1. Definition to use

The standard enthalpy of formation $\Delta H_f^\circ$ of a compound is the enthalpy change when 1 mole of the compound is formed from its constituent elements in their standard states under standard conditions.

So, for a reaction to have $\Delta H_\text{reaction}^\circ = \Delta H_f^\circ,$ it must satisfy:

• reactants are the elements,
• each element is in its standard state,
• exactly 1 mole of product is formed.

2. Check option A

$2C(g) + 3H_2(g) \to C_2H_6(g)$

• Product formed = 1 mole of $C_2H_6$ ✅
• But carbon is taken as $C(g)$, whereas the standard state of carbon is graphite, $C(s,\text{graphite})$ ❌

Therefore, this is not a standard enthalpy of formation reaction.

3. Check option B

$2H_2(g) + O_2(g) \to 2H_2O(l)$

• Reactants are elements in standard states ✅
• But product formed is 2 moles of $H_2O(l)$ ❌

Standard enthalpy of formation must correspond to formation of 1 mole of compound: $H_2(g) + \frac{1}{2}O_2(g) \to H_2O(l)$

So option B is not equal to standard enthalpy of formation.

4. Check option C

$\frac{3}{2}O_2(g) \to O_3(g)$

• Oxygen is an element in its standard state: $O_2(g)$ ✅
• Product formed = 1 mole of $O_3(g)$ ✅

Thus this is the standard formation reaction of ozone.

So, C is correct.

5. Check option D

$\frac{1}{8}S_8(s) + O_2(g) \to SO_2(g)$

• Sulfur in standard state is $S_8(s)$ ✅
• Oxygen in standard state is $O_2(g)$ ✅
• Product formed = 1 mole of $SO_2(g)$ ✅

Thus this is a standard formation reaction for sulfur dioxide.

So, D is correct.

6. Final selection

The reactions for which standard enthalpy of reaction equals standard enthalpy of formation are:

$\boxed{C,\ D}$

7. Comparison with stored correct answer

Stored correct answer: C, D

My derived answer: C, D

They match.