For reaction A(g) + B(g) we start with 2 moles of A and B each. At equilibrium `0.8` moles of AB is formed. Then how much of A changes to ABA. `20%`B. `40%`C. `60%`D. `4%`

For reaction A(g) + B(g) we start with 2 moles of A and B each. At equ

1.	For reaction A(g) + B(g) we start with 2 moles of A and B each. At equilibrium `0.8` moles of AB is formed. Then how much of A changes to ABA. `20%`B. `40%`C. `60%`D. `4%`
Answer» Correct Answer - B `{:(,,A(g),+,B(g),hArr,AB(g)),(,"Initial mole",2,,2,,0),(,"At equilibrium",2-x,,2-x,,x=0.8):}` `:. %` of A changed to AB = `(0.8)/(2)xx100=40%`

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For reaction A(g) + B(g) we start with 2 moles of A and B each. At equilibrium `0.8` moles of AB is formed. Then how much of A changes to ABA. `20%`B. `40%`C. `60%`D. `4%`

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