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

Options

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

Explanation

To calculate the value of ΔG for the reaction \( \text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g) \) at a temperature of 1000 K, we can use the following equation:

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

### Step 1: Identify the values needed for the calculation
- \( \Delta G^\circ = 27.2 \) kcal
- Temperature (T) = 1000 K
- Partial pressure of \( \text{CO}_2 \) = 0.003 atm
- R (universal gas constant) = 1.987 cal/(mol·K) = 0.001987 kcal/(mol·K)

### Step 2: Calculate the reaction quotient (Q)
For the reaction, the reaction quotient \( Q \) is given by the ratio of the partial pressures of the products to the reactants. Since \( \text{CaO} \) is a solid, it does not appear in the expression for \( Q \):

\[ Q = \frac{P_{\text{CO}_2}}{1} = P_{\text{CO}_2} = 0.003 \text{ atm} \]

### Step 3: Calculate \( RT \ln Q \)
Now, we can calculate \( RT \ln Q \):

1. Calculate \( RT \):
\[ RT = (0.001987 \text{ kcal/(mol·K)}) \times (1000 \text{ K}) = 1.987 \text{ kcal/mol} \]

2. Calculate \( \ln Q \):
\[ \ln Q = \ln(0.003) \approx -5.809 \]

3. Now, calculate \( RT \ln Q \):
\[ RT \ln Q = 1.987 \text{ kcal/mol} \times (-5.809) \approx -11.53 \text{ kcal/mol} \]

### Step 4: Calculate ΔG
Now we can substitute the values into the equation for ΔG:

\[ \Delta G = \Delta G^\circ + RT \ln Q \]
\[ \Delta G = 27.2 \text{ kcal} - 11.53 \text{ kcal} \approx 15.67 \text{ kcal} \]

### Final Answer
\[ \Delta G \approx 15.67 \text{ kcal} \]

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