text{ kJ/K·mol} = 0.01 \text{ kJ/K·mol} \]

### Step 5: Calculate ΔG using the Gibbs free energy equation
The Gibbs free energy change is given by the equation:
\[ \Delta G = \Delta H - T \Delta S \]

Substituting the values we have:
\[ \Delta G = 10 \text{ kJ/mol} - (300 \text{ K} \times 0.01 \text{ kJ/K·mol}) \]
\[ \Delta G = 10 \text{ kJ/mol} - 3 \text{ kJ/mol} \]
\[ \Delta G = 7 \text{ kJ/mol} \]

### Final Answer
Thus, the value of ΔG for the reaction \( \frac{1}{2}A + B \rightleftharpoons \frac{3}{2}C \) is:
\[ \Delta G = 7 \text{ kJ/mol} \]

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