Question

$\text{Two reactions are given below:}$ \[ 2\text{Fe}_{(s)} + \frac{3}{2}\text{O}_{2(g)} \rightarrow \text{Fe}_2\text{O}_{3(s)}, \Delta H^\circ = -822 \, \text{kJ/mol} \] \[ \text{C}_{(s)} + \frac{1}{2}\text{O}_{2(g)} \rightarrow \text{CO}_{(g)}, \Delta H^\circ = -110 \, \text{kJ/mol} \] $\text{Then enthalpy change for the following reaction}$ \[ 3\text{C}_{(s)} + \text{Fe}_2\text{O}_{3(s)} \rightarrow 2\text{Fe}_{(s)} + 3\text{CO}_{(g)} \]

Answer 1

Correct Answer: 492

To find the enthalpy change (\(\Delta H\)) for the reaction, we can use Hess’s law and manipulate the given reactions.

Given Reactions:

The first reaction is exothermic with a release of 822 kJ. The second reaction is also exothermic with a release of 110 kJ.

Manipulating Reactions:

To find the enthalpy change for:

\[ 3C(g) + Fe_2O_3(s) \rightarrow 2Fe(g) + 3CO(g), \]

we will reverse the first reaction and adjust the second.

Reversed first reaction:

\[ Fe_2O_3(s) \rightarrow 2Fe(g) + \frac{3}{2}O_2(g), \quad \Delta H = +822 \, \text{kJ} \]

Second reaction remains as it is:

\[ 3C(g) + \frac{3}{2}O_2(g) \rightarrow 3CO(g), \quad \Delta H = -330 \, \text{kJ} \]

Adding the Reactions:

Now we sum the enthalpy changes:

\[ \Delta H = \Delta H_{\text{reversed}} + \Delta H_{\text{second}} = 822 + (-330) = 492 \, \text{kJ} \]

Thus, the enthalpy change for the reaction is: 492 kJ