Four moles of hydrogen iodide were taken in a 10 litre flask kept at 800K. When equilibrium was attained, the mixture was found to contain 0.42 mole of iodine. Calculate the equilibrium constant for dissociation of HI.
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Four moles of hydrogen iodide were present initially. 0.42 mole of iodine were formed from 2×0.42=0.84 moles of HI. 0.42 moles of hydrogen were also formed. 4−0.84=3.16 moles of HI were present at equilibrium.
The equilibrium constant Kc=[H2][I2][HI]2 Kc=[ 10 L 0.42 mol ][ 10 L 0.42 mol ][ 10 L 3.16 mol ]2 Kc=[0.042][0.042][0.316]2 Kc=56.6