For the reaction, `CaCO_(3)(s) hArr CaO(s) + CO_(2)(g)` partial pressure of `CO_(2)` at 1000 K is 0.003 atm. `Delta G^(@)`= 27.2 kcal. Calculate the value of `Delta G`

Options

A. 12.6k cal
B. 15.6 k cal
C. 13.4 kcal
D. 14.2 kcal

Explanation

To calculate the value of ΔG for the reaction:

\[ \text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g) \]

given the partial pressure of CO₂ at 1000 K is 0.003 atm, and ΔG° = 27.2 kcal, we will use the following equation:

\[ \Delta G = \Delta G^\circ + RT \ln Q \]

where:
- \( R \) is the universal gas constant,
- \( T \) is the temperature in Kelvin,
- \( Q \) is the reaction quotient.

### Step-by-Step Solution:

1. **Identify the Reaction Quotient (Q)**:
For the reaction, the reaction quotient \( Q \) is given by the partial pressure of the gaseous product (CO₂) since solids do not appear in the expression for \( Q \):
\[ Q = P_{\text{CO}_2} = 0.003 \, \text{atm} \]

2. **Convert ΔG° to the same units as R**:
The standard Gibbs free energy change \( \Delta G^\circ \) is given as 27.2 kcal. We need to convert this to calories since the gas constant \( R \) in this case will be used in calories per mole per Kelvin:
\[ \Delta G^\circ = 27.2 \, \text{kcal} \times 1000 \, \text{cal/kcal} = 27200 \, \text{cal} \]

3. **Use the appropriate value for R**:
The value of \( R \) in calories is:
\[ R = 1.987 \, \text{cal/(mol K)} \]

4. **Calculate ΔG using the equation**:
Substitute the values into the equation:
\[ \Delta G = \Delta G^\circ + RT \ln Q \]
First, calculate \( RT \):
\[ RT = 1.987 \, \text{cal/(mol K)} \times 1000 \, \text{K} = 1987 \, \text{cal/mol} \]

Now calculate \( \ln Q \):
\[ \ln(0.003) \approx -5.809 \]

Now substitute these values into the ΔG equation:
\[ \Delta G = 27200 \, \text{cal} + 1987 \, \text{cal/mol} \times (-5.809) \]
\[ \Delta G = 27200 \, \text{cal} - 11555.5 \, \text{cal} \]
\[ \Delta G \approx 15644.5 \, \text{cal} \]

5. **Convert ΔG back to kcal**:
\[ \Delta G \approx \frac{15644.5 \, \text{cal}}{1000} \approx 15.6445 \, \text{kcal} \]

### Final Answer:
\[ \Delta G \approx 15.6 \, \text{kcal} \]