`DeltaH^(Theta)underset(f)' 298K` of methanol is given by the chemical equation

Options

A. `CH_(4(g))+(1)/(2)O_(2(g))rarrCH_(3)OH_((g))`
B. `C ("graphite")+(1)/(2)O_(2(g))+2H_(2(g))rarrCH_(3)OH_((l))`
C. `C ("diamond")+(1)/(2)O_(2(g))+2H_(2(g))rarrCH_(3)OH_((l))`
D. `CO_((g))+2H_(2(g))rarrCH_(3)OH_((l))`

Explanation

To determine the standard enthalpy of formation (ΔH_f°) of methanol (CH₃OH) at 298 K, we need to analyze the given chemical equations and identify which one represents the formation of methanol from its constituent elements in their standard states.

### Step-by-Step Solution:

1. **Understanding the Definition of Standard Enthalpy of Formation**:
The standard enthalpy of formation (ΔH_f°) of a compound is defined as the heat change that results when one mole of the compound is formed from its elements in their standard states.

2. **Identifying the Correct Reaction**:
We need to look for a reaction that forms methanol (CH₃OH) from its constituent elements: carbon (C), hydrogen (H₂), and oxygen (O₂). The standard states of these elements are:
- Carbon: Graphite (C)
- Hydrogen: Gas (H₂)
- Oxygen: Gas (O₂)

3. **Analyzing the Given Chemical Equations**:
We have four chemical equations to analyze. We need to check each one to see if it meets the criteria of forming methanol from its elements in their standard states.

- **Option 1**: Involves methane (CH₄) - Incorrect, as methane is a compound, not an element.
- **Option 2**:
\[ C_{(s)} + 2H_{2(g)} + \frac{1}{2}O_{2(g)} \rightarrow CH_{3}OH_{(l)} \]
This reaction shows carbon in its standard state (graphite), hydrogen gas, and oxygen gas forming methanol - Correct.
- **Option 3**: Involves carbon monoxide (CO) - Incorrect, as CO is a compound.
- **Option 4**: Involves diamond (C) - Incorrect, as diamond is not the standard state of carbon.

4. **Conclusion**:
The only equation that correctly represents the formation of methanol from its constituent elements in their standard states is **Option 2**.

### Final Answer:
The correct equation for the standard enthalpy of formation of methanol (ΔH_f°) at 298 K is:
\[ C_{(s)} + 2H_{2(g)} + \frac{1}{2}O_{2(g)} \rightarrow CH_{3}OH_{(l)} \]