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Question

# The mixture of one mole of CO2 and one mole of H2 attains equilibrium at a temperature of 2500C and total pressure of 0.1 atm for the change CO2(e)+H2(e)⇌CO2(e)+H20e, calculate Kp is the analysis of final reaction mixture shows 0.16 volume fraction of CO ?

A
0.46
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B
0.63
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C
0.22
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D
0.82
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Solution

## The correct option is C 0.22 CO(g)+H2(g)⇌CO2(g)+H2O(g)Initially 1mole 1mole 0 0At eqm 1−x 1−x x xTotal moles at equilibrium= 1−x+1−x+x+x=2PCO2=PH2=1−x2×0.1 and PCO=PH2O=x2×0.1Since mole fraction of CO=0.16∴x2=0.16⇒x=0.32KP=PCO×PH2OPCO2×PH2=(x2×0.1)2(1−x2×0.1)2 =(x1−x)2=(0.321−0.32)2 =0.22

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