JEE Main 2021 — Chemical Equilibrium Question with Solution
From: JEE Main 2021 (Online) 17th March Evening Shift
Question
Consider the reaction
The temperature at which KC = 20.4 and KP = 600.1, is ____________ K. (Round off to the Nearest Integer). [Assume all gases are ideal and R = 0.0831 L bar K1 mol1]
The temperature at which KC = 20.4 and KP = 600.1, is ____________ K. (Round off to the Nearest Integer). [Assume all gases are ideal and R = 0.0831 L bar K1 mol1]
Enter your answer
Show full solutionCorrect answer: 354
Correct answer
354
Step-by-step explanation
N2O4(g) 2NO2(g)
ng = 2 1 = 1
KP = KC(RT)ng
600.1 = 20.4 (0.0831 T)1
T = = 354 K
ng = 2 1 = 1
KP = KC(RT)ng
600.1 = 20.4 (0.0831 T)1
T = = 354 K
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