Q24 / 90 · JEE Main 2022 (25 Jun Shift 1) · Chemistry

The standard entropy change for the reaction

4Fes+3O2g→2Fe2O3s is -550 J K-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)

Correct answer: 300

Solution

4Fes+3O2g→3Fe2O3s  ΔSr=-550 J/K

ΔSr=-165 kJ/mole

At equilibrium ΔG=0

T=ΔHΔS at equilibrium

=-165×103-550=300 K

Hence, the correct answer is 300.

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