The heat of combustion of sucrose `C_(12) H_(22) O_(11)(s)` at constant volume is -1348.9 kcal `mol^(-1)` at `25^(@)C`, then the heat of reaction at constant pressure, when steam is producced, is

Options

A. `-1348.9` kcal
B. 1342.344 kcal
C. 1250 kcal
D. `-1250` kcal

Answer

To solve the problem, we need to determine the heat of reaction at constant pressure when steam is produced from the combustion of sucrose (C₁₂H₂₂O₁₁). We will follow these steps:

### Step 1: Write the balanced combustion reaction for sucrose.
The balanced equation for the combustion of sucrose in the presence of oxygen is:

\[ C_{12}H_{22}O_{11}(s) + O_2(g) \rightarrow CO_2(g) + H_2O(l) \]

### Step 2: Determine the change in moles (Δn).
In the combustion reaction, we need to find the change in the number of moles of gas (Δn). The products are carbon dioxide (CO₂) and water (H₂O).

Assuming complete combustion, we can balance the equation:

\[ C_{12}H_{22}O_{11}(s) + 12 O_2(g) \rightarrow 12 CO_2(g) + 11 H_2O(l) \]

In this balanced equation:
- Reactants: 12 moles of O₂
- Products: 12 moles of CO₂

Thus, the change in moles (Δn) is:

\[ Δn = \text{moles of gaseous products} - \text{moles of gaseous reactants} = 12 - 12 = 0 \]

### Step 3: Relate ΔH and ΔU.
According to thermodynamic relations, the heat of reaction at constant pressure (ΔH) is related to the heat of reaction at constant volume (ΔU) by the equation:

\[ ΔH = ΔU + ΔnRT \]

Since Δn = 0, we have:

\[ ΔH = ΔU \]

### Step 4: Substitute the given values.
We know from the problem that the heat of combustion at constant volume (ΔU) is -1348.9 kcal/mol. Therefore:

\[ ΔH = ΔU = -1348.9 \text{ kcal/mol} \]

### Conclusion:
The heat of reaction at constant pressure, when steam is produced, is:

\[ ΔH = -1348.9 \text{ kcal/mol} \]