Download Equilibrium Review Problems N2(g) + H2(g) NH3(g) 1. When 3.29

Equilibrium Review Problems
1
3
N2(g) +
H2(g) ⇌ NH3(g)
2
2
1. When 3.29 moles of nitrogen gas, N2 , and 2.95 moles of hydrogen gas, H2 , are placed in a 5.00 L evacuated container and heated
to 773 K, ammonia gas, NH3 , is formed according to the equation above. When equilibrium is established 0.565 mole of NH 3(g) is
present.
(a) Write the expression for the equilibrium constant, Kc , for the reaction represented above.
(b) Calculate the equilibrium concentrations, in mol L1 , of the following gases in the container at 773 K.
(i)
N2(g)
(ii)
H2(g)
(c) Calculate the value of the following equilibrium constants at 773 K.
(i)
Kc
(ii)
Kp
When the same reaction is carried out at 298 K, the number of moles of NH 3(g) present at equilibrium is much larger than 0.565
mole.
(d) Is the forward reaction endothermic or exothermic? Justify your answer.
(e) Predict the sign of the standard entropy change, S° , for the reaction. Explain.
(f) The value of the standard free energy change, G° , for the reaction is 16.5 kJ mol1. Calculate the value of the equilibrium
constant, K , at 298 K.
NH3(aq) + H2O(l0 ⇌ NH4+(aq) + OH(aq)
Kb = 1.80  10-5
2. Ammonia reacts with water as indicated in the reaction above.
(a) Write the equilibrium constant expression for the reaction represented above.
(b) Calculate the pH of a 0.150 M solution of NH3
A 100 mL sample of the 0.150 M solution of NH3 is titrated with 0.100 M HCl
(c) Determine the volume of HCl required to reach the equivalence point.
(d) Calculate the pH of the solution at the equivalence point of the titration.
To the above solution (at the equivalence point) 250 mL of 0.060 M Pb(NO3)2 are added
(e) Calculate the molar concentration of the Pb2+ ions in the solution.
(f) Will a precipitate form? Justify you answer. The Ksp for the salt in question is 1.6  105
MgF2(s)  Mg2+(aq) + 2 F-(aq)
In a saturated solution of MgF2 at 18ºC, the concentration of Mg2+ is 1.2110-3 molar. The equilibrium is represented by the equation
above.
(a) Write the expression for the solubility-product constant, Ksp, and calculate its value at 18ºC.
(b) Calculate the equilibrium concentration of Mg2+ in 1.000 liter of saturated MgF2 solution at 18ºC to which 0.100 mole of solid KF
has been added. The KF dissolves completely. Assume the volume change is negligible.
(c) Predict whether a precipitate of MgF2 will form when 100.0 milliliters of a 3.0010-3-molar Mg(NO3)2 solution is mixed with 200.0
milliliters of a 2.00l0-3-molar NaF solution at 18ºC. Calculations to support your prediction must be shown.
(d) At 27ºC the concentration of Mg2+ in a saturated solution of MgF2 is 1.1710-3 molar. Is the dissolving of MgF2 in water an
endothermic or an exothermic process? Give an explanation to support your conclusion.
C(s) + H2O(g)  CO(g) + H2(g)
Hº = +131kJ
A rigid container holds a mixture of graphite pellets (C(s)), H2O(g), CO(g), and H2(g) at equilibrium. State whether the number of moles
of CO(g) in the container will increase, decrease, or remain the same after each of the following disturbances is applied to the original
mixture. For each case, assume that all other variables remain constant except for the given disturbance. Explain each answer with a
short statement.
(a) Additional H2(g) is added to the equilibrium mixture at constant volume.
(b) The temperature of the equilibrium mixture is increased at constant volume.
(c) The volume of the container is decreased at constant temperature.
(d) The graphite pellets are pulverized.
A sample of solid U308 is placed in a rigid 1.500 L flask. Chlorine gas, Cl 2(g), is added, and the flask is heated to 862˚C. The equation
for the reaction that takes place and the equilibrium-constant expression for the reaction are given below.
U308(s) + 3 Cl2(g),  3 UO2Cl2(g) + O2(g)
KP 
(PUO2 Cl2 )3 (PO2 )
(PCl2 )3
When the system is at equilibrium, the partial pressure of Cl2(g) is 1.007 atm and the partial pressure of UO2Cl2(g) is 9.734×10-4 atm
(a) Calculate the partial pressure of O2(g) at equilibrium at 862˚C.
(b) Calculate the value of the equilibrium constant, KP, for the system at 862˚C.
(c) Calculate the Gibbs free-energy change, ∆G˚, for the reaction at 862˚C.
(d) State whether the entropy change, ∆S˚ for the reaction at 862˚C is positive, negative, or zero. Justify your answer.
(e) State whether the enthalpy change, ∆H˚, for the reaction at 862˚C is positive, negative, or zero. Justify your answer.
(f) After a certain period of time, 1.000 mol of O 2(g) is added to the mixture in the flask. Does the mass of U 308(s) in the flask
increase, decrease, or remain the same? Justify your answer.
36. CuO(s) + H2(g) <===> Cu(s) + H2
- 2.0 kilojoules
When the substances in the equation above are at equilibrium at pressure P and temperature T, the equilibrium can be shifted to
favor the products by
(A) increasing the pressure by means of a moving piston at constant T
(B) increasing the pressure by adding an inert gas such as nitrogen
(C) decreasing the temperature
(D) allowing some gases to escape at constant P and T
(E) adding a catalyst
57. For the reaction A(g)  B(g) + C(g), the equilibrium constant, Kp, is 2 x 10¯4 at 25 °C. A mixture of the three gases at 25 °C
is placed in a reaction flask and the initial pressures are PA = 2 atm, PB = 0.5 atm, and PC = 1 atm, At the instant of mixing, which
of the following is true for the reaction as written?
76. HgO(s) + 4 I¯ + H2O  HgI42
increase the concentration of HgI42¯?
(A) Increasing the concentration of OH¯
(B) Adding 6 M HNO3
(C) Increasing the mass of HgO present
(D) Increasing the temperature
(E) Adding a catalyst
29. In which of the following systems would the number of moles of the substances present at equilibrium NOT be shifted by a
change in the volume of the system at constant temperature?
(A) CO(g) + NO(g) <===> CO2(g) + 1/2 N2(g)
(B) N2(g) + 3 H2(g) <===> 2 NH3(g)
(C) N2(g) + 2 O2(g) <===> 2 NO2(g)
(D) N2O4(g) <===> 2 NO2(g)
(E) NO(g) + O3(g) <===> NO2(g) + O2(g)
48. PCl3(g) + Cl2(g)  PCl5(g) + energy
Some PCl3 and Cl2 are mixed in a container at 200 °C and the system reaches equilibrium according to the equation above.
Which of the following causes an increase in the number of moles of PCl 5 present at equilibrium?
I. Decreasing the volume of the container
II. Raising the temperature
III. Adding a mole of He gas at constant volume
(A) I only
(B) II only
(C) I and III only
(D) II and III only
(E) I, II, and III
51. 4 HCl(g) + O2(g)  2 Cl2(g) + 2 H2O(g) Equal numbers of moles of HCl and O2 in a closed system are allowed to reach
equilibrium as represented by the equation above. Which of the following must be true at equilibrium?
I. [HCl] must be less than [Cl2].
II. [O2] must be greater than [HCl].
III. [Cl2] must equal [H2O].
(A) I only
(B) II only
(C) I and III only
(D) II and III only
(E) I, II, and III
73 R. 2 SO2(g) + O2(g)  2 SO3(g) When 0.40 mole of SO3 and 0.60 mole of O2 are placed in an evacuated 1.00-liter flask,
the reaction represented above occurs. After the reactants and the product reach equilibrium and the initial temperature is
restored, the flask is found to contain 0.30 mole of S0 3. Based on these results, the expression for the equilibrium constant, K c, of
the reaction is
(A) (0.30)2 / [(0.45)(0.10)2]
(B) (0 30)2 / [(0.60)(0.40)2]
(C) (2 x 0.30) / [(0.45)(2 x 0.10)]
(D) (0.30) / [(0.45)(0.10)]
(E) (0.30) / [(0.60)(0.40)]
54. 2NO(g) + O2(g) 2 NO2
for the reaction represented above?
A) Decreasing the temperature
B) Increasing the temperature
C) Decreasing the volume of the reaction vessel
D) Increasing the volume of the reaction vessel
E) Adding a catalyst
decrease in the value of Keq
CBBEADAB
Answer:
(a) Ksp = [Mg2+][F-]2 = (1.2110-3)(2.4210-3)2
= 7.0910-9
(b) X = concentration loss by Mg2+ ion
2X = concentration loss by F- ion
[Mg2+] = (1.2110-3 - X) M
[F-] = (0.100 + 2.4210-3 - 2X) M
since X is a small number then (0.100 + 2.4210-3 - 2X)  0.100
Ksp = 7.0910-9 = (1.2110-3 - X)(0.100)2
X = 1.209291410-3
[Mg2+]= 1.2110-3-1.2092910-3 = 7.0910-7M
(c) [Mg2+] = 3.0010-3M  100.0 mL/300.0 mL = 1.0010-3M
[F-] = 2.0010-3 M  200.0 mL/300.0 mL = 1.3310-3 M
trial Ksp = (1.0010-3)(1.3310-3)2 = 1.7810-9
trial Ksp < = 7.0910-9,  no ppt.
(d) @ 18ºC, 1.2110-3 M MgF2 dissolves
@ 27ºC, 1.1710-3 M MgF2 dissolves
MgF2  Mg2+ + 2 F- + heat
dissolving is exothermic; if heat is increased it forces the equilibrium to shift left (according to LeChatelier’s Principle) and less
MgF2 will dissolve.
Answer
(a) CO will decrease. An increase of hydrogen gas molecule will increase the rate of the reverse reaction which consumes CO. A
LeChatelier Principle shift to the left.
(b) CO will increase. Since the forward reaction is endothermic (a H > 0) an increase in temperature will cause the forward reaction
to increase its rate and produce more CO. A LeChatelier Principle shift to the right.
(c) CO will decrease. A decrease in volume will result in an increase in pressure, the equilibrium will shift to the side with fewer gas
molecules to decrease the pressure, , a shift to the left.
(d) CO will remain the same. Once at equilibrium, the size of the solid will affect neither the reaction rates nor the equilibrium nor the
concentrations of reactants or products.
Answer:
(a) using the equation, the ratio is 3:1, UO2Cl2:O2;
4 3
PO2 
PUO2 Cl2
3

9.734 104
= 3.245×10-4 atm
3
4
(9.734 10 ) (3.245 10 )
= 2.931×10-13
3
(1.007)
(b)
KP 
(c)
∆G˚ = –RTlnK = -(8.31)(862 + 273) ln (2.931×10 -13) = 272 kJ
(d) positive; a solid (low entropy) and 3 gases are converting into a mixture of gases (high entropy)
(e) positive; ∆G˚ = ∆H˚ – T∆S˚; ∆H˚ = ∆G˚ + T∆S˚; from (c) and (d), ∆G˚ and ∆S˚ are both positive, making ∆H˚ positive
(f) increase; adding oxygen gas will cause a LeChâtelier Principle shift to the left, producing more reactants