Which is true for the combustion of sucrose `(C_(12)H_(22)O_(11))` at `25^(@)C` :-

Options

A. `Delta H gt Delta E`
B. `Delta H lt Delta E`
C. `Delta H = Delta E`
D. None

Explanation

To determine which statement is true for the combustion of sucrose (C₁₂H₂₂O₁₁) at 25°C, we can analyze the combustion reaction and apply the first law of thermodynamics.

### Step-by-Step Solution:

1. **Write the Combustion Reaction**:
The balanced chemical equation for the combustion of sucrose is:
\[ C_{12}H_{22}O_{11} (s) + 12 O_{2} (g) \rightarrow 12 CO_{2} (g) + 11 H_{2}O (l) \]

2. **Identify the States of Matter**:
- Sucrose is in solid form (s).
- Oxygen is in gaseous form (g).
- Carbon dioxide is in gaseous form (g).
- Water is in liquid form (l).

3. **Calculate the Change in Moles of Gas (\(\Delta N_G\))**:
\[ \Delta N_G = \text{(moles of gaseous products)} - \text{(moles of gaseous reactants)} \]
- Moles of gaseous products = 12 (from 12 CO₂)
- Moles of gaseous reactants = 12 (from 12 O₂)
\[ \Delta N_G = 12 - 12 = 0 \]

4. **Apply the First Law of Thermodynamics**:
According to the first law of thermodynamics, the change in enthalpy (\(\Delta H\)) is related to the change in internal energy (\(\Delta E\)) by the equation:
\[ \Delta H = \Delta E + \Delta N_G RT \]
Since \(\Delta N_G = 0\), the equation simplifies to:
\[ \Delta H = \Delta E \]

5. **Conclusion**:
From the above analysis, we conclude that for the combustion of sucrose at 25°C, the change in enthalpy (\(\Delta H\)) is equal to the change in internal energy (\(\Delta E\)). Therefore, the correct statement is that \(\Delta H = \Delta E\).

### Final Answer:
The correct option is **C**: \(\Delta H = \Delta E\).