UNIT 11 Practice Test
Equilibrium
Page 1 of 13
Do NOT write on this test. $0.10/page lost or damaged fee.
1.
In which of the following does the reaction go farthest to completion?
A. K = 105
B. K = 10–5
C. K = 1000
D. K = 100
E. K = 1
2.
The equilibrium constant, Kc, for the reaction
H2(g) + I2(g) → 2 HI(g)
is 55 at 425˚C. The value for the equilibrium constant will be changed if
A. concentrations are given in atmospheres instead of mol/L.
B. a catalyst is added to the reaction system.
C. the size of a reaction vessel is doubled.
D. the temperature is changed to 400˚C.
E. additional HI is added to the reaction system.
3.
Which of the following statements is true in a reaction system at equilibrium?
A. The number of collisions per unit time between reactants is equal to the number of collisions per unit time
between products.
B. Reactants are reacting to form products at the same rate as products are reacting to form reactants.
C. The product of the concentrations of the reactants is equal to the product of the concentrations of the products.
D. All concentrations of reactants and products are the same.
E. No products are being formed, and no reactants are being formed.
4.
Which of the following correctly describes the equilibrium constant for the gas-phase reaction between H2 and O2 to
form gaseous H2O
 H 2O 
 H 2 O2 
2
H 2O 

Kc 
 H 2 O2 
2
H 2O 

Kc 
2
 H 2  O2 
 H O2 
Kc  2
 H 2O 
K c   H 2O 
Kc 
A.
B.
C.
D.
E.
UNIT 11 Practice Test
Equilibrium
5.
The reaction of a mixture of SO2 and O2 at a given temperature is represented by the equation
SO2(g) + ½ O2(g) → SO3(g)
When the equilibrium is established, which of the following ratios is constant, regardless of the initial concentrations
of SO2 and O2
A.
B.
C.
D.
E.
6.
 SO3 
 SO2 O2 
 SO2 O2 
 SO3 
 SO3 
 SO2   12 O2 
2
 SO3 
2
 SO2  O2 
 SO3 
1
SO2 O2 
2
For the reaction system
CoO(s) + H2(g) → Co(s) + H2O(g)
at 550˚C, K = 67. The equilibrium constant expression is
A.
B.
C.
D.
E.
7.
Page 2 of 13
CoO  H 2 
Co H 2O 
Co H 2O 
CoO  H 2 
Co  H 2O 
H2 
H2 
 H 2O 
 H 2O 
H2 
Carbon tetrachloride reacts with oxygen at high temperatures to produce chlorine and carbonyl chloride.
Kc = 1.9 x 1019
2 CCl4(g) + O2(g) → 2 COCl2(g) + 2 Cl2(g)
What is Kc for the following:
COCl2(g) + Cl2(g) → CCl4(g) + ½ O2(g)
A.
B.
C.
D.
E.
5.2 x 10–20
2.3 x 10–10
4.4 x 109
–1.9 x 1019
1.9 x 1019
UNIT 11 Practice Test
Equilibrium
Page 3 of 13
8.
For the following reaction systems at 500 K
Kp = 1.7 x 10–2
I. 2 NOCl(g) → 2 NO(g) + Cl2(g)
Kp = 5.9 x 10–5
II. 2 NO2(g) → 2 NO(g) + O2(g)
Kp = 1.3 x 10–5
III. 2 SO(g) → 2 SO2(g) + O2(g)
the reaction's tendency to go to completion increases in the order
A. I < II < III.
B. III < II < I.
C. II < I < III.
D. II < III < I.
E. III < I < II.
9.
Consider the equilibrium N2(g) + 3 H2(g) → 2 NH3(g) at a certain temperature. An equilibrium mixture in a 4.00-L
vessel contains 1.60 mol NH3, 0.800 mol N2, and 1.20 mol H2. What is the value of Kc?
A. 9.00
B. 29.6
C. 3.37
D. 17.1
E. 7.41
10.
For the reaction system:
2HI(g) → H2(g) + I2(g)
Kc = 0.020 at 720 K.
If the initial concentrations of HI, H2, and I2 are all 1.50x10–3 M at 720 K, which one of the following statements is
correct?
A. The system is at equilibrium.
B. The concentrations of HI and I2 will increase as the system is approaching equilibrium.
C. The concentrations of H2 and HI will decrease as the system is approaching equilibrium.
D. The concentration HI will increase as the system is approaching equilibrium.
E. The concentrations of H2 and I2 will increase as the system is approaching equilibrium.
11.
In an experiment, 0.100 mol H2 and 0.100 mol I2 are mixed in a 1.00-L container and the reaction forms HI. If Kc =
50.0 for this reaction, what is the equilibrium concentration of HI?
I2(g) + H2(g) → 2 HI(g)
A. 0.156 M
B. 0.0780 M
C. 7.07 M
D. 0.200 M
E. 0.0384 M
12.
Exactly 1.0 mol N2O4 is placed in an empty 1.0-L container and is allowed to reach equilibrium described by the
equation
N2O4(g) → 2 NO2(g)
If at equilibrium the N2O4 is 20% dissociated, what is the value of the equilibrium constant (in units of mol/L) for the
reaction under these conditions?
A. 0.05
B. 0.2
C. 0.5
D. 20
E. 400
UNIT 11 Practice Test
Equilibrium
13.
Page 4 of 13
For the equilibrium
PCl5(g) → PCl3(g) + Cl2(g)
Kc = 4.0 at 228˚C. Pure PCl5 is added to the reaction system. At equilibrium there is 0.040 M PCl5. What is the
concentration of PCl3 in this system?
A. 0.16 M
B. 0.20 M
C. 0.40 M
D. 0.80 M
E. 1.6 M
14.
Hydrogen iodide undergoes decomposition according to the equation
2HI(g) → H2(g) + I2(g)
The equilibrium constant at 425˚C for this system is 0.018. If 1.0 mol each of H2, I2, and HI were placed together in a
1-L container at 425˚C, then
A. the concentration of HI would decrease.
B. the value of K would increase to 1.0.
C. because of reaction, the total number of molecules would decrease.
D. because of reaction, the total number of molecules would increase.
E. the concentration of H2 would decrease.
The information that follows is for the following two questions, 15 & 16.
At a given temperature, 0.300 mol NO, 0.200 mol Cl2, and 0.500 mol ClNO were placed in a 25.0-L container. The
following equilibrium is established:
2 ClNO(g) → 2 NO(g) + Cl2(g)
15.
At equilibrium, 0.600 mol ClNO are present. How many moles of Cl2 are present at equilibrium?
A. 0.050
B. 0.100
C. 0.150
D. 0.200
E. 0.250
16.
What is the value of the equilibrium constant Kc?
A. 4.45 x 10–4
B. 6.67 x 10–4
C. 0.111
D. 0.167
E. 1500
17.
Which of the following equilibria would not be affected by pressure changes at constant temperature?
A. FeO(s) + CO(g) → Fe(s) + CO2(g)
B. CaCO3(s) → CaO(s) + CO2(g)
C. H2O(g) → H2(g) + ½ O2(g)
D. 2 NO(g) + O2(g) → 2 NO2(g)
E. PCl5(l) → PCl3(g) + Cl2(g)
UNIT 11 Practice Test
Equilibrium
18.
Page 5 of 13
Consider the reaction
∆H˚ = 84.3 kJ
S2Cl2(l) + CCl4(l) → CS2(g) + 3 Cl2(g)
If the above reactants and products are contained in a closed vessel and the reaction system is at equilibrium, the
number of moles of CS2 can be increased by
A. adding some S2Cl2 to the system.
B. removing some CCl4 from the system.
C. decreasing the size of the reaction vessel.
D. increasing the temperature of the reaction system.
E. adding some Cl2 to the system.
19.
The reaction of carbon monoxide and iodine pentoxide as represented by the equation
5 CO(g) + I2O5(g) → I2(g) + 5 CO2(g)
is endothermic. The yield could be increased by
A. increasing the pressure.
B. decreasing the pressure.
C. increasing the temperature.
D. decreasing the temperature.
E. decreasing the volume of the reaction vessel.
20.
Which of the following equilibria would not be affected by volume changes at constant temperature?
A. 2 CO(g) + O2(g) → 2 CO2(g)
B. 2 NO2(g) → N2O4(g)
C. 2 NO(g) + 3 F2(g) → 2 F3NO(g)
D. BF3(g) + NH3(g) → H3NBF3(s)
E. O3(g) + NO(g) → NO2(g) + O2(g)
21.
For the reaction system H2(g) + I2(g) → 2 HI(g), K = 55 at 425˚C.
For the system HI(g) → ½ H2(g) + ½ I2(g), K at 425˚C is
A. –55
B. 55
C. 0.13
D. 0.018
E. 3.3 x 10–4
22.
For which of the following systems at equilibrium and at constant temperature will decreasing the volume cause the
equilibrium to shift to the right?
A. N2(g) + 3 H2(g) → 2 NH3(g)
B. 2 H2O(g) → 2H2(g) + O2(g)
C. 2 NO2(g) → 2 NO(g) + O2(g)
D. NH4Cl(s) → NH3(g) + HCl(g)
E. H2(g) + Cl2(g) → 2 HCl(g)
23.
In which of the following reactions does an instantaneous increase in the volume of the reaction vessel favor
formation of the products?
A. MgO(s) + CO2(g) → MgCO3(s)
B. PCl5(g) → PCl3(g) + Cl2(g)
C. H2(g) + I2(g) → 2 HI(g)
D. N2(g) + O2(g) → 2 NO(g)
E. N2(g) + 3 H2(g) → 2 NH3(g)
UNIT 11 Practice Test
Equilibrium
24.
For which of the following reactions are the numerical values of Kp and Kc the same?
A. 2 CO(g) + O2(g) → 2 CO2(g)
B. PCl5(g) → PCl3(g) + Cl2(g)
C. N2(g) + 3 H2(g) → 2 NH3(g)
D. 2 H2O(g) → 2H2(g) + O2(g)
E. H2(g) + I2(g) → 2 HI(g)
Page 6 of 13
UNIT 11 Practice Test
Equilibrium
Answer Key:
1.
A
2.
D
3.
B
4.
C
5.
D
6.
E
7.
B
8.
B
9.
B
10.
D
11.
A
12.
B
13.
C
14.
E
15.
C
16.
B
17.
A
18.
D
19.
C
20.
E
21.
C
22.
A
23.
B
24.
E
Page 7 of 13
UNIT 11 Practice Test
Equilibrium
Page 8 of 13
Answer Justifications:
1.
A
By going “farthest to completion,” the statement implies that the reaction creates a lot of products. Since the
equilibrium constant is determined by taking a ratio of products to reactants, the higher the equilibrium constant
shows the further completion of the reaction. This means that the farther completion is Response A.
2.
D
Only one variable alters the value of the equilibrium constant – that is the temperature.
3.
B
Response B is true since it states the definition of dynamic equilibrium.
If Response A had stated that the number of collisions at the correct energy per unit time between reactants is equal to
the number of collisions at the correct energy per unit time between products, then it would have been correct since it
followed the Collision Theory.
Response C and D are completely untrue as the probability that the quantity (and multiplied version, “product”) of the
concentrations of the reactants is equivalent to the quantity (and multiplied version, “product”) concentrations of the
products is slim.
Response E is definitively what will be macroscopically observed, even though the reaction continues to occur both in
the forward reaction and the reverse.
4.
C
First, determine the gas-phase reaction between H2 with O2 and H2O. The balanced reaction looks like:
2 H2 + O2 → 2 H2O
Then, write the expression of the equilibrium constant, K, by placing products in the numerator, with the coefficient
as its exponent, and placing reactants in the denominator, with their coefficients as their exponents:
 H 2O 
Kc 
2
 H 2  O2 
2
5.
D
The only ratio that is ever constant is the equilibrium constant, presenting a ratio between created products to
remaining reactants. Therefore, the only ratio that is constant is the expression of the equilibrium constant.
This should appear as:
 SO3 
 SO2  O2 
However, the reaction can occur in coefficients of multiples, such as:
2 SO2(g) + O2(g) → 2 SO3(g)
So, as long as the ratio provides a stoichiometric relationship to the reaction, the ratio remains constant.
 SO3    SO3  , the stoichiometric relationship is constant.
2
2
 SO2 O2   SO2  O2 
2
Since
6.
E
For any reaction, developing an equilibrium constant expression first requires that the reaction’s equation is balanced.
This equation is balanced.
Next, only aqueous solutions and gaseous species can be entered into the equilibrium constant expression, not solid or
liquid species. Therefore, CoO and Co can not be included in the equilibrium constant expression.
This leaves only the gaseous water and hydrogen – Response E.
UNIT 11 Practice Test
Equilibrium
7.
Page 9 of 13
B
First, compare the original reaction with the reaction being provided. The original reaction has coefficients that are
two times larger than those of the new reaction. In addition, the new reaction is occurring in the opposite direction of
that of the original reaction. This means that the equilibrium constant of the new reaction is the square-root and
inverse of the original equilibrium constant:
K new 
1
 2.3x1010
19
1.9 x10
8.
B
The tendency of a reaction to go to completion implies that the reaction has headed far to right, creating a larger
number of products and reducing the number of reactants. This would, in turn, create an equilibrium constant value
that was larger. Therefore, the reaction that has headed in the forward direction the most is I. The reaction that has
gone to completion the least, with the lowest Kp, is III. This places III lower than II, which is lower than I.
9.
B
First, write the equilibrium expression for the reaction:
 NH 3 
Kc 
3
 N 2  H 2 
2
Next, fill in the mole values corresponding to each species after converting them to molarities:
2
2
1.60mol

0.400

4.00
L


Kc 

 29.6
3
3
 0.800mol
 1.20mol

0.2000.300
4.00 L  
4.00 L 

10.
D
In order to make an interpretation of the system and how it may or may not approach equilibrium, solve for the
quotient. Remember that the equation for the quotient is equivalent to the equation for the equilibrium constant:
Q
 H 2  I 2 
2
 HI 
Plug in the concentrations for the corresponding species:
1.50 x103  1.50 x103 
 1.00
Q
2
1.50 x103 
This shows that Q is larger than Kc, meaning that there are currently too many products than there should be at
equilibrium. This will cause the reaction to reverse and create more reactants. This makes response A incorrect since
the reaction is not at equilibrium. The responses B and C suggest that the quantities of the reactants and products will
be increasing; since the Q value shows that only the reactants will increase, neither of these responses is correct.
Finally, response D suggests that the reaction will head to the left, while response E suggests that the reaction will
head to the right. Since this reaction will head to the left, response D is correct.
UNIT 11 Practice Test
Equilibrium
11.
Page 10 of 13
A
In order to solve this question, an ICE chart must be created.
+
H2(g)
→
I2(g)
I
0.100mol
0.100mol
1.00 L
1.00 L
2 HI(g)
0
C
–x
–x
+2x
E
(0.100M – x)
(0.100M – x)
(2x)
Now, plug these quantities into the equilibrium constant expression:
 HI   50.0 
2 x
Kc 
 H 2  I 2 
 0.1  x  0.1  x 
2
2
50.0  0.01  0.2 x  x 2   4 x 2
0.5  10 x  50 x 2  4 x 2
0  46 x 2  10 x  0.5
solver; x = 0.0780 M, 0.139M
Both numbers appear correct, however, if the 0.139 value were plugged into the expression for H2 or I2 (0.1 – x), it
would create a negative value (0.1 – 0.139 = –0.039). Therefore, the correct x value is 0.0780.
Now, plug this value into the expression for HI: (2x) = (2 * 0.078) = 0.156 M
12.
B
In order to solve this question, an ICE chart must be created.
→
2 NO2(g)
N2O4(g)
0
I
1.0mol
1.0 L
C
–x
+2x
= – (1.0M)(0.2)
= +(2)(0.2)
= – 0.2
= +0.4
E
(1.0 – 0.2)
(0 + 0.4)
= 0.80 M
= 0.40M
Now, plug these quantities into the equilibrium constant expression:
 NO2    0.4
Kc 
 N 2O4  0.8
2
2
 0.2
**This is a good example of a multiple choice question that you could do without a calculator.
13.
C
In order to solve this question, an ICE chart must be created.
→
PCl3(g)
+
Cl2(g)
PCl5(g)
I
Plenty of reactant
0
0
C
–x
+x
+x
E
0.040 M
x
x
Now, plug these quantities into the equilibrium constant expression:
Kc 
 PCl3 Cl2   4.0   x  x 
 PCl5 
0.040
0.16 = x2
0.40 = x = [PCl3] = [Cl2]
**This is a good example of a multiple choice question that you could do without a calculator.
UNIT 11 Practice Test
Equilibrium
14.
Page 11 of 13
E
Because the quantities of the reactants and products are provided, but are not necessarily at equilibrium, it is a good
opportunity to utilize a quotient, Q. Remember that the expression for the quotient is identical to the expression of the
equilibrium constant, but the values that can be entered into the quotient can be at any point during the reaction. Set
up the quotient expression and then plug in the concentrations, not moles:
2
1.0mol 
1L 


 1.0
Q
 H 2  I 2  1.0mol 1L  1.0mol 1L 
 HI 
2
Now, compare this value to the equilibrium constant. Since Q > K, there are currently too many products so the
reaction will proceed in the reverse direction.
Response B makes no sense, since the value of K will not change unless the temperature changes.
Neither response C nor D make sense since if the reaction goes to the right, it would mean that every two molecules of
HI that are reacted, one molecule of hydrogen and one molecule of iodine would be created, and vice versa for the
reaction going to the left. This means that this reaction would maintain the number of total molecules regardless of
the direction of shift.
Response A suggests that the reaction is headed in the forward direction, which has just been shown otherwise. And
finally, response E suggests that the reaction is headed in the reverse direction, which it is.
15.
C
In order to solve this question, an ICE chart can be created. Notice, the ICE chart will be completed in moles.
→
2 NO(g)
+
Cl2(g)
2 ClNO(g)
I
0.500 mol
0.300 mol
0.200 mol
C
+2x
–2x
–x
E
0.600 mol
(0.3 – 2x)
(0.2 – x)
Notice that the quantity of ClNO has increased from initial to equilibrium. This shows that the amount of ClNO that
was changed is a positive 0.1 moles. Since the balanced equation provides a 2 ClNO : 2 NO : 1 Cl2 ratio, the addition
of 0.1 mole to the ClNO means that the Cl2 will be reduced by 0.05 mole.
Cl2 = 0.2 – 0.05 = 0.15 moles
16.
B
Completing the ICE chart from question 15 above,
→
2 NO(g)
+
Cl2(g)
2 ClNO(g)
I
0.500 mol
0.300 mol
0.200 mol
C
+2x
–2x
–x
E
0.600 mol
(0.3 – 0.1)
(0.2 – 0.05)
= 0.2 mol
= 0.15 mol
Plug the equilibrium mole quantities into the equilibrium constant expression as molarities:
2
0.2mol
 0.15mol
 0.008 2 0.006

  6.67 x104
25
L
25
L



 
Kc 
2
2
0.6mol

 0.024
25L 

17.
A
Reactions that involve gases tend to be affected by pressure changes, while reactions that do not have any gases will
definitively not be affected by pressure changes. Notice that no reaction is entirely void of gases.
Reactions that involved gases with equivalent moles of gas on both sides of the reaction will have rates that are
affected by a pressure change equally. So look at how many gaseous particles appear on each side of the reactions:
Response A: 1 gaseous reactant to 1 gaseous product
Response B: 0 gaseous reactants to 1 gaseous product
Response C: 1 gaseous reactant to 1 ½ gaseous products
Response D: 3 gaseous reactants to 2 gaseous products
Response E: 1 gaseous reactant to 2 gaseous products
UNIT 11 Practice Test
Equilibrium
Page 12 of 13
The only reaction that has equivalent moles of gas on both sides of the reaction is response A. Therefore, it will not
be affected by altering the pressure.
18.
D
In order to increase the number of CS2 moles, the equilibrium reaction must be stressed so that it causes a shift toward
the CS2, which is on the product’s side. This can be caused by removing the other product (Cl2), by increasing the
temperature for this endothermic reaction, and by decreasing the pressure/ increasing the volume of the system. The
only response that follows one of these stresses is response D.
Notice that altering the quantities of reactants (as in responses A or B) does not cause any shift, since the reactants do
not have concentrations, as they are liquids.
19.
C
The “yield,” remember, means the creation of products. Therefore, the equilibrium reaction must be stressed so that it
shifts to the product’s side. This can be caused by adding more reactant or by increasing the temperature for this
endothermic reaction. The only response that follows one of these stresses is response C.
Notice that altering the pressure of this reaction (as in responses A or B, or similarly by response E) does not cause
any shift, since there are 6 moles of gaseous reactants and 6 moles of gaseous products.
20.
E
Reactions that involve gases tend to be affected by pressure changes, while reactions that do not have any gases will
definitively not be affected by pressure changes. The same is true of volume changes. Notice that no reaction is
entirely void of gases.
Reactions that involved gases with equivalent moles of gas on both sides of the reaction will have rates equally
affected by a volume change. So look at how many gaseous particles appear on each side of the reactions:
Response A: 3 gaseous reactants to 2 gaseous products
Response B: 2 gaseous reactants to 1 gaseous product
Response C: 5 gaseous reactants to 2 gaseous products
Response D: 2 gaseous reactants to 0 gaseous products
Response E: 2 gaseous reactants to 2 gaseous products
The only reaction that has equivalent moles of gas on both sides of the reaction is response E. Therefore, it will not be
affected by altering the volume.
21.
C
First, notice that the first reaction presented has the same temperature as the second reaction. Next, compare the
second reaction to the first reaction.
The second reaction is reversed, compared to the first reaction. This means that the new K value will be the inverse of
the original K value.
The second reaction is also ½ the coefficients of the first reaction. This means that the new K value will be the square
root of the original K value.
Putting this all together,
K new 
22.
1
1

 0.134
55
K old
A
Reactions that involve gases tend to be affected by volume changes. If a decrease in volume causes a shift to the
right, the number of gaseous particles on the right side of the reaction must be lower than the number of gaseous
particles on the left side. So look at how many gaseous particles appear on each side of the reactions:
Response A: 4 gaseous reactants to 2 gaseous products
Response B: 2 gaseous reactants to 3 gaseous productx
Response C: 2 gaseous reactants to 3 gaseous products
Response D: 0 gaseous reactants to 2 gaseous products
Response E: 2 gaseous reactants to 2 gaseous products
UNIT 11 Practice Test
Equilibrium
Page 13 of 13
The only reaction that has fewer moles of gas on the product’s side is response A. Therefore, it is the only one that
will shift right when the volume is decreased.
23.
B
Reactions that involve gases tend to be affected by volume changes. If an increase in volume causes formation of
products, the number of gaseous particles on the right side of the reaction must be higher than the number of gaseous
particles on the left side. So look at how many gaseous particles appear on each side of the reactions:
Response A: 1 gaseous reactant to 0 gaseous products
Response B: 1 gaseous reactant to 2 gaseous products
Response C: 2 gaseous reactants to 2 gaseous products
Response D: 2 gaseous reactants to 2 gaseous products
Response E: 4 gaseous reactants to 2 gaseous products
The only reaction that has more moles of gas on the product’s side is response B. Therefore, it is the only one that
will shift right when the volume is increased.
24.
E
Reactions that involve gases relate their pressure equilibrium constant to their concentration equilibrium constant
through the equation:
Kp = Kc (RT) ∆n
In order for Kp and Kc to be equal, the quantity (RT) ∆n must be removed by having a ∆n = 0. This only occurs when
the number of gaseous product moles is equal to the number of gaseous reactant moles. So look at how many gaseous
particles appear on each side of the reactions:
Response A: 3 gaseous reactants to 2 gaseous products
Response B: 1 gaseous reactant to 2 gaseous products
Response C: 4 gaseous reactants to 2 gaseous products
Response D: 2 gaseous reactants to 3 gaseous products
Response E: 2 gaseous reactants to 2 gaseous products
The only reaction that has equal gaseous moles on the both sides of the reaction is response E. Therefore, it is the
only one that will have equal Kp and K values.