Question

The equilibrium constant of the reaction, ` A_(2) (g) + B_(2) (g) hArr 2AB (g) " at " 100^(@) C " is " 50.` . If a one litre flask containing one mole of `A_(2) ` is connected to a two litre flask containing two moles of `B_(2)` how many moles of AB will be formed at 373 K ?

Answer 1

` {: (,A_(2),+,B_(2),hArr,2AB),(" Intial amounts:",1 "mole",,2 "mole",,0),("Amounts at eqm:",1-x,,2-x,,2x),(" Molar concs at eqm.",(1-x)/3,,(2-x)/3,,(2x)/3 "mol"L^(1)):}`
` K = ([AB]^(2))/([A_(2)][B_(2)]) or 50 = (2x^(2))/(((1-x)/3)((2-x)/3))`
On solving , we get `x= 0*955 ` mole
Hence, no. of moles of AB formed at eqm. `= 2 xx 0* 955 = 1*91 ` moles