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Equilibrium Intro Disturbing Equilibrium
Disturbing Equil'm Answers Pretest 3.1 Law of Chem Equil'm Pretest 3.2
Acids and Bases

Solutions to Law of Chemical Equilibrium

1.

a. K= [Ag(NH3)2+1]/([Ag+1][NH3]2)

  1. K= [NO2]2/[N2O4]
  2. K = [H2]3[N2]/[NH3]2
  3. K = [CH3CO2-1][H+1]/[CH3CO2H]
  4. K = [H+1][OH-1] ; do not include water ( liquid)
  5. K = [Cu+2]/[Ag+1]2 ; do not include solids

2.

  1. K = ([CO][H2O2])/([H2][CO2])

TRIAL

K ( at 900 C) so it should give the same value

1

1.29

2

1.29

3

1.29

3.

[Cu+1]2/[Cu+2] = 1 X 10-6

Isolating for [Cu+2], the reactant, we obtain:

[Cu+2] =1 000 000[Cu+1]2 . So obviously there is a lot more reactant.

4.

D

5.

K = 0.020

6.

3A = 2B + C

Substance

A

B

C

Initial moles/L

5.0/1 =5

0

0

changing moles/L

5 – 4 = 1.0

B/A = 2/3, so B = (2/3)*1.0 = 0.67

C/A = 1/3, so C =

(1/3)*1.0 = 0.33

moles/L at equilibrium

4.0 /1=4

0.67

0.33

a. From table, B = 0.67 moles/L(1.0 L) = 0.67 moles and C = 0.33 moles/L(1.0 L) = 0.33 moles.

  1. K = [B]2[C]/[A]3 = [0.67 ]2[0.33 ]/[4 ]3 = 0.0023.
  2. No.
  3. Two moles is more than the equil’m number that we calculated, so the answer is no. A catalyst will cause only 0.67 moles of B to form faster. To obtain more, you would have to manipulate temperature or start with more A.
  4. 3A = 2B + C
  5. Substance

    A

    B

    C

    Initial moles/L

    5.0 moles/1.0 L

    0

    0

    changing moles/L

    5 – 2 = 3.0

    B/A = 2/3, so B = (2/3)*3.0 = 2.0

    C/A = 1/3, so C =

    (1/3)*3.0 = 1.0

    moles/L at equilibrium

    2.0/1.0L

    2.0

    1.0

  6. K = [B]2[C]/[A]3 = [2.0]2[1.0 ]/[ 2.0 ]3 = 0.50.
  7. Endothermic; a higher temperature allowed more products to form, raising the K

7.

2 NH3 = 3 H2 + N2

Substance

NH3

H2

N2

Initial moles/L

5.0/2.0 = 2.5

0

0

changing moles/L

NH3/H2 = 2/3, so NH3 = 2/3( 0.75) = 0.50

1.5/2.0 L = 0.75 moles/L

0.5 /2.0 L = 0.25 moles/L

moles/L at equilibrium

2.5 - 0.50 = 2.0

1.5/2.0 L = 0.75 moles/L

0.5/2.0 L = 0.25 moles/L

K = [H2]3[N2]/[NH3]2 = [0.75]3[0.25]/[2.0 ]2 = 0.026

 

b. 2 NH3 = 3 H2 + N2

Substance

NH3

H2

N2

Initial moles/L

4.0/2.0L = 2.0

0

0

changing moles/L

1.5/2.0 = 0.75 moles/L

H2/ NH3 =3/2 , so H2 =3/2 (0.75) = 1.125

N2/ NH3 =1/2, so N2 = ½(0.75) = 0.375

moles/L at equilibrium

2.0 - 0.75 = 1.25

1.125

0.375

[N2] = 0.375 moles/L

n = CV

=0.375 moles/L (2.0 L) = 0.75 moles

K = [H2]3[N2]/[NH3]2 = [1.125]3[0.375]/[1.25 ]2 = 0.34

8.a. 3A = 2B + C

Substance

A

B

C

Initial moles/L

8.0

0

0

changing moles/L

A/B = 3/2, so A =

(3/2)* 4.0= 6.0

4.0

2.0

moles/L at equilibrium

8.0 - 6.0 = 2.0

4.0

2.0

K = [B]2[C]/[A]3 = [4.0 ]2[2.0 ]/[ 2.0 ]3 = 4.0.

 

b. 3A = 2B + C

Substance

A

B

C

Initial moles/L

16/2.0 L= 8

0

0

changing moles/L

8– 2.0 = 6.0

B/A = 2/3, so B = (2/3)* 6.0 = 4.0

C/A = 1/3, so C =

(1/3)*6.0= 2.0

moles/L at equilibrium

4.0/2 = 2.0

4.0

2.0

K = [B]2[C]/[A]3 = [4.0]2[2.0]/[2.0 ]3 = 4.0.

Temperature was identical! Same K was obtained.

Answers for #s 9 and 10 and extra practice 11--> 15