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

JEE Main · 2024 · 30 Jan 2024 · Shift 2 · Chemical Thermodynamics

Two reactions are given below:

2 Fe_{(s)} + \frac{3}{2} O_{2 (g)} \to Fe_{2} O_{3 (s)}, Δ H^{o} = - 822 kJ / mol

C_{(s)} + \frac{1}{2} O_{2 (g)} \to CO_{(g)}, Δ H^{o} = - 110 kJ / mol

Then enthalpy change for following reaction

3 C_{(s)} + Fe_{2} O_{3 (s)} \to 2 Fe_{(s)} + 3 CO_{(g)}

Answer: 492

By Hess's law, the target reaction

3 C_{(s)} + Fe_{2} O_{3 (s)} \to 2 Fe_{(s)} + 3 CO_{(g)}

is built from tripling the second given reaction and reversing the first. Tripling

C_{(s)} + \frac{1}{2} O_{2 (g)} \to CO_{(g)}

(

Δ H = - 110 kJ/mol

) gives

Δ H = 3 (- 110) = - 330 kJ

; reversing

2 Fe_{(s)} + \frac{3}{2} O_{2 (g)} \to Fe_{2} O_{3 (s)}

(

Δ H = - 822 kJ/mol

) gives

Δ H = + 822 kJ

. Adding these two,

Δ H_{3} = 3 (- 110) - (- 822) = - 330 + 822 = 492 kJ/mol

JEE Main · 2024 · 30 Jan 2024 · Shift 2 · Integer answer