Given
(A) $$2CO(g) + O_2(g) \to 2CO_2(g)$$, $$\Delta H_1^0 = -x$$ kJ mol$$^{-1}$$
(B) $$C_{graphite} + O_2(g) \to CO_2(g)$$, $$\Delta H_2^0 = -y$$ kJ mol$$^{-1}$$
The $$\Delta H^0$$ for the reaction $$C_{graphite} + \frac{1}{2}O_2(g) \to CO(g)$$ is

We need to find $$\Delta H^\circ$$ for $$C_{\text{graphite}} + \frac{1}{2}O_2(g) \to CO(g)$$ using Hess's law.

reactions.

(A) $$2CO(g) + O_2(g) \to 2CO_2(g)$$, $$\Delta H_1^\circ = -x$$ kJ/mol

(B) $$C_{\text{graphite}} + O_2(g) \to CO_2(g)$$, $$\Delta H_2^\circ = -y$$ kJ/mol

Write the target reaction.

Target: $$C_{\text{graphite}} + \frac{1}{2}O_2(g) \to CO(g)$$, $$\Delta H = ?$$

Express the target as a combination of given reactions.

We need to obtain CO on the right side. From reaction (A), CO appears on the left. If we reverse half of (A):

$$CO_2(g) \to CO(g) + \frac{1}{2}O_2(g)$$, $$\Delta H = +\frac{x}{2}$$

Adding this to reaction (B):

$$C + O_2 \to CO_2$$, $$\Delta H = -y$$

$$CO_2 \to CO + \frac{1}{2}O_2$$, $$\Delta H = +\frac{x}{2}$$

Sum: $$C + \frac{1}{2}O_2 \to CO$$

Calculate $$\Delta H$$.

$$ \Delta H = -y + \frac{x}{2} = \frac{x - 2y}{2} $$

The correct answer is Option 3: $$\dfrac{x - 2y}{2}$$.