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 question regarding the reaction \( C_2H_5OH(l) + 3 O_2(g) \rightarrow 2 CO_2(g) + 3 H_2O(l) \) and to find the relationship between \( \Delta H \) and \( \Delta E \), we can follow these steps:

### Step 1: Write the reaction and identify the components
The given reaction is:
\[ C_2H_5OH(l) + 3 O_2(g) \rightarrow 2 CO_2(g) + 3 H_2O(l) \]

### Step 2: Identify the gaseous reactants and products
- **Reactants**:
- \( C_2H_5OH(l) \) (liquid)
- \( 3 O_2(g) \) (gas)

- **Products**:
- \( 2 CO_2(g) \) (gas)
- \( 3 H_2O(l) \) (liquid)

### Step 3: Calculate \( \Delta N_g \)
\( \Delta N_g \) is defined as the change in the number of moles of gaseous products minus the change in the number of moles of gaseous reactants.

- **Gaseous products**:
- \( 2 \) moles of \( CO_2(g) \)

- **Gaseous reactants**:
- \( 3 \) moles of \( O_2(g) \)

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

### Step 4: Use the relationship between \( \Delta H \) and \( \Delta E \)
The relationship between the change in enthalpy (\( \Delta H \)) and the change in internal energy (\( \Delta E \)) is given by:
\[ \Delta H = \Delta E + \Delta N_g RT \]
Where:
- \( R \) is the universal gas constant
- \( T \) is the temperature in Kelvin

### Step 5: Substitute \( \Delta N_g \) into the equation
Substituting \( \Delta N_g = -1 \) into the equation:
\[ \Delta H = \Delta E + (-1)RT \]
This simplifies to:
\[ \Delta H = \Delta E - RT \]

### Conclusion
The correct relationship for the given reaction is:
\[ \Delta H = \Delta E - RT \]