The change in Gibbs free energy of the system along provides a criterion for the spontaneity of a process at constant temperature and pressure. A change in the free energy of a sytem at constant temperature and pressure will be: `DeltaG_("system") = DeltaH_("system") - T DeltaS_("system")`
The free energy for a reaction having `Delta H= 31400` cal, `DeltaS= 32` cal `K^(-1) mol^(-1)` at `1000^(@)C` is

Options

A. a. `-9336` cal
B. b. `-7386` cal
C. c `-1936` cal
D. d. `+9336` cal

Answer

To solve the problem, we need to calculate the change in Gibbs free energy (ΔG) using the formula:

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

### Step 1: Identify the given values
We are given:
- ΔH = 31,400 cal
- ΔS = 32 cal K⁻¹ mol⁻¹
- Temperature (T) = 1000 °C

### Step 2: Convert temperature from Celsius to Kelvin
To convert the temperature from Celsius to Kelvin, we use the formula:

\[ T(K) = T(°C) + 273.15 \]

Substituting the given temperature:

\[ T = 1000 + 273.15 = 1273.15 \, K \]

For simplicity, we can round this to:

\[ T \approx 1273 \, K \]

### Step 3: Calculate TΔS
Now we calculate the product of temperature and entropy change (TΔS):

\[ T \Delta S = 1273 \, K \times 32 \, \text{cal K}^{-1} \text{mol}^{-1} \]

Calculating this gives:

\[ T \Delta S = 1273 \times 32 = 40,736 \, \text{cal} \]

### Step 4: Substitute values into the Gibbs free energy equation
Now we can substitute ΔH and TΔS into the Gibbs free energy equation:

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

Substituting the values we have:

\[ \Delta G = 31,400 \, \text{cal} - 40,736 \, \text{cal} \]

### Step 5: Perform the calculation
Now we perform the subtraction:

\[ \Delta G = 31,400 - 40,736 = -9,336 \, \text{cal} \]

### Final Answer
Thus, the change in Gibbs free energy (ΔG) for the reaction is:

\[ \Delta G = -9,336 \, \text{cal} \]