For a reaction `2X(s)+2Y(s)rarr 2Cl(l)+D(g)`
The `q_(p)` at `27^(@)C` is -28 K Cal. `mol^(-1)`.
The `q_(V)` is ___________ K. Cal. `mol^(-1)` :-

Options

A. `-27.4`
B. `+27.4`
C. `-28.6`
D. `28.6`

Answer

To solve the problem, we need to find the value of \( q_V \) (the heat exchanged at constant volume) given the value of \( q_p \) (the heat exchanged at constant pressure) for the reaction:

\[ 2X(s) + 2Y(s) \rightarrow 2Cl(l) + D(g) \]

### Step-by-Step Solution:

1. **Identify Given Values**:
- \( q_p = -28 \, \text{kcal/mol} \) at \( 27^\circ C \).

2. **Understand the Relationship**:
- The relationship between \( q_p \) and \( q_V \) is given by the equation:
\[ \Delta H = \Delta E + \Delta n_g RT \]
- Here, \( \Delta H \) is the change in enthalpy (which is \( q_p \)), \( \Delta E \) is the change in internal energy (which is \( q_V \)), \( \Delta n_g \) is the change in the number of moles of gas, \( R \) is the ideal gas constant, and \( T \) is the temperature in Kelvin.

3. **Calculate \( \Delta n_g \)**:
- From the reaction, we identify the gaseous products and reactants:
- Products: 1 mole of \( D(g) \)
- Reactants: 0 moles of gas (both \( X \) and \( Y \) are solids).
- Therefore, \( \Delta n_g = \text{moles of products} - \text{moles of reactants} = 1 - 0 = 1 \).

4. **Convert Temperature to Kelvin**:
- Convert \( 27^\circ C \) to Kelvin:
\[ T = 27 + 273 = 300 \, K \]

5. **Use the Ideal Gas Constant**:
- The value of \( R \) in kcal is \( 0.001987 \, \text{kcal/(mol K)} \) (or approximately \( 2 \, \text{cal/(mol K)} \)).
- For our calculations, we will use \( R = 0.001987 \, \text{kcal/(mol K)} \).

6. **Substitute Values into the Equation**:
- Substitute \( \Delta H \), \( \Delta n_g \), \( R \), and \( T \) into the equation:
\[ -28 = \Delta E + (1)(0.001987)(300) \]

7. **Calculate the Right Side**:
- Calculate \( (1)(0.001987)(300) \):
\[ 0.001987 \times 300 = 0.5961 \, \text{kcal} \]

8. **Rearrange to Find \( \Delta E \)**:
- Rearranging the equation gives:
\[ \Delta E = -28 - 0.5961 = -28.5961 \, \text{kcal/mol} \]

9. **Final Answer**:
- Therefore, the value of \( q_V \) is:
\[ q_V \approx -28.6 \, \text{kcal/mol} \]

### Summary:
The value of \( q_V \) is approximately \( -28.6 \, \text{kcal/mol} \).