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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?`
2 years ago

Arun
23045 Points
```							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 RegardsArun (askIITians forum expert)
```
2 years ago
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