Question

# The reversible chemical reaction A+B⇌C+D has the following equilibrium constant: Kc=[C][D][A][B]=7.5 Part A Initially, only A...

The reversible chemical reaction A+B⇌C+D has the following equilibrium constant: Kc=[C][D][A][B]=7.5 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? Part B What is the final concentration of D at equilibrium if the initial concentrations are [A] = 1.00 M and [B] = 2.00 M ?

A+B⇌C+D

Kc=[C][D][A][B]=7.5

[A] = [B] = 2 M

Find [A] when in equilibrium

in the equilibrium

[A] = [B] = 2-x

[C] = [D] = +x

substitute in K

Kc=[C][D][A][B]=7.5

(x)^2 /(2-x)^2 = 7.5

x/(2-x) = sqrt(7.5)

x = (2-x) * 2.74

x = 5.477 -2.74x

3.7x = 5.477

x = 1.48

substitute in [A]

[A] = 2-x = 2-1.48 = 0.52

B)

Find D when

[A] = 1 M

[B] = 2 M

then in equilibrium

[A] = 1 -x

[B] = 2 -x

[C] = [D] = x

Substitute in K

Kc=[C][D][A][B]=7.5

7.5 = (x^2)/((2 -x)( 1 -x))

solve for x

7.5(2-3x+x^2) =x^2

7.5(2-3x+x^2) - x^2 =

6.5x^2-22.5x+15 = 0

x = 0.90 or 2.56

x = 0.90 since 2.5 is impossible

then

[D] = x = 0.90

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