Which of the following equations corresponds to the definition of enthalpy of formation at 298 K?

Options

A. `C_("graphite") + 2H_2(g) + 1/2 O_(2)(l) rarr CH_(3)OH(g)`
B. `C_("diamond") + 2H_2(g) + 1/2 O_(2)(g) rarr CH_(3)OH(l)`
C. `C_("graphite") + 4H_2(g) + O_(2)(g) rarr 2CH_(3)OH(l)`
D. `C_("graphite") + 2H_2(g) + 1/2 O_(2)(g) rarr CH_(3)OH(l)`

Explanation

To determine which equation corresponds to the definition of enthalpy of formation at 298 K, we need to follow these steps:

### Step 1: Understand the Definition of Enthalpy of Formation
The enthalpy of formation (\( \Delta H_f^\circ \)) of a compound is defined as the heat change that occurs when one mole of the compound is formed from its elements in their standard states at 298 K and 1 atm pressure.

### Step 2: Identify the Components of the Reaction
For the formation of methanol (\( CH_3OH \)), we need to identify the standard states of its constituent elements:
- Carbon (C) should be in its standard state, which is graphite.
- Hydrogen (H) should be in its standard state, which is diatomic hydrogen gas (\( H_2(g) \)).
- Oxygen (O) should also be in its standard state, which is diatomic oxygen gas (\( O_2(g) \)).

### Step 3: Write the Balanced Chemical Equation
The balanced equation for the formation of one mole of methanol from its elements is:
\[ C(s) + 2H_2(g) + \frac{1}{2}O_2(g) \rightarrow CH_3OH(l) \]
Here, we have:
- 1 mole of \( CH_3OH \) produced.
- Carbon in the form of graphite (solid).
- Hydrogen in the form of \( H_2 \) (gas).
- Oxygen in the form of \( O_2 \) (gas).

### Step 4: Analyze the Given Options
Now, we need to analyze the provided options to see which one matches the balanced equation we derived:
- **Option A**: Contains \( O_2 \) in liquid state (incorrect).
- **Option B**: Contains carbon in diamond form (incorrect).
- **Option C**: Produces 2 moles of \( CH_3OH \) (incorrect).
- **Option D**: Contains carbon in graphite form, \( H_2 \) in gas state, \( O_2 \) in gas state, and produces 1 mole of \( CH_3OH \) (correct).

### Conclusion
The correct equation that corresponds to the definition of enthalpy of formation at 298 K is:
\[ C(s) + 2H_2(g) + \frac{1}{2}O_2(g) \rightarrow CH_3OH(l) \]
This corresponds to **Option D**.

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