Q . THE EQUILIBRIUM :- H 2 (g) + CO(g) --------- --------- H 2 O(g) + CO(g) IS ESHTABLISHED IN A EVACUATED VESSEL AT 723 K STARTING WITH 0.1 mol OF H 2 AND 0.2 mol OF CO 2 . IF THE EQUILIBRIUM MIXTURE CONTAINS 10 MOLE PERCENT OF WATER VAPOUR , CALCULATE K p , GIVEN THAT EQUILIBRIUM PRESSURE IS 0.35 atm.

Question

Q . THE EQUILIBRIUM :- H 2 (g) + CO(g) --------- ---------> H 2 O(g) + CO(g) IS ESHTABLISHED IN A EVACUATED VESSEL AT 723 K STARTING WITH 0.1 mol OF H 2 AND 0.2 mol OF CO 2 . IF THE EQUILIBRIUM MIXTURE CONTAINS 10 MOLE PERCENT OF WATER VAPOUR , CALCULATE K p , GIVEN THAT EQUILIBRIUM PRESSURE IS 0.35 atm.

Naveen Kumar · Answer

For the reaction,
H2(g) + CO2(g)------------------> H2O(g) + CO(g)
0.1mol 0.2mol 0 0 …..............initially

0.1-x 0.2-x x x ….............at equilibrium

So total number of moles of all gaseous species at equilibrium= (0.1-x)+(0.2-x)+x+x
=0.3mol
mol % of water vapour at equilibrium=10(given)
⇒(x/0.3)*100=10
⇒x=0.03mol
Partial pressure of H2O (PH2O)=x*R*T/V
SimilarlyPCO=x*R*T/V
PH2=(0.1-x)*R*T/V
PCO2=(0.1-x)*R*T/V
V=volume of mixture

Kp=(PH2O*PCO)/(PH2*PCO2)
=x*x/(0.1-x)*(0.1-x)
=x2/(0.1-x)2
=0.076

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Q . THE EQUILIBRIUM :- H2 (g) + CO(g) --------- ---------> H2O(g) + CO(g) IS ESHTABLISHED IN A EVACUATED VESSEL AT 723 K STARTING WITH 0.1 mol OF H2 AND 0.2 mol OF CO2.IF THE EQUILIBRIUM MIXTURE CONTAINS 10 MOLE PERCENT OF WATER VAPOUR , CALCULATE Kp , GIVEN THAT EQUILIBRIUM PRESSURE IS 0.35 atm.