One mole of ethanol is produced reacting graphite , `H_2 and O_2` together . The standard enthalpy of formation is `-277.7 " kJ mol"^(-1)` Calculate the standard enthalpy of the reaction when 4 moles of graphite is involved.

Options

A. `-227.7` kJ/mol
B. `-555.4` kJ/mol
C. `-138.85` kJ/mol
D. `-69.42` kJ/mol

Answer

To calculate the standard enthalpy of the reaction when 4 moles of graphite are involved in the production of ethanol, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Reaction**:
The formation of ethanol (C2H5OH) from its elements can be represented as:
\[ 2C (graphite) + 3H_2 + \frac{1}{2}O_2 \rightarrow C_2H_5OH \]

2. **Standard Enthalpy of Formation**:
The standard enthalpy of formation (\( \Delta H_f^\circ \)) of ethanol is given as:
\[ \Delta H_f^\circ = -277.7 \, \text{kJ/mol} \]
This value corresponds to the formation of 1 mole of ethanol.

3. **Scaling the Reaction**:
If the reaction produces 1 mole of ethanol from 2 moles of graphite, we need to determine the enthalpy change when 4 moles of graphite are used. Since the stoichiometry of graphite in the reaction is 2 moles for 1 mole of ethanol, using 4 moles of graphite means we are doubling the amount of graphite.

4. **Calculating the New Enthalpy Change**:
Since 4 moles of graphite correspond to 2 moles of ethanol (because 2 moles of graphite produce 1 mole of ethanol), we can scale the enthalpy change accordingly:
\[ \Delta H = 2 \times (-277.7 \, \text{kJ/mol}) = -555.4 \, \text{kJ} \]

5. **Final Result**:
Therefore, the standard enthalpy of the reaction when 4 moles of graphite are involved is:
\[ \Delta H = -555.4 \, \text{kJ} \]