The free energy for a reaction having `DeltaH=31400` cal, `DeltaS=32" cal "K^(-1)mol^(-1)` at `1000^(@)C` is:

Options

A. `-9336` cal
B. `-7386` cal
C. `-1936` cal
D. `+9336` cal

Explanation

To find the Gibbs free energy change (ΔG) for the reaction, we can use the formula:

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

### Step 1: Identify the values given in the problem.
- ΔH (enthalpy change) = 31400 cal
- ΔS (entropy change) = 32 cal K⁻¹ mol⁻¹
- Temperature (T) = 1000°C

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

\[ T(K) = T(°C) + 273 \]

So,

\[ T = 1000 + 273 = 1273 \text{ K} \]

### Step 3: Substitute the values into the Gibbs free energy formula.
Now we substitute the values into the formula:

\[ \Delta G = 31400 \text{ cal} - (1273 \text{ K} \times 32 \text{ cal K}^{-1} \text{ mol}^{-1}) \]

### Step 4: Calculate the product of T and ΔS.
Now calculate \( T \Delta S \):

\[ T \Delta S = 1273 \text{ K} \times 32 \text{ cal K}^{-1} \text{ mol}^{-1} = 40736 \text{ cal} \]

### Step 5: Calculate ΔG.
Now substitute this back into the ΔG equation:

\[ \Delta G = 31400 \text{ cal} - 40736 \text{ cal} \]

\[ \Delta G = -9336 \text{ cal} \]

### Step 6: State the final answer.
Therefore, the Gibbs free energy change (ΔG) for the reaction at 1000°C is:

\[ \Delta G = -9336 \text{ cal} \]