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 find the value of \( q_V \) (which is equal to \( \Delta E \)) for the given reaction, we can follow these steps:

### Step 1: Write down the reaction and known values
The reaction is:
\[ 2X(s) + 2Y(s) \rightarrow 2C(l) + D(g) \]

Given:
- \( q_p \) (or \( \Delta H \)) at \( 27^\circ C \) is \( -28 \, \text{kcal/mol} \).

### Step 2: Identify the relationship between \( \Delta H \) and \( \Delta E \)
From thermodynamics, we know that:
\[ \Delta H = \Delta E + \Delta n_g RT \]
where:
- \( \Delta n_g \) is the change in the number of moles of gas,
- \( R \) is the universal gas constant,
- \( T \) is the temperature in Kelvin.

### Step 3: Calculate \( \Delta n_g \)
In the reaction:
- Reactants: 0 moles of gas (2 solids)
- Products: 1 mole of gas (D)

Thus,
\[ \Delta n_g = 1 - 0 = 1 \]

### Step 4: Convert temperature to Kelvin
The temperature in Celsius is \( 27^\circ C \). To convert to Kelvin:
\[ T = 27 + 273 = 300 \, K \]

### Step 5: Use the value of \( R \)
The value of \( R \) in kcal is:
\[ R = 2 \times 10^{-3} \, \text{kcal/mol K} \]

### Step 6: Substitute values into the equation
Now, substituting the values into the equation:
\[ \Delta H = \Delta E + \Delta n_g RT \]
\[ -28 = \Delta E + (1)(2 \times 10^{-3})(300) \]

### Step 7: Calculate \( RT \)
Calculating \( RT \):
\[ RT = 2 \times 10^{-3} \times 300 = 0.6 \, \text{kcal/mol} \]

### Step 8: Rearrange to find \( \Delta E \)
Now, rearranging the equation to solve for \( \Delta E \):
\[ \Delta E = \Delta H - RT \]
\[ \Delta E = -28 - 0.6 \]
\[ \Delta E = -28.6 \, \text{kcal/mol} \]

### Final Answer
Thus, the value of \( q_V \) is:
\[ q_V = -28.6 \, \text{kcal/mol} \]

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