At `986^(@)`C the value of equilibrium constant `(K_c)` for the reaction, `CO(g)+H_2O(g)toCO_2(g)+H_2(g)` is 0.63At this temperature, 1 mol of `H_2O`(g) is allowed to react with 3 mol of CO(g). This results in the above equilibrium in which the observed pressure is 2.0 atm. Find the number of moles of `H_2`(g) produced at equilibrium and the partial pressure of each gas

Explanation

Suppose, a moles of CO reacts at equilibrium.
`CO(g)+H_2O(g) Initial number of moles : 3 1 0 0
Number of moles at
Equilibrium: `3-prop" "1-prop" "prop" "prop`
Total number of moles at equilibrium
`=3-prop+1-prop+prop+prop=4`
At equilibrium : Pco`=((3-prop)/(4))xx2=1/2(3-prop)` atm
`P_(H_2O)=((1-prop)/4)xx21/2(1-prop)atm`
`P_(CO_2)=prop/4xx2=prop/2=P_(H_2)`
`K_p=(P_(CO_2)xxP_(H_2))/(P_(CO)xxP_(H_2O))=(prop/2xxprop/2)/(((3-prop)/(2))((1-prop)/(2)))=(prop^2)/((2-prop)(1-prop))`
For the given reaction : `K_p=K_c` or `/_\n=0`
`:.(prop^2)/((3-prop)(1-prop))=0.63` or `prop=0.683`
`:.` Number of moles of `H_2` at equilibrium=0.683 mol
Pco=`1/2(3-0.683)atm=1.16atm`
`P_(H_2O)=1/2(1-0.683)atm=0.16atm`
`P_(CO_2)=1/2P_(H_2)=1/2xx0.683=0.34atm`