Unit 7 Equilibrium Homework AP
Directions: For each question, provide your work and final answer below the question. Without work, no
credit will be given. Check significant figures and units.
C(s) + H2O(g) ↔ CO(g) + H2(g)
ΔHº = +131kJ
1. 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.
2. Ammonium hydrogen sulfide is a crystalline solid that decomposes as follows:
NH4HS(s)  NH3(g) + H2S(g)
(a) Some solid NH4HS is placed in an evacuated vessel at 25ºC. After equilibrium is attained, the total
pressure inside the vessel is found to be 0.659 atmosphere. Some solid NH4HS remains in the vessel
at equilibrium. For this decomposition, write the expression for KP and calculate its numerical value
at 25ºC.
(b) Some extra NH3 gas is injected into the vessel containing the sample described in part (a). When
equilibrium is reestablished at 25ºC, the partial pressure of NH3 in the vessel is twice the partial
pressure of H2S. Calculate the numerical value of the partial pressure of NH3 and the partial pressure
of H2S in the vessel after the NH3 has been added and the equilibrium has been reestablished.
(c) In a different experiment, NH3 gas and H2S gas are introduced into an empty 1.00 liter vessel at
25ºC. The initial partial pressure of each gas is 0.500 atmospheres. Calculate the number of moles
of solid NH4HS that is present when equilibrium is established.
Unit 7 Equilibrium Homework AP
Directions: Read the passage below and answer the summary questions that follow in complete sentences.
Haber Process
Princeton.edu
Early in the twentieth century several chemists tried and failed to produce ammonia from atmospheric nitrogen.
The enormous technical problems associated with the process were first solved by German chemist Fritz Haber
(with the invaluable help of Robert Le Rossignol, who developed and built the necessary high-pressure
devices). They first demonstrated their success in the summer of 1909, producing ammonia from air drop by
drop, at the rate of about a cup every two hours. The process was purchased by the German chemical company
BASF, which assigned Carl Bosch the difficult task of scaling up Haber's tabletop machine to industrial-level
production. Haber and Bosch were later awarded Nobel prizes, in 1918 and 1931 respectively, for their work in
overcoming the chemical and engineering problems posed by the use of large-scale, continuous-flow, highpressure technology. Ammonia was first manufactured using the Haber process on an industrial scale in 1913 in
BASF's Oppau plant in Germany. During World War I, production was shifted from fertilizer to explosives,
particularly through the conversion of ammonia into a synthetic form of Chile saltpeter, which could then be
changed into other substances for the production of gunpowder and high explosives (the Allies had access to
large amounts of saltpeter from natural nitrate deposits in Chile that belonged almost totally to British
industries; Germany had to produce its own). It has been suggested that without this process, Germany would
not have fought in the war, or would have had to surrender years earlier.
Nitrogen (N2) is very unreactive because the molecules are held together by strong triple bonds. The Haber
process relies on catalysts that accelerate the scission of this triple bond. Two opposing considerations are
relevant to this synthesis: the position of the equilibrium and the rate of reaction. At room temperature, the
equilibrium is strongly in favor of ammonia, but the reaction doesn't proceed at a detectable rate. The obvious
solution is to raise the temperature, but because the reaction is exothermic, the equilibrium constant (using atm
units) becomes 1 around 150° or 200 °C. Above this temperature, the equilibrium quickly becomes quite
unfavorable at atmospheric pressure. Thus one might suppose that a low temperature is to be used and some
other means to increase rate. However, the catalyst itself requires a temperature of at least 400 °C to be
efficient.
Pressure is the obvious choice to favor the forward reaction because there are 4 moles of reactant for every 2
moles of product, and the pressure used (around 200 atm) alters the equilibrium concentrations to give a
profitable yield. Economically, though, pressure is an expensive commodity. Pipes and reaction vessels need to
be strengthened, valves more rigorous, and there are safety considerations of working at 200 atm. In addition,
running pumps and compressors takes considerable energy. Thus the compromise used gives a single pass yield
of around 15%.
Another way to increase the yield of the reaction would be to remove the product (i.e. ammonia gas) from the
system. In practice, gaseous ammonia is not removed from the reactor itself, since the temperature is too high;
but it is removed from the equilibrium mixture of gases leaving the reaction vessel. The hot gases are cooled
enough, whilst maintaining a high pressure, for the ammonia to condense and be removed as liquid. Unreacted
hydrogen and nitrogen gases are then returned to the reaction vessel to undergo further reaction.
Unit 7 Equilibrium Homework AP
The most popular catalysts are based on iron promoted with K2O, CaO, SiO2, and Al2O3. The original Haber–
Bosch reaction chambers used osmium as the catalyst. The reaction mechanism, involving the heterogeneous
catalyst, is believed to involve the following steps:
N2 (g) → N2 (adsorbed)
N2 (adsorbed) → 2 N (adsorbed)
H2(g) → H2 (adsorbed)
H2 (adsorbed) → 2 H (adsorbed)
N (adsorbed) + 3 H(adsorbed)→ NH3 (adsorbed)
NH3 (adsorbed) → NH3 (g)
Experimental evidence points to reaction 2 as being the slow, rate-determining step. This is not unexpected
since the bond broken, the nitrogen triple bond, is the strongest of the bonds that must be broken.
Equilibrium
Scienceclarified.com
Hemoglobin, a protein containing iron, is the material in red blood cells responsible for transporting oxygen to
the cells. Each hemoglobin molecule attaches to four oxygen atoms, and the equilibrium conditions of the
hemoglobin-oxygen interaction can be expressed thus: Hb (aq) + 4O2(g) ⇋ Hb(O2)4 (aq) , where "Hb" stands
for hemoglobin. As long as there is sufficient oxygen in the air, a healthy equilibrium is maintained; but at high
altitudes, considerable changes occur. At significant elevations above sea level, the air pressure is lowered, and
thus it is more difficult to obtain the oxygen one needs. The result, in accordance with Le Châtelier's principle,
is a shift in equilibrium to the left, away from the oxygenated hemoglobin. Without adequate oxygen fed to the
body's cells and tissues, a person tends to feel light-headed. When someone not physically prepared for the
change is exposed to high altitudes, it may be necessary to introduce pressurized oxygen from an oxygen tank.
This shifts the equilibrium to the right. For people born and raised at high altitudes, however, the body's
chemistry performs the equilibrium shift—by producing more hemoglobin, which also shifts equilibrium to the
right.
When someone is exposed to carbon monoxide gas, a frightening variation on the normal hemoglobin-oxygen
interaction occurs. Carbon monoxide "fools" hemoglobin into mistaking it for oxygen because it also bonds to
hemoglobin in groups of four, and the equilibrium expression thus becomes: Hb(aq) + 4CO(g) ⇋ Hb(CO)4(aq).
Instead of hemoglobin, what has been produced is called carboxyhemoglobin, which is even redder than
hemoglobin. Therefore, one sign of carbon monoxide poisoning is a flushed face. The bonds between carbon
monoxide and hemoglobin are about 300 times as strong as those between hemoglobin and oxygen, and this
means a shift in equilibrium toward the right side of the equation—the carboxyhemoglobin side. It also means
that K for the hemoglobin-carbon monoxide reaction is much higher than for the hemoglobin-oxygen reaction.
Due to the affinity of hemoglobin for carbon monoxide, the hemoglobin puts a priority on carbon monoxide
bonds, and hemoglobin that has bonded with carbon monoxide is no longer available to carry oxygen. Carbon
monoxide in small quantities can cause headaches and dizziness, but larger concentrations can be fatal. To
reverse the effects of the carbon monoxide, pure oxygen must be introduced to the body. It will react with the
carboxyhemoglobin to produce properly oxygenated hemoglobin, along with carbon monoxide:
Hb(CO)4(aq) + 4O2(g) ⇋ Hb(O2)4(aq) + 4CO(g). The gaseous carbon monoxide thus produced is dissipated
when the person exhales.
Unit 7 Equilibrium Homework AP
Questions:
1. Use the elementary steps given in the catalyst mechanism to create the overall reaction for the
production of ammonia from its elements.
2. Identify any intermediates in the reaction mechanism.
3. Explain why the N22N step is the rate determining step in terms of energy and bonds.
4. Explain how changes in temperature affect the reaction equilibrium.
5. Explain how changes in pressure affect the reaction equilibrium.
6. Explain how changes in ammonia concentration affect the reaction equilibrium.
7. Why do people feel light headed on planes?
8. Explain carbon monoxide poisoning.
9. In terms of Le Chatelier’s principle, explain how to reverse the effects of CO poisoning.