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Two moles HI were heated in a sealed tube at 440C till the equilibrium was reached.HIwas found to be 22% decomposed.The equilibrium constant for dissociation is?

Two moles HI were heated in a sealed tube at 440C till the equilibrium was reached.HIwas found to be 22% decomposed.The equilibrium constant for dissociation is?

Grade:11

1 Answers

Arun
25750 Points
6 years ago
Dear Student
 
2HI(g) ⇐⇒ H2(g) + I2(g) 

K = [H2][ I2] / [HI]^2 
22% of 2 moles of HI decompose = 0.22(2) = 0.44 
0.44 moles HI x 1 mole H2 / 2 moles HI = 0.22 
moles of I2 = moles of H2 
moles of HI remaining = 2.0 – 0.44 = 1.56 
The 
K = 0.22(0.22) / (1.56)^2 
K = 0.0199
 
Regards
Arun (askIITians forum expert)

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