Solutions to Law of Chemical Equilibrium

1.

a. K= [Ag(NH₃)₂⁺¹]/([Ag⁺¹][NH₃]²)

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

2.

1. K = ([CO][H₂O₂])/([H₂][CO₂])

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.

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.

2. K = [B]²[C]/[A]³ = [0.67 ]²[0.33 ]/[4 ]³ = 0.0023.
3. No.
4. 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.
5. 3A = 2B + C

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.

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

b. 2 NH₃ = 3 H₂ + N₂

[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

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

b. 3A = 2B + C

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