`K_(c )` for `CO(g)+H_(2)O(g) hArr CO_(2)(g)+H_(2)(g)` at `986^(@)C` is `0.63`. A mixture of `1` mol `H_(2)O(g)` and `3` mol `CO_(2)(g)` is allowed to react to come to an equilibrium. The equilibrium pressure is `2.0` atm.
a. Hoe many moles of `H_(2)` are present at equilibrium ?
b. Calculate partial pressure of each gas at equilibrium.

Explanation

`{:(,CO(g),+,H_(2)O,hArr,CO_(2)(g),+,H_(2)(g)),("Intial moles",3,,1,,0,,0),(" Moles at eqm.",3-x,,1-x,,x,,x):}`
Total no. of moles at equilibrium `= 3-x x+1 - x + x + x = 4 `
` K_(c) = (x xxx)/((3-x)(1-x)) , i.e., 0*63 = (x^(2))/(3 + x^(2) - 4x ) `
On solving , it gives `x = 0*681 " " ( x =-(bpm sqrt(b^(2) - 4ac))/(2a) )`
` :. " Moles of " H_(2) " present at eqm . " = 0* 681 "mole" `
Total pressure at eqm. = 2 atm ( Given )
Total moles at eqm . = 4
` :. p_(CO) = ( 3- 0*681 )/4 xx 2 = 1*16 "atm " , p_(H_(2)O) = (1-0*681)/4 xx 2 = 1*16 " atm " , p_(CO_(2)) = p_(H_(2)) = (0* 681)/4 xx 2 = 0* 34 "atm" `