Calculate the free energy change for the following reaction at 300 K.
`2CuO_((s)) rarr Cu_(2)O_((s))+(1)/(2)O_(2(g))` Given `Delta H = 145.6 kJ mol^(-1)` and `Delta S = 116.JK^(-1) mol^(-1)`

Options

A. `110.8"kJ mol"^(-1)`
B. `221.5"kJ mol"^(-1)`
C. `55.4"kJ mol"^(-1)`
D. `145.6"kJ mol"^(-1)`

Answer

To calculate the free energy change (ΔG) for the reaction at 300 K, we can use the Gibbs free energy equation:

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

### Step 1: Identify the given values
- ΔH = 145.6 kJ/mol
- ΔS = 116 J/K·mol
- T = 300 K

### Step 2: Convert ΔS from J/K·mol to kJ/K·mol
Since ΔH is given in kJ, we need to convert ΔS into the same unit (kJ):
\[ \Delta S = 116 \text{ J/K·mol} \times \frac{1 \text{ kJ}}{1000 \text{ J}} = 0.116 \text{ kJ/K·mol} \]

### Step 3: Substitute the values into the Gibbs free energy equation
Now we can substitute the values into the equation:
\[ \Delta G = 145.6 \text{ kJ/mol} - (300 \text{ K} \times 0.116 \text{ kJ/K·mol}) \]

### Step 4: Calculate TΔS
Calculate the product of temperature and entropy change:
\[ T \Delta S = 300 \text{ K} \times 0.116 \text{ kJ/K·mol} = 34.8 \text{ kJ/mol} \]

### Step 5: Calculate ΔG
Now substitute this value back into the equation for ΔG:
\[ \Delta G = 145.6 \text{ kJ/mol} - 34.8 \text{ kJ/mol} = 110.8 \text{ kJ/mol} \]

### Final Answer
Thus, the free energy change (ΔG) for the reaction at 300 K is:
\[ \Delta G = 110.8 \text{ kJ/mol} \]

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