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Equilibrium

Question
CBSEENCH11006435

Dihydrogen gas used in Haber’s process is produced by reacting methane from natural gas with high temperature steam. The first stage of two stage reaction involves the formation of CO and H2. In second stage, CO formed in first stage is reacted with more steam in water gas shift reaction,
CO left parenthesis straight g right parenthesis space plus space straight H subscript 2 straight O left parenthesis straight g right parenthesis space space rightwards harpoon over leftwards harpoon space space space CO subscript 2 left parenthesis straight g right parenthesis space plus space straight H subscript 2 left parenthesis straight g right parenthesis
If a reaction vessel at 400°C is charged with an equimolar mixture of CO and steam such that pCO = 4·0 bar, what will be the partial pressure of H2 at equilibrium ? Kp = 10·1 at 400°C. 


 

Solution

Suppose the partial pressure of H2 at equilibrium  = p bar
                         CO left parenthesis straight g right parenthesis space plus space straight H subscript 2 straight O space space space rightwards harpoon over leftwards harpoon space space space space space CO subscript 2 left parenthesis straight g right parenthesis space plus space straight H subscript 2 left parenthesis straight g right parenthesis
Initial pressure      4.0 bar   4.0bar =   -            -
At eqm.          
                         (4 -p)     (4-p) =      p            p
Applying the law of chemical equilibrium,
     straight K subscript straight p space equals space fraction numerator straight p subscript CO subscript 2 end subscript space cross times space straight p subscript straight H subscript 2 end subscript over denominator straight p subscript CO cross times space straight p subscript straight H subscript 2 straight O end subscript end fraction space space space space... left parenthesis 1 right parenthesis    
Putting the values in expression (1), we have
   straight K subscript straight p space equals space fraction numerator straight p space cross times space straight p over denominator left parenthesis 4 minus straight p right parenthesis thin space left parenthesis 4 minus straight p right parenthesis end fraction
space space space space space equals space fraction numerator straight p squared over denominator left parenthesis 4 minus straight p right parenthesis squared end fraction space equals space 0.1 space space space space space space space left square bracket given right square bracket
therefore space space space space fraction numerator 4 over denominator 4 minus straight p end fraction space equals space square root of 0.1 end root space equals space 0.316
therefore space space space space space straight p space equals space 1.264 space minus space 0.316 straight p
space or space space space 1.316 straight p space equals space 1.264 space or space space straight p space equals space 0.96 space bar