Heat of reaction is defined as the amount of heat absorbed or evolved at a given temperaturewhen the reactants have combined to form the products is represented by a balanced chemcial equation. If the heat is denofed by q then the numerical value of q depends on the manner in which the reaction is performed for the two methods of conducting chemical reactions in calorimeters.
Constant volume W = 0 and `q_(v) = Delta E` Bomb calorimeter
Constant pressure W = - V `Delta`P, therefore `q_(P) = Delta E + P Delta V rar (V. Delta P)`
The heat of combustion of maltose at constant volume in the above question if water vapour are not condensed

Options

A. `-1350` kcal/mol
B. `-1343` kcal/mol
C. `-1370` kcal/mol
D. `-1400` kcal/mol

Answer

To find the heat of combustion of maltose at constant volume, we will use the relationship between the heat at constant pressure (ΔH) and the change in internal energy (ΔE). The relevant equations are:

1. At constant volume: \( q_v = \Delta E \)
2. At constant pressure: \( q_p = \Delta H = \Delta E + P \Delta V \)

Given that water vapor is not condensed, we will assume that the reaction is performed under constant volume conditions.

### Step-by-Step Solution:

1. **Identify Given Values:**
- \( q_p = \Delta H = -1350 \, \text{kcal/mol} \)
- \( n_g = 1 \, \text{mol} \) (number of moles)
- \( R = 2 \, \text{kcal/(K·mol)} \) (universal gas constant)
- \( T = 298 \, \text{K} \) (temperature)

2. **Use the Relationship Between ΔE and ΔH:**
\[ \Delta E = q_p - P \Delta V \]

3. **Calculate PΔV:**
- We know that \( P \Delta V = n_g R T \).
- Substitute the values:
\[ P \Delta V = 1 \, \text{mol} \times 2 \, \text{kcal/(K·mol)} \times 298 \, \text{K} = 596 \, \text{kcal} \]

4. **Substitute Values into the ΔE Equation:**
\[ \Delta E = -1350 \, \text{kcal/mol} - 596 \, \text{kcal} \]

5. **Calculate ΔE:**
\[ \Delta E = -1350 - 596 = -1946 \, \text{kcal/mol} \]

### Final Result:
The heat of combustion of maltose at constant volume is:
\[ \Delta E = -1946 \, \text{kcal/mol} \]