For the reaction `C_2H_(5)OH(l) + 3 O_(2)(g) rightarrow 2CO_(2)(g) + 3 H_(2)O(l)`, which one is true ?

Options

A. `Delta H = DeltaE -RT`
B. `Delta H = DeltaE + RT`
C. `Delta H = DeltaE + 2RT`
D. `Delta H = DeltaE - 2RT`

Answer

To solve the problem regarding the reaction:

\[ C_2H_5OH(l) + 3 O_2(g) \rightarrow 2 CO_2(g) + 3 H_2O(l) \]

we need to analyze the relationship between the change in enthalpy (\(\Delta H\)) and the change in internal energy (\(\Delta E\)) for this reaction.

### Step-by-Step Solution:

1. **Identify the Reaction Components**:
- Reactants: Ethanol (\(C_2H_5OH\)) in liquid form and 3 moles of oxygen gas (\(O_2\)).
- Products: 2 moles of carbon dioxide gas (\(CO_2\)) and 3 moles of water in liquid form (\(H_2O\)).

2. **Determine the Change in Moles of Gas**:
- Count the total moles of gaseous products and reactants.
- Gaseous products: 2 moles of \(CO_2\).
- Gaseous reactants: 3 moles of \(O_2\).
- Therefore, the change in the number of moles of gas (\(\Delta n_g\)) is:
\[ \Delta n_g = \text{(moles of gaseous products)} - \text{(moles of gaseous reactants)} = 2 - 3 = -1 \]

3. **Apply the Relationship Between \(\Delta H\) and \(\Delta E\)**:
- The relationship is given by:
\[ \Delta H = \Delta E + P \Delta V \]
- We can express \(P \Delta V\) in terms of the change in moles of gas:
\[ P \Delta V = \Delta n_g \cdot R \cdot T \]
- Substituting \(\Delta n_g = -1\):
\[ P \Delta V = -1 \cdot R \cdot T = -RT \]

4. **Substitute into the Enthalpy-Internal Energy Relationship**:
- Now substituting \(P \Delta V\) back into the equation for \(\Delta H\):
\[ \Delta H = \Delta E - RT \]

5. **Conclusion**:
- The correct relationship for the given reaction is:
\[ \Delta H = \Delta E - RT \]
- Thus, the true statement regarding the reaction is that \(\Delta H\) is equal to \(\Delta E\) minus \(RT\).

### Final Answer:
\[ \Delta H = \Delta E - RT \]