HI(g)⇌21H2(g)+21I2(g)K1=[HI][H2]1/2[I2]1/2H2(g)+I2(g)⇌2HI(g)K2=[H2][I2][HI2 From Eqs (i) and (ii) ∵∴K12=K21K1=8.0K2=K121=821=641
NEET UG 2008 — Chemistry Inorganic Chemistry
The value of equilibrium constant of the reaction HI(g)⇌21H2(g)+21I2 is 8.0 The equilibrium constant of the reaction H2(g)+I2(g)⇌2HI(g) will be
Held on 30 Apr 2008 · Verified 9 Jul 2026.
161
641
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81
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