Consider the following reaction at certain temperature :
`H_(2) (g) + Cl_(2) (g) hArr 2HCl (g)`
The mixing of 1 mol of `H_(2)` with 4 moles of `Cl_(2)` from x moles of HCl at equilibrium. If we add 5 moles of `H_(2)` at equilibrium then another `2x` moles of HCl are produced. Then find `K_("eq")` for above reaction.

Answer

4

Explanation

`{:(,H_(2) (g),+,Cl_(2) (g),hArr,2HCl (g),,),("Initially",1,,4,,-,,),("Equilibrium",1 - (x)/(2),,4 - (x)/(2),,x,,),("New Equilibrium",6 - (3x)/(2),,4 - (3x)/(2),,3x,["After adding 5 moles of" H_(2)],):}`
`K_(C) = (x^(2))/((1 - (x)/(2))(4 - (x)/(2))) = ((3x)^(2))/((4 - (3x)/(2)) (6 - (3x)/(2)))`
`x = 1.6`
`K_(eq) = (1.6 xx 1.6)/((1 - (1.6)/(2))(4 - (1.6)/(2)))`
`K_(eq) = 4`