Equilibrium: Please solve the question in the image. Ans. is 0.66 M bu

1.	Equilibrium: Please solve the question in the image. Ans. is 0.66 M but I need the solution. THANKS A LOTWILL MARK BRAINLIESTTROLLERS WILL BE REPORTED IMMEDIATELY
Answer» Answer: The reaction STARTS when the tap is open . So, when the tap OPENS the volume of the actual mixture will be = 1+3 = 4 lt. so, the initial concentration of a is = moles of a /vol. of solution (in lt) = 2/4 = 0.5 mole the initial concentration of b is = moles of b/vol.of solution (in lt) = 4/4 = 1 mole a2 + b2 - 2ab at t=0 = 0.5 1 0 at t=equilibrium = 0.5 -X 1-x 2X kc = (2x)^2/(0.5-x)(1-x) GIVEN that kc = 4 solve this you get x = 0.33 but as you see that at equilibrium the concentration of ab is 2x hence , the required answer is 20.33 = 0.66 moles*

Equilibrium: Please solve the question in the image. Ans. is 0.66 M but I need the solution. THANKS A LOTWILL MARK BRAINLIESTTROLLERS WILL BE REPORTED IMMEDIATELY

Answer»

Answer:

The reaction STARTS when the tap is open .

So, when the tap OPENS the volume of the actual mixture will be = 1+3 = 4 lt.

so, the initial concentration of a is = moles of a /vol. of solution (in lt)

= 2/4 = 0.5 mole

the initial concentration of b is = moles of b/vol.of solution (in lt)

= 4/4 = 1 mole

a2 + b2 - 2ab

at t=0 = 0.5 1 0

at t=equilibrium = 0.5 -X 1-x 2X

kc = (2x)^2/(0.5-x)(1-x)

GIVEN that kc = 4

solve this you get x = 0.33

but as you see that at equilibrium the concentration of ab is 2x hence ,

the required answer is 2*0.33 = 0.66 moles