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Equilibrium

Question
CBSEENCH11006390

A sample of HI(g) is placed in a flask at a pressure of 0·2 atm. At equilibrium, the partial pressure of HI(g) is 0·04 atm. What is Kp for the given equilibrium?

                      2 HI space left parenthesis straight g right parenthesis space space rightwards harpoon over leftwards harpoon space space space space space space straight H subscript 2 left parenthesis straight g right parenthesis space plus space straight I subscript 2 left parenthesis straight g right parenthesis

Solution

The given reversible reaction is 
                               
space space space space space space space space space space space space space space space space space space space space space space 2 HI left parenthesis straight g right parenthesis space rightwards harpoon over leftwards harpoon space space space space space space space space straight H subscript 2 left parenthesis straight g right parenthesis straight I space space space plus space straight I subscript 2 left parenthesis straight g right parenthesis
nitial space pressure space 0.2 space atm space space space space space space space space space space space space minus space space space space space space space space space space space space space minus
At space equilibrium space left parenthesis 0.2 minus straight p right parenthesis atm space space space space space space space straight p divided by 2 space atm space space space space space space straight p divided by 2 space atm
space space space space space space space space space space space space space space space space space space space space 0.04 space atm space space space space space space space space space space 0.08 space atm space space space space space 0.08 space atm
     (Decrease in the pressure of HI = 0.2 - 0.04  = 0.16 atm).
Applying the law of chemical equilibrium
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Putting the values in expression (1), we have
              straight K subscript straight p space equals space fraction numerator 0.08 space atm space cross times space 0.08 space atm over denominator left parenthesis 0.04 space atm right parenthesis squared end fraction space equals space 4.0