3.1.7 Practice: General Equilibrium
Practice
AP Chemistry Sem
Points Possible:25
Name:
Rodolfo Marquez
Valencia
Date: 3/15/2021
You will need to use a calculator and a periodic table to complete this activity.
1. The formation of water from its elements at 1200 K can be written as a series of
coupled reactions: (1 point
Suppose the system has the following concentrations
[H2] = 0.0041 M
[O2] = 0.0062 M
[H2O] = 2.3 M
Predict the direction in which the system will shift to reach equilibrium. Justify your
answer
Answer: With these concentrations that are given, the equilibrium direction will shift to
the right. This is because if more product is added to a reaction, the equilibrium will shift
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to the right to be in favor of the reactants.
2. The gure below shows two possible methods for the production of acetyl CoA from
acetate as part of the biosynthesis of fatty acids in the body. The rst method is a direct
reaction, and the second method involves coupled reactions
Direct Reaction
Coupled Reactions
Step 1
Step 2
a. The standard Gibbs free energy for the direct reaction, ΔG°', is 32 kJ/mol. (Note that
fi
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fi
the prime symbol is used for biological systems.) Calculate the equilibrium constant for
the reaction at a temperature of 298 K. Is the reaction favorable or unfavorable?
Explain. (1 point
Answer: Kc = 2.46 * 10-6, ∆Gº > 0 ; Therefore, the reaction is not favorable because it is
an endergonic reaction, meaning that the process needs some type of energy added to
move forward and create a reaction.
b. Calculate the equilibrium constant for each step of the coupled reactions. ΔG1°' is
about 11 kJ/mol for the rst reaction, and ΔG2°' is about –10 kJ/mol for the second
reaction. Is each step favorable or unfavorable? Explain. (1 point
Kc = 0.012, ∆Gº > 0 ; Therefore, the reaction is not favorable because it is an
endergonic reaction which requires added energy to proceed.
Kc = 56.6, ∆Gº < 0 ; Therefore, the reaction is favorable because it is an exergonic
reaction which releases energy and the reactants have more energy than the products.
c. Determine the net reaction for the coupled reaction, its Gibbs free energy value, and
its overall equilibrium constant. (1 point
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fi
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Answer: Acetyl CoA net reaction; Kc = 0.67, ∆Gº > 0 ; The reaction is not favorable.
d. Predict whether the direct method or the coupled reaction method would be more
likely to occur in the body to produce acetyl CoA from acetate. What might happen to
make the reaction even more likely to occur? (1 point
Answer: Acetyl CoA is produced much easier through the coupled reaction method.
Metabolic pathways make it more likely to occur because they are favored when acetate
concentrations increase.
3. The decomposition of copper(I) oxide to produce copper is unfavorable at a certain
temperature. (1 point
When the reaction is coupled with the oxidation of carbon to produce carbon monoxide,
however, the production of copper from copper(I) oxide is favorable
Explain why this is possible
Answer:
Kc = [CS2][H2]4 / [CH4][H2S]2
Kc = a*(4a)4/[(0.5-a)(0.75-a)2
a = 0.44/4 —> 0.11
Kc = 3.8*10-2
This is possible because ∆G depends on the ∆H and ∆S. So, if ∆H < 0 and ∆S > 0, the
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reaction is favorable.
4. Carbonyl bromide decomposes to produce carbon monoxide and bromine gas
according to the following equation
After the system reaches equilibrium at a certain temperature, a small amount of Br2 is
introduced, causing an increase in concentration of COBr2. Explain why this occurs. (1
point
Answer: When the concentration of a product increases, the shift of the equilibrium and
the backward directions indicates that the reaction has an increase in the concentration
of COBr2.
5. The mineral magnetite (Fe3O4) oxidizes to form hematite (Fe2O3) according to the
following equation
At 298 K, the equilibrium constant is Kc = 1.02 × 1086
a. Suppose the reaction takes place in a closed container. What effect would an
increase in the volume of the container have on Kc? (1 point
Answer: By increasing the volume of the container, the value of Kc remains the same
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b. Consider the reverse reaction
What is the equilibrium constant for this reaction at 298 K? (1 point
Answer: K = [O2] ; Since there is one mole of O2 in the reaction, the equilibrium constant
at 298 K is 1.
6. The equilibrium constant based on concentration is Kc = 190 for the reaction of
carbon monoxide with hydrogen gas at 1000 K
a. What can you infer about the relative rates of the forward and reverse reactions when
the gases have the following concentrations at this temperature? (1 point
Answer:
Qc = [CH4][H2O]/[CO][H2]3
Qc = [1.4][0.32]/[1.7][0.83]
Qc =0.3
Since Qc < Kc, the reaction will proceed in the forward direction. So, the relative rate of
the forward reaction is higher than the rate of the reverse reaction.
b. Use Le Châtelier's principle to explain the difference in reaction rates described in
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part a. (1 point
Answer: The reaction is not under equilibrium. So, Le Châtelier’s principle is not
application to conditions in part a.
7. The oxidation of sulfur dioxide is described by the following equation
At T = 950 K, Kp = 1.0 × 101. Predict whether the reaction will proceed toward the
product or reactants as equilibrium is approached when the gases have the following
partial pressures or concentrations at this temperature. In each case, justify your
prediction
a.
(1 point
Answer:
Qp = [PSO3]2/[PSO2]2[PO2
Qp = [0.00637]2/[0.0437]2[0.0237
Qp = 9.0 * 10-1
Kp/Qp = 11.1
Kp/Qp >
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1
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1
Kp > Qp
Since Kp is greater than Qp, the reaction will proceed toward the product because there
is less pressure for the products.
b
(1 point
Answer: The reaction will proceed toward the product again because there are more
moles of product in total than there are in reactants.
8. Consider the following reaction
The table shows the equilibrium concentrations at a certain temperature. *I referred to
the one on the course platform because I don’t know what happened here*
Concentration (mol/L)
Br2
1.82 × 10–2
CH4
4.33 × 10–3
HBr
3.27 × 102
CBr4
2.02 × 101
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Species
Calculate the equilibrium constant, K, at this temperature. (1 point
K= products/reactants *will not include solid or water in this equilibrium constant*
K = [HBr]4[CBr4]/[Br2]4[CH4
K = [3.27*102]4[2.02*101]/[1.82*10-2]4[4.33*10-3
K = 4.86 * 1020
9. Ethanol (C2H5OH) can be produced from the hydrocarbon ethylene (C2H4) for
industrial use by the following equation
At 600.0 K, the equilibrium constant based on pressure is Kp = 1.83 × 102. Gaseous
C2H4 and H2O are placed in a 1.2 L closed ask at 600.0 K. At equilibrium, the ask
contains 0.0062 mol of C2H4 and 0.041 mol of H2O
a. Determine the equilibrium constant, Kc, assuming the pressure of the gases is in
atmospheres. (1 point
Answer: Kc = 1.83 * 102 * 0.0821 * 600 = 9014.5
b. What is the partial pressure of C2H5OH at equilibrium? (1 point
Partial pressure of C2H4: 0.0062*0.0821*600/1.2 = 0.2545
Partial pressure of H2O: 0.041*0.0821*600/1.2 = 1.6830
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Partial pressure of C2H5OH: 1.83 * 102 * 0.25451 * 1.68305 = 78.4 atm
10. Answer the following questions about concentrations of chemical species at
equilibrium. Explain the estimate you make in each case
a. The equilibrium constant is K = 3.4 × 1090 for the reactio
at a certain temperature. Is the concentration of Cl2 very large or very small at
equilibrium? (1 point
Answer: The concentration of Cl2 is very small because the equilibrium constant is very
low, which essentially makes all concentrations very small.
b. The equilibrium constant is K = 0.274 × 1042 for the reactio
at a certain temperature. Is the concentration of C2H6 very large or very small at
equilibrium? (1 point
Answer: Again, the concentration of C2H6 is very large because the equilibrium constant
is astronomically large, which makes all concentrations very large
c. The equilibrium constant is K = 4.30 × 10–13 for the reactio
at a certain temperature. Is the concentration of NO very large or very small at
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equilibrium? (1 point
Answer: In this case, the concentration of NO will be very large because the equilibrium
constant is very large. It makes all of the concentrations large.
d. The equilibrium constant is K = 1.38 × 10–84 for the reactio
at a certain temperature. Is the concentration of GeO2 very large or very small at
equilibrium? (1 point
Answer: The same goes for this one. The concentration fo GeO2 is very small because
the equilibrium constant is very small. It is relative with all concentration
11. Predict the direction of the shift in equilibrium of the reaction, if any, resulting from
each of the following actions. In each case, write the equilibrium constant expression
and explain your reasoning for the direction of the equilibrium shift
a. The reaction of carbon monoxide and hydrogen proceeds to equilibrium
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Argon is then introduced into the system. (1 point
Equilibrium Constant: [CH3OH]/[Ar][CO][H2]2
Reasoning: If the volume of a system is constant, there will be no change in any way.
So, the equilibrium will not shift to any direction. Therefore, no effect will take place.
b. The production of glucose by photosynthesis, an endothermic reaction, proceeds to
equilibrium
The temperature is then increased. (1 point
Equilibrium Constant: [C6H12O6][O2]6 / [CO2]6[H2O]6
Reasoning: This reaction is a synthesis of glucose, which means it’s an endothermic
reaction. Endothermic reaction tend to go in a forward direction at a high temperature.
So, when increasing the temperature, energy will be more available in the surrounding
and the system can absorb more energy to proceed in the forward direction. When the
temperature rises, the equilibrium will shift to the forward direction.
c. A reaction between sulfur dioxide and oxygen occurs in a closed container with a
piston
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After the reaction reaches equilibrium, the piston is pressed down as shown. (1 point
Equilibrium Constant: [SO3]2 / [SO2]2 [O2
Reasoning: When the piston is pressed down, the volume of the system will reduce in
size. Therefore, the equilibrium will shift in the forward direction.
12. Consider the synthesis reaction shown below
a. Assume the reactants and the product are gases. Suppose K = 4.3 × 10–4 at T = 350
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ºC and the initial concentrations are
[X] = 2.0 × 10–2 M
[Y] = 1.0 × 10–3
[Z] = 1.5×10–2 M
Describe the stress in the system and how it will be relieved. Show your calculations. (1
point
Q = [Z] / [X][Y]
Q = [1.5*10-2] / [2.0*10-2][1.0*10-3
Q = 0.75 * 103
Q = 75
K = 4.3 *10-4 (given
Since K < Q, this means that we’ll have less products than reactants. Therefore, the
reaction will proceed in the backward direction.
b. Do the values of K and Q change as the system moves toward equilibrium? Explain.
(1 point
Answer: The value of K will remain constant. However, the value of Q is 750 and will
continue to change until it becomes equal to K, otherwise known as until reaching
equilibrium. However, since Q is greater than K, the value of Q decreases, and the
reaction changed to being in favor of the reactants to reach equilibrium. Therefore,
since the reaction is changing towards equilibrium, the of Q will decreases and the
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value of K increases until it reaches equilibrium.