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. `-110` KJ
B. `-284` KJ
C. `-394` KJ
D. `-504` KJ

Answer

To find the heat of combustion when one mole of graphite burns, we can use the standard enthalpy of formation values provided for carbon monoxide (CO) and carbon dioxide (CO2).

### Step-by-Step Solution:

1. **Identify the combustion reaction of graphite**:
The combustion of graphite (C) in the presence of oxygen (O2) can be represented by the following balanced equation:
\[ C + O_2 \rightarrow CO_2 \]
This reaction represents the complete combustion of graphite to form carbon dioxide.

2. **Use the standard enthalpy of formation for CO2**:
The standard enthalpy of formation (ΔH_f) for CO2 is given as -394 kJ/mol. This value indicates the heat released when one mole of CO2 is formed from its elements in their standard states.

3. **Relate the combustion of graphite to the formation of CO2**:
The heat of combustion of graphite (ΔH_comb) can be directly related to the standard enthalpy of formation of CO2:
\[ \Delta H_{\text{comb}} \text{ (for graphite)} = \Delta H_f \text{ (for CO2)} \]
Therefore, the heat of combustion when one mole of graphite is burned is equal to the standard enthalpy of formation of CO2.

4. **Conclude the result**:
Since the ΔH_f for CO2 is -394 kJ/mol, the heat of combustion when one mole of graphite burns is:
\[ \Delta H_{\text{comb}} = -394 \text{ kJ/mol} \]

### Final Answer:
The heat of combustion when one mole of graphite burns is **-394 kJ/mol**.