AB211−α0.535⇌Aα0.465+2B2α0.9
1.9×25=nT×0.08206×300
nT=1.93=1+2α
α=0.465
Kp=(1.930.535×1.9)(1.930.465×19)(1.930.93×1.9)2
≃73×10−2atm2
JEE Main 2021 — Chemistry Physical Chemistry
A homogeneous ideal gaseous reaction AB2(g)⇌A(g)+2B(g) is carried out in a 25litre flask at 27∘C. The initial amount of AB2 was 1 mole and the equilibrium pressure was 1.9atm. The value of Kp is x×10−2. The value of x is (Integer answer) [R=0.08206dm3atmK−1mol−1]
Held on 26 Feb 2021 · Verified 6 Jul 2026.
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