The given diagram represents the enthalpy changes for the formation of CO(g) from C(graphite) and O\u2082(g) via an intermediate step of CO\u2082(g).

According to Hess's Law, the total enthalpy change for a reaction is independent of the pathway taken. Therefore, the enthalpy change for the direct formation of CO(g) from C(graphite) and O\u2082(g) is equal to the sum of the enthalpy changes for the two-step process.

Step 1: Identify the reactions and their enthalpy changes.

1. C(graphite) + O\u2082(g) \u2192 CO\u2082(g); \u0394H\u2081 = -393.5 kJ/mol

2. CO(g) + \u00bd O\u2082(g) \u2192 CO\u2082(g); \u0394H\u2082 = -283.0 kJ/mol

We need to find \u0394H for the reaction: C(graphite) + \u00bd O\u2082(g) \u2192 CO(g)

Step 2: Apply Hess's Law.

To get the target reaction, we can use reaction (1) as is, and reverse reaction (2).

1. C(graphite) + O\u2082(g) \u2192 CO\u2082(g); \u0394H\u2081 = -393.5 kJ/mol

2'. CO\u2082(g) \u2192 CO(g) + \u00bd O\u2082(g); \u0399\u0394H\u2082' = -(\u0394H\u2082) = -(-283.0 kJ/mol) = +283.0 kJ/mol

Adding these two reactions:

C(graphite) + O\u2082(g) + CO\u2082(g) \u2192 CO\u2082(g) + CO(g) + \u00bd O\u2082(g)

Simplifying, we get:

C(graphite) + \u00bd O\u2082(g) \u2192 CO(g)

Step 3: Calculate the total enthalpy change.

\u0394H = \u0394H\u2081 + \u0394H\u2082'

\u0394H = -393.5 kJ/mol + 283.0 kJ/mol

\u0394H = -110.5 kJ/mol

Option Analysis:

• A) +10.5 kJ: Incorrect. This value is obtained if the signs are incorrectly handled or if the wrong reactions are combined.
• B) -11.05 kJ: Incorrect. This value is numerically incorrect and has a misplaced decimal.
• C) -110.5 kJ: Correct. This is the calculated enthalpy change for the formation of CO(g).
• D) -10.5 J: Incorrect. The magnitude is wrong, and the unit is J instead of kJ. Enthalpy changes for reactions are typically in kJ.

Correct Answer: (C)