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If in a question in equilibrium chapter,we have asked to find out the equilibrium constant for a reaction like compound A gives rise to two compounds 2B+C given that number of moles of compound A is 4 moles but at equilibrium only 50% of compound A is dissociated into 2B and C. Then how will we find equilibrium constant (Kc).

If in a question in equilibrium chapter,we have asked to find out the equilibrium constant for a reaction like compound A gives rise to two compounds 2B+C given that number of moles of compound A is 4 moles but at equilibrium only 50% of compound A is dissociated into 2B and C. Then how will we find equilibrium constant (Kc).

Grade:11

1 Answers

Vinayak Saxena
12 Points
6 years ago
A--->2B+C Then equilibrium constant Kc is given as {[B]^2 [C]/[A]}Now according to stichiometry if 1 mole of A is consumed them 2 mole of B is formed and 1 mole of C is Therefore now 50% of A is consumed i.e 2 moles thus 4 mole of B & 2 mole of C is formed therefore Kc={(4)^2×2/2 }=Here (4)^2 refers square of 4 =16&[B] is concentration of B at equilibrium Thank u

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