If `DeltaE` is the heat of reaction for `C_(2)H_(5)OH (l)+3O_(2)(g) to 2CO_(2) (g) + 3H_(2)O (l)` at constant volume, the `DeltaH` (heat of reaction at constant pressure), then the correct relation is:

Options

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

Explanation

To solve the problem, we need to establish the relationship between the heat of reaction at constant pressure (ΔH) and the heat of reaction at constant volume (ΔE).

### Step-by-Step Solution:

1. **Understand the Reaction**:
The given reaction is:
\[ C_{2}H_{5}OH (l) + 3O_{2}(g) \rightarrow 2CO_{2} (g) + 3H_{2}O (l) \]
This is a combustion reaction where ethanol reacts with oxygen to produce carbon dioxide and water.

2. **Identify the Relationship**:
The relationship between ΔH and ΔE is given by the equation:
\[ \Delta H = \Delta E + \Delta N_{g}RT \]
where:
- ΔH = heat of reaction at constant pressure
- ΔE = heat of reaction at constant volume
- ΔN_{g} = change in the number of moles of gas
- R = universal gas constant
- T = temperature in Kelvin

3. **Calculate ΔN_{g}**:
To find ΔN_{g}, we need to determine the number of moles of gaseous products and reactants:
- **Reactants**: 3 moles of \(O_{2}\) (g)
- **Products**: 2 moles of \(CO_{2}\) (g)
- Water \(H_{2}O\) is in liquid form and does not count towards ΔN_{g}.

Therefore,
\[ \Delta N_{g} = \text{(moles of gaseous products)} - \text{(moles of gaseous reactants)} = 2 - 3 = -1 \]

4. **Substitute ΔN_{g} into the Equation**:
Now we can substitute ΔN_{g} into the relationship:
\[ \Delta H = \Delta E + (-1)RT \]
This simplifies to:
\[ \Delta H = \Delta E - RT \]

5. **Conclusion**:
The correct relation between ΔH and ΔE for the given reaction is:
\[ \Delta H = \Delta E - RT \]

### Final Answer:
The correct relation is:
\[ \Delta H = \Delta E - RT \]