JEE Main 2020 — Chemical Equilibrium Question with Solution
From: JEE Main 2020 (Online) 6th September Morning Slot
Question
The variation of equilibrium constant with
temperature is given below :
The values of Ho, Go at
T1 and Go at T2 (in kJ mol–1) respectively, are close to :
[Use R = 8.314 J K–1 mol–1]
| Temperature | Equilibrium Constant |
|---|---|
| T1 = 25oC | K1 = 10 |
| T2 = 100oC | K2 = 100 |
The values of Ho, Go at
T1 and Go at T2 (in kJ mol–1) respectively, are close to :
[Use R = 8.314 J K–1 mol–1]
Choose an option
Show full solutionCorrect option: A
Correct answer
A28.4, –5.71 and –14.29
Step-by-step explanation
ln =
ln(10) =
= 28.37 kJ/mol
Go = –RT ln K
T1 = 25oC K1 = 10
Go at T1 = –8.314 × 298 × 2.303 × log 10
= –5.71 kJ/mol
Go at T2 = –8.314 × 373 × 2.303 × log(100)
= –14.29 kJ/mol
ln(10) =
= 28.37 kJ/mol
Go = –RT ln K
T1 = 25oC K1 = 10
Go at T1 = –8.314 × 298 × 2.303 × log 10
= –5.71 kJ/mol
Go at T2 = –8.314 × 373 × 2.303 × log(100)
= –14.29 kJ/mol
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