Given :
`C("diamond")+O_(2)rarrCO_(2),DeltaH=-395 kJ`
`C("graphite")+O_(2)rarrCO_(2),DeltaH=-393 kJ`
The enthalpy of formation of diamond from graphite is

Options

A. `+ 2.0 kJ`
B. `-1.5 kJ`
C. `-788 kJ`
D. 788 kJ

Explanation

To find the enthalpy of formation of diamond from graphite, we can use Hess's law, which states that the total enthalpy change for a reaction is the sum of the enthalpy changes for the individual steps.

### Step-by-Step Solution:

1. **Write the Given Reactions:**
- For diamond:
\[ C(\text{diamond}) + O_2 \rightarrow CO_2, \quad \Delta H = -395 \, \text{kJ} \]
- For graphite:
\[ C(\text{graphite}) + O_2 \rightarrow CO_2, \quad \Delta H = -393 \, \text{kJ} \]

2. **Identify the Target Reaction:**
- We want to find the enthalpy change for the reaction:
\[ C(\text{diamond}) \rightarrow C(\text{graphite}) \]

3. **Manipulate the Given Reactions:**
- To find the enthalpy change for the formation of diamond from graphite, we can subtract the enthalpy change of the graphite reaction from that of the diamond reaction:
\[ C(\text{diamond}) + O_2 \rightarrow CO_2 \quad (\Delta H = -395 \, \text{kJ}) \quad \text{(1)} \]
\[ C(\text{graphite}) + O_2 \rightarrow CO_2 \quad (\Delta H = -393 \, \text{kJ}) \quad \text{(2)} \]

4. **Subtract the Reactions:**
- We will subtract reaction (2) from reaction (1):
\[ C(\text{diamond}) + O_2 - (C(\text{graphite}) + O_2) \rightarrow CO_2 - CO_2 \]
- This simplifies to:
\[ C(\text{diamond}) - C(\text{graphite}) = \Delta H \]

5. **Calculate the Enthalpy Change:**
- The enthalpy change can be calculated as:
\[ \Delta H = (-395 \, \text{kJ}) - (-393 \, \text{kJ}) = -395 + 393 = -2 \, \text{kJ} \]

6. **Final Result:**
- The enthalpy of formation of diamond from graphite is:
\[ \Delta H = -2 \, \text{kJ} \]