$\text{Two reactions are given below:}$ \[ 2\text{Fe}_{(s)} + \frac{3}{2}\text{O}_{2(g)} \rightarrow \text{Fe}_2\text{O}_{3(s)}, \Delta H^\circ = -822 \, \text{kJ/mol} \] \[ \text{C}_{(s)} + \frac{1}{2}\text{O}_{2(g)} \rightarrow \text{CO}_{(g)}, \Delta H^\circ = -110 \, \text{kJ/mol} \] $\text{Then enthalpy change for the following reaction}$ \[ 3\text{C}_{(s)} + \text{Fe}_2\text{O}_{3(s)} \rightarrow 2\text{Fe}_{(s)} + 3\text{CO}_{(g)} \]
Correct Answer: 492
Approach Solution - 1
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
Approach Solution -2
Given the standard enthalpy changes for two reactions, we are to find the enthalpy change for the reaction: \( 3\text{C}_{(s)} + \text{Fe}_2\text{O}_{3(s)} \rightarrow 2\text{Fe}_{(s)} + 3\text{CO}_{(g)} \).
Concept Used:
Hess's Law states that the total enthalpy change for a reaction is the same, regardless of the number of steps or the path taken. To find the enthalpy change for a target reaction, we can algebraically manipulate the given thermochemical equations (reversing them and/or multiplying them by a coefficient) and then add them together. When a reaction is reversed, the sign of \( \Delta H \) is changed. When a reaction is multiplied by a coefficient, its \( \Delta H \) is multiplied by the same coefficient.
Step-by-Step Solution:
Step 1: Write down the given reactions and their enthalpy changes.
\[ \begin{aligned} \text{(1)} \quad & 2\text{Fe}_{(s)} + \frac{3}{2}\text{O}_{2(g)} \rightarrow \text{Fe}_2\text{O}_{3(s)}, & \Delta H^\circ_1 &= -822 \, \text{kJ/mol} \\ \text{(2)} \quad & \text{C}_{(s)} + \frac{1}{2}\text{O}_{2(g)} \rightarrow \text{CO}_{(g)}, & \Delta H^\circ_2 &= -110 \, \text{kJ/mol} \end{aligned} \]
Step 2: Identify the target reaction and plan the manipulation of the given equations.
Target: \( 3\text{C}_{(s)} + \text{Fe}_2\text{O}_{3(s)} \rightarrow 2\text{Fe}_{(s)} + 3\text{CO}_{(g)} \)
- \( \text{Fe}_2\text{O}_{3(s)} \) is a reactant in the target but a product in reaction (1). Therefore, reverse reaction (1).
- The coefficient of \( \text{Fe}_2\text{O}_{3(s)} \) in the reversed reaction (1) is 1, which matches the target.
- We need 3 moles of \( \text{CO}_{(g)} \) as a product. Reaction (2) produces 1 mole of CO, so multiply it by 3.
- Check that all other species (\( \text{C}_{(s)}, \text{Fe}_{(s)}, \text{O}_{2(g)} \)) cancel out.
Step 3: Perform the manipulations on the given equations.
Reverse reaction (1): \[ \text{Fe}_2\text{O}_{3(s)} \rightarrow 2\text{Fe}_{(s)} + \frac{3}{2}\text{O}_{2(g)}, \quad \Delta H^\circ_3 = +822 \, \text{kJ/mol} \] Multiply reaction (2) by 3: \[ 3\text{C}_{(s)} + \frac{3}{2}\text{O}_{2(g)} \rightarrow 3\text{CO}_{(g)}, \quad \Delta H^\circ_4 = 3 \times (-110) = -330 \, \text{kJ/mol} \]
Step 4: Add the manipulated equations and cancel out common terms.
\[ \begin{aligned} &\text{Fe}_2\text{O}_{3(s)} \rightarrow 2\text{Fe}_{(s)} + \frac{3}{2}\text{O}_{2(g)}, & \Delta H &= +822 \, \text{kJ} \\ + \, &3\text{C}_{(s)} + \frac{3}{2}\text{O}_{2(g)} \rightarrow 3\text{CO}_{(g)}, & \Delta H &= -330 \, \text{kJ} \\ \hline &3\text{C}_{(s)} + \text{Fe}_2\text{O}_{3(s)} \rightarrow 2\text{Fe}_{(s)} + 3\text{CO}_{(g)}, & \Delta H_\text{rxn} &= ? \end{aligned} \]
The \( \frac{3}{2}\text{O}_{2(g)} \) cancels out as it appears on both sides.
Step 5: Calculate the enthalpy change for the target reaction by summing the enthalpy changes of the manipulated equations.
\[ \Delta H_\text{rxn} = (+822 \, \text{kJ}) + (-330 \, \text{kJ}) = +492 \, \text{kJ} \]
Therefore, the enthalpy change for the reaction \( 3\text{C}_{(s)} + \text{Fe}_2\text{O}_{3(s)} \rightarrow 2\text{Fe}_{(s)} + 3\text{CO}_{(g)} \) is +492 kJ/mol.
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