JEE Chemistry 2024 Question with Solution

Given:

• $2\text{Fe}_{(s)} + \frac{3}{2}\text{O}_{2(g)} \rightarrow \text{Fe}_2\text{O}_{3(s)}$ with $\Delta H = -822 \, \text{kJ/mol}$
• $\text{C}_{(s)} + \frac{1}{2}\text{O}_{2(g)} \rightarrow \text{CO}_{(g)}$ with $\Delta H = -110 \, \text{kJ/mol}$

Find: Enthalpy change for

$$3\text{C}_{(s)} + \text{Fe}_2\text{O}_{3(s)} \rightarrow 2\text{Fe}_{(s)} + 3\text{CO}_{(g)}$$

Using Hess's law, reverse the first reaction and multiply the second reaction by $3$.

Reversed first reaction:

$$\text{Fe}_2\text{O}_{3(s)} \rightarrow 2\text{Fe}_{(s)} + \frac{3}{2}\text{O}_{2(g)}$$

$\Delta H = +822 \, \text{kJ}$

Second reaction multiplied by $3$:

$$3\text{C}_{(s)} + \frac{3}{2}\text{O}_{2(g)} \rightarrow 3\text{CO}_{(g)}$$

$\Delta H = 3 \times (-110) = -330 \, \text{kJ}$

Adding these two reactions, $\frac{3}{2}\text{O}_{2(g)}$ cancels from both sides and we get

$$3\text{C}_{(s)} + \text{Fe}_2\text{O}_{3(s)} \rightarrow 2\text{Fe}_{(s)} + 3\text{CO}_{(g)}$$

So,

$$\Delta H = (+822) + (-330) = +492 \, \text{kJ}$$

Therefore, the enthalpy change for the reaction is $+492 \, \text{kJ}$, so the correct option is A.

Given: The standard enthalpy changes are:

$$\begin{aligned} (1)\;& 2\text{Fe}_{(s)} + \frac{3}{2}\text{O}_{2(g)} \rightarrow \text{Fe}_2\text{O}_{3(s)}, & \Delta H_1 &= -822 \, \text{kJ/mol} \\ (2)\;& \text{C}_{(s)} + \frac{1}{2}\text{O}_{2(g)} \rightarrow \text{CO}_{(g)}, & \Delta H_2 &= -110 \, \text{kJ/mol} \end{aligned}$$

Find:

$$3\text{C}_{(s)} + \text{Fe}_2\text{O}_{3(s)} \rightarrow 2\text{Fe}_{(s)} + 3\text{CO}_{(g)}$$

Apply the principle: According to Hess's law, when a reaction is reversed, the sign of $\Delta H$ changes, and when a reaction is multiplied by a number, $\Delta H$ is multiplied by the same number.

Step 1: Reverse reaction $(1)$ because $\text{Fe}_2\text{O}_{3(s)}$ must appear on the reactant side.

$$\text{Fe}_2\text{O}_{3(s)} \rightarrow 2\text{Fe}_{(s)} + \frac{3}{2}\text{O}_{2(g)}$$

$\Delta H = +822 \, \text{kJ/mol}$

Step 2: Multiply reaction $(2)$ by $3$ to produce $3\text{CO}_{(g)}$.

$$3\text{C}_{(s)} + \frac{3}{2}\text{O}_{2(g)} \rightarrow 3\text{CO}_{(g)}$$

$\Delta H = 3 \times (-110) = -330 \, \text{kJ/mol}$

Step 3: Add the manipulated equations.

$$\begin{aligned} &\text{Fe}_2\text{O}_{3(s)} \rightarrow 2\text{Fe}_{(s)} + \frac{3}{2}\text{O}_{2(g)} \\ +\;&3\text{C}_{(s)} + \frac{3}{2}\text{O}_{2(g)} \rightarrow 3\text{CO}_{(g)} \end{aligned}$$

After cancellation of $\frac{3}{2}\text{O}_{2(g)}$, the net reaction becomes

$$3\text{C}_{(s)} + \text{Fe}_2\text{O}_{3(s)} \rightarrow 2\text{Fe}_{(s)} + 3\text{CO}_{(g)}$$

Step 4: Sum the enthalpy changes.

$$\Delta H_{\text{rxn}} = +822 + (-330) = +492 \, \text{kJ}$$

Conclude: Therefore, the enthalpy change is $+492 \, \text{kJ/mol}$ and the correct option is A.