The following reaction is carried out at a particular temperature in a closed vessel of definite volume: `CO_(2)(g)+H_(2)(g)hArrCO(g)+H_(2)O(g)`. Initially, partial pressures of `CO_(2)(g) and H_(2)(g)` are 2 atm and 1 atm respectively and that of `CO_(2)(g)` at equilibrium is 1.4 atm. calculate equilibrium constant of the reaction.

Explanation

For the gien reaction, `K_(p)=(p_(CO)xxp_(H_(2)O))/(p_(CO_(2))xxp_(H_(2)))`
As given, partial pressure of `CO_(2)(g)` at equilibrium `(p_(CO_(2)))=1.4` atm. Hence, decrease in pressure of `CO_(2)(g)` until the equilibrium is reached=(2-.14)atm=0.6atm
According to the equation, 1 mol `CO_(2)(g)` reacts with 1 mol `H_(2)(g)` to form 1 mol CO(g) and 1 mol `H_(2)O(g)`. therefore, if the pressure of `CO_(2)(g)` is reduced by 0.6 atm, then the pressure of `H_(2)(g)` will also be reduced by 0.6 atm, then the pressure of `H_(2)` will also be reduced by 0.6 atm and the pressure of each of CO(g) and `H_(2)O(g)` will be 0.6 atm.
Hence, partial pressures of different constituent at equilibrium will be as follows:
`{:(,CO_(2)(g),+,H_(2)(g),hArr,CO(g),+,H_(2)O(g)),("Pressure (atm):",1.4,,(1-0.6)=0.4,,0.6,,0.6):}`
`therefore K_(p)=(p_(CO)xxpH_(2)O)/(p_(CO_(2))xxp_(H_(2)))=(0.6xx0.6)/(1.4xx0.4)=0.64`