If `Delta E` is the heat of reaction for `C_(2)H_(5)OH_((1))+3O_(2(g)) rarr 2CO_(2(g))+3H_(2)O_((1))` at constant volume, the `Delta H` (Heat of reaction at constant pressure) at constant temperature is

Options

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

Explanation

To solve the problem of finding the relationship between the heat of reaction at constant pressure (ΔH) and the heat of reaction at constant volume (ΔE) for the given reaction:

\[ C_{2}H_{5}OH_{(l)} + 3O_{2(g)} \rightarrow 2CO_{2(g)} + 3H_{2}O_{(l)} \]

we will follow these steps:

### Step 1: Identify the reaction components
We need to identify the gaseous components in the reaction. The reactants are ethanol (C2H5OH) and oxygen (O2), while the products are carbon dioxide (CO2) and water (H2O). Since water is in the liquid state, we will only consider the gaseous components for calculating Δn.

### Step 2: Count the number of gaseous moles
- **Reactants**:
- 3 moles of O2 (g)

- **Products**:
- 2 moles of CO2 (g)

### Step 3: Calculate Δn (Change in number of gaseous moles)
Δn is calculated as the number of moles of gaseous products minus the number of moles of gaseous reactants:

\[ \Delta n = \text{(moles of gaseous products)} - \text{(moles of gaseous reactants)} = 2 - 3 = -1 \]

### Step 4: Use the relation between ΔH and ΔE
The relationship between the change in enthalpy (ΔH) and the change in internal energy (ΔE) at constant temperature is given by the equation:

\[ \Delta H = \Delta E + \Delta n \cdot R \cdot T \]

### Step 5: Substitute Δn into the equation
Now we substitute Δn into the equation:

\[ \Delta H = \Delta E + (-1) \cdot R \cdot T \]

This simplifies to:

\[ \Delta H = \Delta E - RT \]

### Conclusion
Thus, the heat of reaction at constant pressure (ΔH) is related to the heat of reaction at constant volume (ΔE) by the equation:

\[ \Delta H = \Delta E - RT \]

### Final Answer
The correct expression relating ΔH and ΔE at constant temperature is:

\[ \Delta H = \Delta E - RT \]