Two reactions are given below, Thermodynamics Questions | JEE Main Chemistry Chapter Wise Question Papers

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Solution

Q.
Two reactions are given below:

$2 {Fe}_{(s)} + \frac{3}{2} O_{2 (g)} \to {Fe}_{2} O_{3 (s)}, ΔH ° = - 822 kJ / mol$

$C_{(s)} + \frac{1}{2} O_{2 (g)} \to {CO}_{(g)}, ΔH ° = - 110 kJ / mol$

Then enthalpy change for following reaction

$3 C_{(s)} + {Fe}_{2} O_{3 (s)} \to 2 {Fe}_{(s)} + 3 {CO}_{(g)}$ is _________ kJ/mol. [2024]

Ans.
(492)

$2 {Fe}_{(s)} + \frac{3}{2} O_{2} (g) \to {Fe}_{2} O_{3} (s), {ΔH}_{1}^{o} = - 822 kJ / mol - I$

Reverse this equation:

${Fe}_{2} O_{3} (s) \to 2 Fe (s) + \frac{3}{2} O_{2} (g), Δ H_{2}^{o} = 822 kJ / mol - II$

$C (s) + \frac{1}{2} O_{2} (g) \to CO (g), {ΔH}_{3}^{o} = - 110 kJ / mol - III$

Multiply this equation by 3

$3 C (s) + \frac{3}{2} O_{2} (g) \to 3 CO (g), Δ H_{4}^{o} = - 3 \times 110 kJ / mol$

$= - 330 kJ / mol - IV$

Adding II and IV gives us the desired equation:

$3 C (s) + {Fe}_{2} O_{3} (s) \to 2 Fe (s) + 3 CO (g), Δ H_{n e t}^{o}$

By Hess's law:

$Δ H_{n e t}^{o} = Δ H_{2}^{o} + Δ H_{4}^{o} = (822 - 330) kJ / mol = 492 kJ / mol$

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