Question

# The reversible chemical reaction A+B⇌C+D has the following equilibrium constant: Kc=[C][D][A][B]=3.3 Part A Initially, only A...

The reversible chemical reaction

A+B⇌C+D

has the following equilibrium constant:

Kc=[C][D][A][B]=3.3

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] = 1.00 M and [B] = 2.00 M ?

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

K = C*D/(A*B)

K = 3.3

A = 2

B = 2

C = 0

D = 0

find A in equilibrium

A = 2-x

B = 2 -x

C = 0+x

D = 0+x

substitute all in

K = C*D/(A*B)

3.3= (x*x)/(2-x)^2

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

solve

sqrt(3.3)= x/(2-x)

1.8165 = x/(2-x)

0.55050x = 2-x

1.55050x = 2

x = 2/1.55050 = 1.289906

since

A = 2-x= 2-1.289906 = 0.710094 M of A

pretty similar

A = 1-x (this changes!)

B = 2 -x

C = 0+x

D = 0+x

substitute all in

K = C*D/(A*B)

3.3= (x*x)/(2-x)(1-x)

3.3 = x^2 /(2-3x+x^2)

solve

(2-3x+x^2) = 0.303*x^2

0.697 x^2 -3x +2 = 0

x = 0.8246

subsitute in d

D = 0+x = 0.8246 M

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