Equilibrium Constant and Standard Free Energy Change
For the react...
Question
For the reaction A(g)⇌B(g)at495K,ΔrGo=−9.478kJmol−1. If we start this 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). [R8.314Jmol−1K−1;] [ln10=2.303]
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JEE MAIN 2021
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Solution
ΔGo=−RTlnKeq
ΔGo=−9.478kJ/mole
T=495KR=8.314Jmol−1
−9.478×103=−495×8.314×lnKeq
lnKeq=2.303
lnKeq=ln10
SoKeq=10
NowA(g)⇌B(g)
t=0220
t=t22−xx
Keq=[B][C]=x22−x=10
or x =20
So millmoles of B = 20