Given standard enthalpy of formation of `CO(-110 "KJ mol"^(-1))` and `CO_(2)(-394 "KJ mol"^(-1))`. The heat of combustion when one mole of graphite burns is

Options

A. `-504` kJ
B. `-394` kJ
C. `-284` kJ
D. `-110` kJ

Explanation

To determine the heat of combustion when one mole of graphite burns, we can use the given standard enthalpy of formation values for carbon monoxide (CO) and carbon dioxide (CO2). Here’s a step-by-step solution:

### Step 1: Understand the Reaction
When graphite (a form of carbon) burns in the presence of oxygen, it typically forms carbon dioxide (CO2). The combustion reaction can be represented as:
\[ \text{C (graphite)} + \text{O}_2 \rightarrow \text{CO}_2 \]

### Step 2: Write the Enthalpy Changes
We know the standard enthalpy of formation (\( \Delta H_f \)) for the substances involved:
- For CO: \( \Delta H_f (\text{CO}) = -110 \, \text{kJ/mol} \)
- For CO2: \( \Delta H_f (\text{CO}_2) = -394 \, \text{kJ/mol} \)

### Step 3: Relate the Combustion to Formation
The heat of combustion (\( \Delta H_c \)) of graphite can be derived from the enthalpy of formation of CO2. The combustion of graphite to form CO2 can be expressed in terms of the enthalpy of formation of CO2:
\[ \Delta H_c = \Delta H_f (\text{CO}_2) \]

### Step 4: Substitute the Values
Substituting the known value:
\[ \Delta H_c = -394 \, \text{kJ/mol} \]

### Step 5: Conclusion
Thus, the heat of combustion when one mole of graphite burns is:
\[ \Delta H_c = -394 \, \text{kJ/mol} \]

### Final Answer
The heat of combustion when one mole of graphite burns is \(-394 \, \text{kJ/mol}\).

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