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JEE Main 202225 JuneMorning Shift
Exam
JEE Main
Year
2022
Date
25 June
Shift
Morning Shift
Subject
Chemistry
Chapter
Thermodynamics (C)

JEE Main 2022 · 25 June · Morning Shift · Chemistry · Thermodynamics (C)

The standard entropy change for the reaction 4Fe(s) + 3 O _ 2 (g) →...

The standard entropy change for the reaction

4Fe(s)+3O2(g)→2Fe2O3(s)4Fe\left(s\right) + 3{O}_{2}\left(g\right) \to 2{Fe}_{2}{O}_{3}\left(s\right) is −550JK−1- 550J{K}^{- 1} at 298K298K

[Given : The standard enthalpy change for the reaction is −165kJmol−1- 165kJ{mol}^{- 1}]. The temperature in KK at which the reaction attains equilibrium is (Nearest Integer)

Correct answer

0

Step-by-step solution

4Fe(s)+3O2(g)→3Fe2O3(s)ΔSr=−550J/K4Fe\left(s\right) + 3{O}_{2}\left(g\right) \to 3{Fe}_{2}{O}_{3}\left(s\right){\Delta S}_{r} = - 550J / K

ΔSr=−165kJ/mole{\Delta S}_{r} = - 165kJ / mole

At equilibrium ΔG=0\Delta G = 0

T=(ΔHΔS)T = \left(\dfrac{\Delta H}{\Delta S}\right) at equilibrium

=−165×103−550=300K= \dfrac{- 165 \times {10}^{3}}{- 550} = 300K

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