Doris F.
Ch. 15 & 16 equilibrium test [100 points] AP Chemistry
Show your work, include equation / formula, proper units in your work / answer, and proper
number of significant figures in your answer.
1. write the equilibrium constant expression for the following reactions: [20 points]
a. equilibrium involving solutions of sodium chloride & silver nitrate expressed as its net ionic
equation.
NaCl(aq) + AgNO3(aq)  AgCl(s) + NaNO3(aq)
Ag+ + Cl-  AgCl(s)
Net Ionic equation:
Kc =
1
[Ag+] [Cl−]
b. C2H6 (g) + O2 (g)⇄ CO2 (g) + H2O (l)
2 C2H6 (g) + 7 O2 (g)⇄ 4 CO2 (g) + 6 H2O (l)
Kc =
[𝐶𝑂2]^4
[O2]^7[C2H6]^2
c. CaCO3 (s)⇄ CaO (s) + CO2 (g)
Kc =
[𝐶𝑂2]
1
d. NO (g) + O2 (g)⇄ NO2 (g)
2 NO (g) + O2 (g)⇄ 2 NO2 (g)
Kc =
[𝑁𝑂2]^2
[O2] [NO]^2
e. CH3NH2 (a weak base) + water ⇄
CH3NH2 (a weak base) + H2O ⇄ OH-(aq) + CH3NH3+(aq)
Kc =
[𝐶𝐻3𝑁𝐻3+]
[OH−] [CH3NH2]
2. For the reaction:
N2 (g) + H2 (g)⇄ NH3 (g) ,
Balance: N2 (g) + 3 H2 (g)⇄ 2 NH3 (g)
determine and justify / rationalize the change in [H2] and [NH3] due to: [16 points]
a. addition of He to the system
↑ He → nothing happens → chemical reaction @ equilibrium
b. removal of N2
↓ N2 → want to ↑N2 → ↑rate forward → ↓ [NH3] → ↑ [H2] &↑ [N2]
c. reduce the volume of the system
↓ Volume → ↑Pressure → want to ↓ Pressure →↓# of moles on the products’ side →↑rate forward → ↑
[NH3] → ↓[N2] & ↓[H2]
d. cooling the system
∆H = Σ b Products – Σ a Reactants
= [2 NH3 (g)] - [1 N2 + 3 H2]
= [2(-46.19 kJ/mol)] – [1(0 kJ/mol) + 3(0 kJ/mol)]
∆H = -92.38 kJ = exothermic reaction
↓ Temperature → want to ↑ Temperate (to generate heat) because the reaction is exothermic (see
calculation above) → ↑ rate forward → ↑ [NH3] → ↓ [H2] & ↓ [N2]
3. H2 (g) + I2 (g)⇄ 2 HI (g) ; Kc = 51 at 448 ◦C
If the initial:
PH2 = 0.059 atm
PI2 = 0.118 atm
then the equilibrium PHI is ___. [20 points]
i)
(𝑃𝐻𝐼)^2
Kp = (𝑃𝐼2)(𝑃𝐻2)
Kp = Kc (RT)(c+d)-(a+b)
= 51(0.0821 × 721K)(2-1-1)
Kp = 51
ii)
PI atm
PC atm
PE atm
H2 (g) + I2 (g) ⇄ 2 HI (g)
0.059
0.118
0
-x
-x
+2x
0.059-x 0.118-x
2x
(𝑃𝐻𝐼)^2
Kp = (𝑃𝐼2)(𝑃𝐻2) = 51 = (
(2𝑥)^2
0.059−x)(0.118−x)
(0.059-x)(0.118-x) foil  x2 – 0.177 + 0.006962
51 =
(2𝑥)^2
x^2 – 0.177 + 0.006962
(x2 – 0.177 + 0.006962)
51 =
4𝑥^2
x^2 – 0.177 + 0.006962
(x2 – 0.177 + 0.006962)
51x2 – 9.027x + 0.35506 = 4x2
-4x2
-4x2
-------------------------------------47x2 – 9.027x + 0.35506 = 0
Do quadratics:
𝑥=
−𝑏±√𝑏 2 −4𝑎𝑐
𝑥=
2𝑎
−(−9.027)±√(−9.027)2 −4(47×0.35506)
2(47)
x= 0.0552 or x= 0.137
0.059-0.137 = -0.078 for PH2 @ equilibrium & 0.118- 0.137 = -0.019 for P I2 @
equilibrium, so x ≠ 0.137, # @ equilibrium pressure can’t be negative.
so: x= 0.0552 & equilibrium pressure for HI = 2x = 2 (0.0552) = 0.1104 atm.
PHI E = 0.1104 atm
4. What is the pH of a 10.0 mM sodium benzoate (C6H5COONa) solution? [20 points]
C6H5COONa  C6H5COO- + Na+
C6H5COO- +
10×10-3 M
-x
10×10-30- x
[I]M
[C]M
[E]M
Kb =
𝐾𝑤
𝐾𝑎
Kb =
H2O
⇄ C6H5COOH + OH0M
0M
+x
+x
x
x
10^−14
= 6.5×10^−5 = 1.54×10-10
[𝐶6𝐻5𝐶𝑂𝑂𝐻] [ 𝑂𝐻−]
[C6H5COO−]
𝑥^2
1.54×10-10(10×10^−3)−𝑥
=
; since Kb is such a small #, so approx. x ≈ 0
𝑥^2
(10×10-3) 1.54×10-10(10×10^−3) (10×10-3)
√1.54 × 10 − 12 = 𝑥2
1.24×10-6 = x = [OH-]
pOH = -log [OH-]
pOH= 5.91
14 – pOH = pH
14 – 5.91 = 8.09 = pH
5. fill-in the below table and show / label your work [24 points]
[H+]
25.0 mM
d. 1.33×10-13 M
g. 1.78×10-7 M
j. 2.09×10-5 M
10^−14
a) [OH-] =
[𝐻+]
[OH-]
a. 2.0×10-13 M
75.0 mM
h. 5.62×10-8 M
k. 4.79×10-10 M
pH
b. 1.60
e. 12.88
6.75
l. 4.68
10^−14
= 25.0×10^−3 = 2.0×10-13 M
b) pH = -log [H+] = -log (25.0×10-3) = 1.60
c) pOH = -log [OH-] = -log (4.0x10-13) = 12.40
pOH
c. 12.40
f. 1.12
i. 7.25
9.32
10^−14
10^−14
d) [H+] = [𝑂𝐻−] = 75.0×10^−3 = 1.33×10-13 M
e) pH = -log [H+] = -log (1.33×10-13) = 12.88
f) pOH = -log [OH-] = -log (75.0x10-3) = 1.12
g) pH = -log [H+]  10^(-pH) = 10^(-6.75) = 1.78×10-7 M
h) [OH-] =
10^−14
[𝐻+]
10^−14
= 1.78×10^−7 = 5.62×10-8 M
i) ) pOH = -log [OH-] = -log (5.62x10-8) = 7.25
10^−14
10^−14
j) [H+] = [𝑂𝐻−] = 4.79×10^−10 = 2.09×10-5 M (do 2nd to get info for l)
k) pOH = -log [OH-]  10^(-pOH) = 10^(-9.32) = 4.79×10-10 M (do 1st to get info for j)
l) pH = -log [H+] = -log (2.09×10-5) = 4.68 (do 3rd)