Enthalpy of which of the following reactions does not represent the standard enthalpy of formation?

Options

A. `H_2(g)+frac{1}{2}O_2(g)rightarrow H_2O(l)`
B. `H_2(g)+Br_2(l)rightarrow 2HBr(g)`
C. S(rhombic, s) + `O_2(g)rightarrowSO_2(g)`
D. 2C (graphite, s)+`3H_2(g)+frac{1}{2}O_2(g)rightarrow C_2H_5OH(l)`

Answer

To determine which of the given reactions does not represent the standard enthalpy of formation, we first need to understand the definition of standard enthalpy of formation (ΔHf°).

**Definition:**
The standard enthalpy of formation (ΔHf°) is defined as the enthalpy change when one mole of a compound is formed from its elements in their standard states.

Now, let's analyze each reaction step by step:

### Step 1: Analyze the first reaction
**Reaction:**
\[ \text{H}_2(g) + \frac{1}{2} \text{O}_2(g) \rightarrow \text{H}_2\text{O}(l) \]

- **Elements in standard states:** Hydrogen (H₂ gas) and Oxygen (O₂ gas).
- **Product:** Water (H₂O) in liquid form.
- **Moles of product formed:** 1 mole of H₂O.

**Conclusion:** This reaction represents the standard enthalpy of formation.

### Step 2: Analyze the second reaction
**Reaction:**
\[ \text{H}_2(g) + \text{Br}_2(l) \rightarrow 2 \text{HBr}(g) \]

- **Elements in standard states:** Hydrogen (H₂ gas) and Bromine (Br₂ liquid).
- **Product:** Hydrogen bromide (HBr) in gas form.
- **Moles of product formed:** 2 moles of HBr.

**Conclusion:** This reaction does NOT represent the standard enthalpy of formation because it forms 2 moles of HBr instead of 1 mole.

### Step 3: Analyze the third reaction
**Reaction:**
\[ \text{S}(rhombic) + \text{O}_2(g) \rightarrow \text{SO}_2(g) \]

- **Elements in standard states:** Rhombic sulfur (S) and Oxygen (O₂ gas).
- **Product:** Sulfur dioxide (SO₂) in gas form.
- **Moles of product formed:** 1 mole of SO₂.

**Conclusion:** This reaction represents the standard enthalpy of formation.

### Step 4: Analyze the fourth reaction
**Reaction:**
\[ 2 \text{C}(graphite) + 3 \text{H}_2(g) + \frac{1}{2} \text{O}_2(g) \rightarrow \text{C}_2\text{H}_5\text{OH}(l) \]

- **Elements in standard states:** Carbon (graphite), Hydrogen (H₂ gas), and Oxygen (O₂ gas).
- **Product:** Ethanol (C₂H₅OH) in liquid form.
- **Moles of product formed:** 1 mole of ethanol.

**Conclusion:** This reaction represents the standard enthalpy of formation.

### Final Answer:
The reaction that does NOT represent the standard enthalpy of formation is:
**Option 2:** \[ \text{H}_2(g) + \text{Br}_2(l) \rightarrow 2 \text{HBr}(g) \]

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