The standard entropy change for the reaction $4 \mathrm{Fe}(\mathrm{s})+3 \mathrm{O}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{s})$ is $-550 \mathrm{JK}^{-1}$ at 298 K . [Given : The standard enthalpy change for the reaction is $-165 \mathrm{~kJ} \mathrm{~mol}^{-1}$ ]. The temperature in K at which the reaction attains equilibrium is \_\_\_\_. (Nearest Integer)

JEE Main · 2022 · 25 Jun 2022 · Shift 1 · Chemical Thermodynamics

The standard entropy change for the reaction

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

- 550 JK^{- 1}

at 298 K .
[Given : The standard enthalpy change for the reaction is

- 165 kJ mol^{- 1}

]. The temperature in K at which the reaction attains equilibrium is
____. (Nearest Integer)

Answer: 300

Solution

At equilibrium, ΔG = 0, so ΔH - TΔS = 0.

Given: ΔH = -165 kJ/mol = -165 × 10³ J/mol and ΔS = -550 J/K.

Solving: T = ΔH/ΔS = (-165 × 10³)/(-550) = 300 K.

Therefore, the answer is 300.

JEE Main · 2022 · 25 Jun 2022 · Shift 1 · Integer answer

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