JEE Main 2021 — Chemical Equilibrium Question with Solution
From: JEE Main 2021 (Online) 18th March Evening Shift
Question
The gas phase reaction at 400 K has Go = + 25.2 kJ mol-1.
The equilibrium constant KC for this reaction is ________ 102. (Round off to the Nearest Integer).
[Use : R = 8.3 J mol1 K1, ln 10 = 2.3 log10 2 = 0.30, 1 atm = 1 bar]
[antilog (0.3) = 0.501]
The equilibrium constant KC for this reaction is ________ 102. (Round off to the Nearest Integer).
[Use : R = 8.3 J mol1 K1, ln 10 = 2.3 log10 2 = 0.30, 1 atm = 1 bar]
[antilog (0.3) = 0.501]
Enter your answer
Show full solutionCorrect answer: 1.66
Correct answer
1.66
Step-by-step explanation
Using formula,
G = RTln(Kp)
25.2 103 = 8.3 400 2.3 log (Kp)
Kp = 103.3
= 103 0.501
= 5.01 104 Bar1
Also,
m3/mole
L/mol
G = RTln(Kp)
25.2 103 = 8.3 400 2.3 log (Kp)
Kp = 103.3
= 103 0.501
= 5.01 104 Bar1
Also,
m3/mole
L/mol
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