Please help with answer and steps, as I am not understanding this chapter. Thank

ID: 837486 • Letter: P

Question

Please help with answer and steps, as I am not understanding this chapter. Thanks in advance!

The reversible chemical reaction A+B (equilibrium symbol here) C+D has the following equilibrium constant: Kc=[C][D][A][B]=7.2

Part A

Initially, only A and B are present, each at 2.00 M. What is the final concentration of A once equilibrium is reached?

Express your answer to two significant figures and include the appropriate units. A=

Part B

What is the final concentration of D at equilibrium if the initial concentrations are [A] = 1.00M and[B] = 2.00M ?

Express your answer to two significant figures and include the appropriate units. D=

A + B <===> C + D ; Kc = 7.2

initially [A] = [B] = 2M and [C] = [D] = 0 M

let at equilibrium , [A] = [B] = (2-x) M and [C] = [D] = x M

thus, Kc = x2/(2-x)2

or, 7.2 = x2/(2-x)2

or, 7.2 = {x/(2-x)}2

or, sqrt(7.2) = {x/(2-x)}

or, x = 1.546

thus, at equilibrium [A] = [B] = (2-X) = 0.544 M

and [C] = [D] = x M = 1.546 M............(1)

now,

initially [A] = 1 M and [B] = 2M and [C] = [D] = 0 M

let at equilibrium , [A] = (1-x) M and [B] = (2-x) M and [C] = [D] = x M

thus, Kc = x2/{(2-x)*(1-x)}

or, 7.2 = x2/{(2-x)(1-x)}

or, 7.2 = {x2/(x2-3x + 2)}

or, x = 0.898

thus, at equilibrium [A] = (1-x) = 0.102 M and [B] = (2-x) = 1.102 M

and [C] = [D] = x M = 0.898 M

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