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 1: Understand the Reaction
The formation of ethanol (C2H5OH) from graphite (C), hydrogen (H2), and oxygen (O2) can be represented by the following balanced chemical equation:
\[ 2C (s) + 3H_2 (g) + \frac{1}{2}O_2 (g) \rightarrow C_2H_5OH (l) \]

### Step 2: Identify the Enthalpy of Formation
The standard enthalpy of formation (ΔH_f) of 1 mole of ethanol is given as:
\[ ΔH_f = -277.7 \text{ kJ/mol} \]

This means that when 1 mole of ethanol is formed, 277.7 kJ of energy is released.

### Step 3: Determine the Amount of Graphite Used
From the balanced equation, we see that to produce 1 mole of ethanol, 2 moles of graphite are required.

### Step 4: Calculate the Enthalpy Change for 2 Moles of Graphite
Since the formation of 1 mole of ethanol releases -277.7 kJ, the enthalpy change for the reaction involving 2 moles of graphite (which produces 1 mole of ethanol) is:
\[ ΔH = -277.7 \text{ kJ} \]

### Step 5: Calculate the Enthalpy Change for 4 Moles of Graphite
If we use 4 moles of graphite, we need to determine how many moles of ethanol can be produced. Since 2 moles of graphite produce 1 mole of ethanol, 4 moles of graphite will produce:
\[ \frac{4 \text{ moles of C}}{2 \text{ moles of C/mole of ethanol}} = 2 \text{ moles of ethanol} \]

### Step 6: Calculate the Total Enthalpy Change for 2 Moles of Ethanol
The enthalpy change for producing 2 moles of ethanol is:
\[ ΔH = 2 \times (-277.7 \text{ kJ}) = -555.4 \text{ kJ} \]

### Final Answer
The standard enthalpy of the reaction when 4 moles of graphite are involved is:
\[ ΔH = -555.4 \text{ kJ} \]

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