1. We need the enthalpy change for

$3C_{(s)} + Fe_2O_{3(s)} \rightarrow 2Fe_{(s)} + 3CO_{(g)}$

We will use Hess's law.

2. Given reactions:

$2Fe_{(s)} + \frac{3}{2}O_{2(g)} \rightarrow Fe_2O_{3(s)}, \qquad \Delta H^\circ = -822\;\text{kJ}$

$C_{(s)} + \frac{1}{2}O_{2(g)} \rightarrow CO_{(g)}, \qquad \Delta H^\circ = -110\;\text{kJ}$

3. First, reverse the iron oxide formation reaction because in the required reaction $Fe_2O_3$ is a reactant:

$Fe_2O_{3(s)} \rightarrow 2Fe_{(s)} + \frac{3}{2}O_{2(g)}$

So,

$\Delta H^\circ = +822\;\text{kJ}$

4. Multiply the carbon monoxide formation reaction by 3:

$3C_{(s)} + \frac{3}{2}O_{2(g)} \rightarrow 3CO_{(g)}$

So,

$\Delta H^\circ = 3(-110) = -330\;\text{kJ}$

5. Add the two modified equations:

$Fe_2O_{3(s)} \rightarrow 2Fe_{(s)} + \frac{3}{2}O_{2(g)}$

$3C_{(s)} + \frac{3}{2}O_{2(g)} \rightarrow 3CO_{(g)}$

Adding,

$3C_{(s)} + Fe_2O_{3(s)} \rightarrow 2Fe_{(s)} + 3CO_{(g)}$

The $\frac{3}{2}O_2$ cancels out.

6. Therefore,

$\Delta H^\circ = 822 + (-330) = 492\;\text{kJ/mol}$

So the enthalpy change is

$\boxed{492}$

kJ/mol.