JEE Main 2021 — Chemical Equilibrium Question with Solution
From: JEE Main 2021 (Online) 16th March Morning Shift
Question
For the reaction at 495 K, rG = 9.478 kJ mol1.
If we start the reaction in a closed container at 495 K with 22 millimoles of A, the amount of B in the equilibrium mixture is ____________ millimoles.
(Round off to the Nearest Integer). [R = 8.314 J mol1 K1; ln 10 = 2.303]
If we start the reaction in a closed container at 495 K with 22 millimoles of A, the amount of B in the equilibrium mixture is ____________ millimoles.
(Round off to the Nearest Integer). [R = 8.314 J mol1 K1; ln 10 = 2.303]
Enter your answer
Show full solutionCorrect answer: 20
Correct answer
20
Step-by-step explanation
Go = RT ln Keq
9.478 103 = 495 8.314 ln Keq
ln Keq = 2.303 = ln 10
So, Keq = 10
Now, A(g) B(g)
\matrix{ {t = 0} & {22} & 0 \cr {t = t} & {22 - x} & x \cr }
x = 20
So, millimoles of B = 20
9.478 103 = 495 8.314 ln Keq
ln Keq = 2.303 = ln 10
So, Keq = 10
Now, A(g) B(g)
\matrix{ {t = 0} & {22} & 0 \cr {t = t} & {22 - x} & x \cr }
x = 20
So, millimoles of B = 20
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