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 understand what enthalpy of formation means. Here’s a step-by-step solution:

### Step 1: Understand the Definition of Enthalpy of Formation
Enthalpy of formation (ΔH_f) is defined as the change in enthalpy when one mole of a compound is formed from its constituent elements in their standard states at a specified temperature (298 K) and pressure (1 atm).

### Step 2: Identify Key Criteria
1. **One mole of substance**: The equation must produce exactly one mole of the product.
2. **Elements in their standard states**: The reactants must be in their most stable form at 298 K and 1 atm pressure.

### Step 3: Analyze Each Option
- **Option 1**: If the equation involves carbon in its standard state (graphite) and hydrogen in its standard state (gas), but oxygen is in a liquid state, this does not meet the criteria since oxygen must be in its gaseous state.

- **Option 2**: If carbon is in the diamond form, it is not in its standard state (graphite is the standard state for carbon). Therefore, this option is incorrect.

- **Option 3**: If the equation produces two moles of methanol, it does not meet the requirement of forming one mole of substance. Hence, this option is also incorrect.

- **Option 4**: If the equation has carbon in its standard state (graphite), hydrogen in its gaseous state, and oxygen in its gaseous state, and produces one mole of methanol, this option meets both criteria.

### Step 4: Conclusion
Based on the analysis, **Option 4** corresponds to the definition of enthalpy of formation at 298 K.