Show that a =√k/c , where a is degree of dissociation , k is the equ

1.	Show that a =√k/c , where a is degree of dissociation , k is the equlibrium constant . c is the concentration of the electrolyte
Answer» Answer: According to law of ACTIVE mass Explanation: CONSIDER the reaction A + B -------> C + D Initial moles of A and B are one Initial moles of C and D are zero At equilibrium moles of A and B are 1 — a At equilibrium moles of C and D are a Equilibrium CONCENTRATION of A is 1—a/V Equilibrium concentration of B is 1—a/V Equilibrium concentration of C is a/V Equilibrium concentration of D is a/V according to law K = [C][D]/[A][B] = a²/(a—1)² at equilibrium 1-a ≈ 1 therefore K = a²C a= √K/C

Show that a =√k/c , where a is degree of dissociation , k is the equlibrium constant . c is the concentration of the electrolyte

Answer»

Answer:

According to law of ACTIVE mass

Explanation:

CONSIDER the reaction

A + B -------> C + D

Initial moles of A and B are one

Initial moles of C and D are zero

At equilibrium moles of A and B are 1 — a

At equilibrium moles of C and D are a

Equilibrium CONCENTRATION of A is 1—a/V

Equilibrium concentration of B is 1—a/V

Equilibrium concentration of C is a/V

Equilibrium concentration of D is a/V

according to law

K = [C][D]/[A][B]

= a²/(a—1)²

at equilibrium 1-a ≈ 1

therefore

K = a²C

a= √K/C