CHEM 1412
Practice Exam I (Ch. 11, 12, 13): Zumdahl & Zumdahl
Multiple Choices: Please select one best answer.
Some questions include the application-type questions: similar concept but asked
reversely or in the other way.
Q 1. Consider “like dissolves like” rule, which of the following pair are soluble to each
other?
(A) CH3COCH3 and H2O
(B) CCl4 and NaCl
(C) C6H12O6 and C6H6
(D) C6H12O6 and CCl4
(E) C2H5OH and Br2
Hint: For 9th ed., see Section 11.3: structure effects: p.p. 517-518. Be sure to review
CHEM 1411 Molecular Polarity and Intermolecular Forces. “Like dissolves like” rule is
the summary for molecular solubility. “Like dissolves like” rule states that the polar
solute dissolves in polar solvent; while the nonpolar solute dissolves in nonpolar
solvent.
Which of the following pair are insoluble to each other?
(A) C2H5OH and H2O
(B) CH3COOH and C2H5OH
(C) CH3COOH and H2O
(D) CH3OCH3 and H2O
(E) CH3OH and CCl4
Hint: For 9th ed., see Section 11.3: structure effects or “Like dissolves like” rule.
Q 2. Which of the following action will increase the solubility of a gas in water?
(A) Increasing pressure
(B) Increasing temperature
(C) Decreasing the molar mass
(D) Decreasing the polarity
Hint: For 9th ed., see Section 11.3: p.p. 518-519: Pressure effects or say the
Henry’s Law: C = kP only works for gases. Pressure effects do not apply to liquids and
solids. Factors that affect solubility include temperature, pressure, and polarity of
molecule (or the structure).
Q 3. A 1.00 M Ba(OH)2 solution has a density of 1.50 g/mL. What is its molality (m)?
(A) 0.3
(B) 0.4
(C) 0.6
(D) 0.7
(E) 0.8
th
Hint: For 9 ed., see Section 11.1: p. 513 Interactive Example 11.2
Review from CHEM 1411:
What is the molarity of a solution made by dissolving 58.45 g NaCl in enough water to
give 1000 mL of solution?
(A) 0.5 M
(B) 1.0 M
(C) 1.5 M
(D) 2.0 M
(E) 2.5 M
th
Hint: For 9 ed., see Section 11.1: p. 512 Interactive Example 11.1.
Q 4. Which of the following aqueous solution has the lowest boiling point?
(A) 0.2 m KI
(B) 0.1 m K2SO4
(C) 0.2 m C6H12O6
(D) 0.2 m CH3COOH
th
Hint: For 9 ed., see Section 11.5 & section 11.7, Table 11.6 & Interactive Example 11.8:
Tb = Kb  msolute  i = Tb, solution – Tb, solvent
Tb, solution = Tb + Tb, solvent
1
So the greater the Tb , the higher the boiling point of solution;
The smaller the Tb , the lower the boiling point of solution.
Here i is the Van Hoff’s factor; Kb is affected by solvent only; solvent is water. Be sure to
review CHEM 1411 nonelectrolytes, strong electrolytes and weak electrolytes and the
writing of net ionic equation. For nonelectrolyte, i = 1; for strong electrolyte, i = sum of
coefficients of the product ions in the balanced net ionic equation.
Here (A) m x i = 0.2 x 2 = 0.4; (B) m x i = 0.1 x 3 = 0.3; (C) m x i = 0.2 x 1 = 0.2; (D) 0.2 x 1 =
0.2 < m x i < 0.2 x 2 = 0.4
Which of the following solution has the lowest freezing point?
(A) 1.0 m KI
(B) 0.5 m K2SO4
(C) 1.0 m C6H12O6
(D) 1.0 m CH3COOH
th
Hint: For 9 ed., see Section 11.5 & section 11.7, Table 11.6 & Interactive Example 11.9:
Tf = Kf  msolute  i = Tf, solvent – Tf, solution
Tf, solution = Tf, solvent – Tf
So the greater the Tf , the lower the freezing point of solution;
The smaller the Tf , the higher the boiling point of solution.
Here i is the Van Hoff’s factor; Kf is affected by solvent only; solvent is water. Be sure to
review CHEM 1411 nonelectrolytes, strong electrolytes and weak electrolytes and the
writing of net ionic equation. For nonelectrolyte, i = 1; for strong electrolyte, i = sum of
coefficients of the product ions in the balanced net ionic equation.
Here (A) m x i = 1.0 x 2 = 2.0; (B) m x i = 0.5 x 3 = 1.5; (C) m x i = 1.0 x 1 = 1.0; (D) 1.0 x 1 =
1.0 < m x i < 1.0 x 2 = 2.0
Q 5. Which of the following concentration units will be affected by temperature? Note
that increasing temperature results in increasing volume.
(A) percent by mass
(B) mole fraction
(C) molarity
(D) molality
Hint: Only the mathematical formulas involve the volume. So memorize the formulas.
What is the percent by mass for a solution made by dissolving 40 g of NaCl in 60 g of
water?
(A) 20%
(B) 40%
(C) 60%
(D) 80%
(E) 100%
th
Hint: For 9 ed., see Section 11.1 p. 511-512: Interactive Example 11.1
Q 6. What is the boiling point of a 0.3 m MgCl2 aqueous solution, assuming that it
dissociates completely? The Kb of water is 0.51oC/m and the boiling point of water is
100oC.
(A) 100.5 oC (B) 100.10 oC (C) 99.8 oC
(D) 99.5 oC
(D) 100.20oC (E) 100.41 oC
th
Hint: For 9 ed., see Section 11.5 & 11.7: p.p. 527-528: Interactive Example 11.8 for
nonelectrolyte solute.
Because Tb = Kb  msolute  i = 0.51 x 0.3 x (1+2) = 0.459 = Tb, solution – Tb, solvent
Thus, Tb, solution = Tb + Tb, solvent = 0.459 + 100 = 100.459  100.5
Q 7. A solution of 0.85 g of an unknown non-dissociating (i.e non-electrolyte) compound
2
dissolved in 100.0 g of benzene freezes at 5.16oC. The Kf of benzene is 5.07oC/m. The
freezing point of benzene is 5.45oC. What is the molar mass (g/mol) of the unknown?
(A) 130
(B) 149
(C) 154
(D) 200
(E) 250
Hint: For 9th ed., see Section 11.5 & 11.7: p.p. 528-530: Interactive Example 11.9 and
Interactive Example 11.10 for nonelectrolyte solute.
Because Tf = Kf  msolute  i = Tf, solvent – Tf, solution  5.45 -5.16 = 5.07 x m x 1
 0.29 = 5.07 x m  m = 0.29 / 5.07 = 0.0572 mol solute/Kg solvent  0.0572 = mole
solute/0.1 kg  mole solute = 0.0572 x 0.1 = 0.00572 = mass in gram /molar mass 
molar mass = mass /mole = 0.85 g/0.00572 mole = 148.60 g/mole
Q 8. The average osmotic pressure of blood is 22.7 atm at 25 oC. What concentration (M)
of glucose, C6H12O6, will be isotonic (p. 534 textbook; i.e. same osmotic pressure) with
blood to be used in IV injection?
(A) 0.31
(B) 0.62
(C) 0.93
(D) 1.24
(E) 1.55
th
Hint: For 9 ed., see Section 11.6: p.p. 532-534: Interactive Example 11.11 & 11.12.
Π = iMRT 22.7 = 1 x M x 0.082 x (25 + 273.15)  M = 22.7 / 0.04090 = 0.928  0.93
The osmotic pressure of blood is 3.5 atm at 25 oC. How much glucose (molar mass 180
g/mol) should be used per liter for an intravenous injection that is to have the same
osmotic pressure as blood?
(A) 54.2 g
(B) 18.4 g
(C) 15.6 g
(D) 20.5 g
(E) 25.8 g
Hint: For 9th ed., see Section 11.6: p.p. 532-534: Interactive Example 11.11 & 11.12.
Π = iMRT 3.5 = 1 x M x 0.082 x (25+273.15)  M = 3.5/0.0409 = 0.143 mole/liter 
Mass = mole x molar mass = 0.143 x 180 = 25.768 gram  25.8 g
Q 9. Given the following data for the reaction 2A + B  D + E
Experiment Initial [A]
Initial [B]
Initial Rate of Formation of D
1
0.080 M
0.034 M
2.2 x 10-4 M.min-1
2
0.080 M
0.017 M
1.1 x 10-4 M.min-1
3
0.16 M
0.017 M
2.2 x 10-4 M.min-1
What are the reaction order for A, for B, and the overall reaction order, respectively?
(A) 0, 1, 2
(B) 1, 1, 2
(C) 2, 1, 3
(D) 1, 1, 1
(E) 1, -1, 0
Hint: For 9th ed., see Section 12.3 p.p. 561-563: Initial Rate Method: Example and
Example 12.1. End-of-Chapter-Exercises: 29, 30 (same as lab: clock reaction), 36, etc.
Q 10-11. The mechanism for reaction H2(g) + I2(g)  2HI(g) has been proposed:
Step 1
I2(g)  2I(g)
(k1, k -1)
(fast)
Step 2
H2(g) + 2I(g)  2HI(g)
(k2)
(slow)
10. Which are respectively the intermediate and the rate-determining step in this
proposed mechanism?
(A) I(g); step 2
(B) I2(g); step 1
(C) HI (g); step 2
(D)I(g); step 1
3
Hint: For 9th ed., see Section 12.5: p.p. 574-577: Example 12.6; End-of-Chapter-Exercise
61.
How to differentiate the intermediate and the catalyst? Both do not appear in the
overall reaction. Catalyst is added as reactant and then reappears as product; catalyst
speeds up the reaction rate by changing the reaction mechanism and lowering the
activation energy; while intermediate appears as product and then is consumed as
reactant; the intermediate can be chemically separated and its structure can be
determined experimentally.
How to differentiate the transition state and the intermediate? The transition state
cannot be separated and thus its structure cannot be determined chemically; instead its
structure can be computer-simulated using both the structures of the reactant and
product.
11. What are the reaction orders for I2, for H2 and the overall reaction, respectively?
(A) 1, 2, 3
(B) 1, 1, 2
(C) 2, 1, 3
(D) 2, -1, 1
(E) 2, 2, 4
Additional Practice:
The following mechanism for the reaction between H2 and CO to form formaldehyde,
H2CO, has been proposed:
Step 1
H2 2H (fast, equilibrium) (k1, k-1)
Step 2
H + CO  HCO (slow)
(k2)
Step 3
H + HCO  H2CO (fast)
(k3)
(a) Which step is the rate-determining step? Answer: Step 2 because the step right
after the equilibrium is always slow or the rate-determining step.
(b) Write the balanced equation for the overall reaction. Answer: H2 + CO  H2CO
(c) Identify any intermediates in this proposed mechanism. Answer: H and HCO
(d) What is the overall reaction order? Note: Intermediate(s) cannot appear in the rate
law. Answer: 3/2
Solution for (d):
R = k2[H][CO]; since H is the intermediate which cannot appear in the rate law and thus
we need to replace it by reactant using the equilibrium concept. Equilibrium in CHEM
1411 saying it’s a dynamic equilibrium. This means, the forward rate = reverse rate.
Thus, k1[H2] = k-1[H]2; so [H] = {k1[H2]/ k-1}1/2;
Therefore, R = k2[H][CO] = k2{(k1[H2]/ k-1)1/2}[CO] = k2(k1/ k-1)1/2[H2]1/2[CO] = k[H2]1/2[CO]
Here, k = k2(k1/ k-1)1/2 and the overall reaction order = ½ + 1 = 3/2
Q 12-13. The reaction 4PH3(g)  P4(g) + 6H2(g) is the first order reaction and its rate
constant is 0.0198 s-1.
4
12. What is the half-life for this reaction?
(A) 41 s
(B) 35 s
(C) 18 s
(D) 65 s
(E) 26 s
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Hint: For 9 ed., see Section 12.4: p. 564: Equation 12.2; Example 12.3.; End-of-Chapter
Exercise 51; t 1/2 = 0.693/k = 0.693/0.0198 s-1 = 35 s
13. If 5.0 moles of PH3 were placed in a 2.0-liter container at that temperature, how
many moles of PH3 would remain after 2.0 minute?
(A) 0.46
(B) 0.25
(C) 0.12
(D) 0.56
(E) 0.36
th
Hint: For 9 ed., see Section 12.4: p. 564: Equation 12.2; Example 12.3.; End-of-Chapter
Exercise 51.; [A]0 = 5.0/2 = 2.5 M  ln [A]/2.5 = - 0.0198 x (2.0 x 60) = - 2.376 
[A]/2.5 = e (-2.376) = 0.09292  [A] = 0.09292 x 2.5 = 0.2323 M 
mole A = M x V = 0.2323 x 2.0 = 0.4646
Q. 14-15. The gas-phase decomposition of NOBr is second order in [NOBr] and its
rate constant is 51 M-1.min-1 at 24oC. Initially there are 0.0092 M NOBr placed in a flask.
14. What is the half-life for this reaction?
(A) 4.5 min
(B) 8.5 min
(C) 2.1 min
(D) 3.4 min
(E) 1.8 min
Hint: For 9th ed., see Sectiuon 12.4: p.p. 568-571 or Summary in Table 12.6;
Example 12.5; t ½ = 1/k[A]0 = 1/(51 x 0.0092) = 1/0.4692 = 2.13 min.
15. How many minutes does it take to use up 0.0055 M of this NOBr at such
temperature?
(A) 4.9
(B) 1.2
(C) 3.2
(D) 6.0
(E) 8.3
Hint: For 9th ed., see Sectiuon 12.4: p.p. 568-571 or Summary in Table 12.6; Example
12.5; “use up 0.0055 M” means the final concentration = 0.0092 – 0.0055 = 0.0037 M.
From 1/[A]t –1/[A]o = kt  1/0.0037 – 1/0.0092 = 51 x t 270.270 – 108.696 = 51 x t 
161.574 = 51 x t  t = 161.574/51 = 3.168 min
Q 16. The gas-phase decomposition of ethyl iodide to give ethylene and hydrogen iodide
is a first-order reaction:
C2H5I  C2H4 + HI
At 298 K, the value of k is 3.32 x 10-10 s-1. When the temperature is raised to 313 K, the
value of k increases to 3.13 x 10-9 s-1. What is the activation energy (in kJ) for this
reaction?
(A) 819
(B) 332
(C) 116
(D) 232
(E) 131
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Hint: For 9 ed., see Section 12.6: p.p. 577-581; Equation 12.11 & Interactive Example
12.8. From ln(k2/k1) = (-Ea/R)  (1/T2 –1/T1) = (Ea/R)  (1/T1 –1/T2)
Then ln (3.13 x 10-9/3.32 x 10-10) = (Ea/8.314)  (1/298 –1/313)
 ln 9.428 = (Ea/8.314) x (0.0001608)  2.244 = Ea x 0.00001934 
Ea = 2.244/0.00001934 = 116011.8134 J  116.011 kJ
Q 17-18. Consider the equilibria:
N2(g) + ½ O2(g)  N2O(g)
N2(g) + O2(g)  2NO(g)
Kc = 2.40 x 10-18
Kc = 4.1 x 10-31
5
17. What is the equilibrium constant for N2O(g) + ½ O2(g)  2NO?
(A) 4.80 x 10-18
(B) 4.1 x 10-31
(C) 4.17 x 1017
-13
35
(D) 1.708x10
(E) 1.74 x 10
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Hint: For 9 ed., see Section 13.2: Interactive Example 13.2(c) & p. 612; End-of-Chapter
Exercises 23, 71 & 72.
N2O(g)  N2(g) + ½ O2(g)
+) N2(g) + O2(g) 2NO(g)
N2O(g) + ½ O2(g)  2NO
K = 1/Kc = 1/(2.40 x 10-18) = 4.17 x 10+17
Kc = 4.1 x 10-31
K = (4.17 x 10+17)x(4.1 x 10-31)= 1.71x10-13
Characteristics of K, equilibrium constant:
For a hypothetical reaction: aA + bB <==> cC + dD, here <==> represents equilibrium.
Assuming that A, B, C, D are either aqueous solutions or gases (i.e. they are neither
liquids nor soilds), then the equilibrium constant K can be established as
K = [C]c[D]d/[A]a[B]b
K = [C]c[D]d/[A]a[B]b
Applications: Given aA + bB <==> cC + Dd
Then
(1) cC + dD <==> aA + bB
(2) xaA + xbB <==> xcC + xdD
K(1) = [A]a[B]b/[C]c[D]d = 1/K
K(2) = [C]xc[D]xd/[A]xa[B]xb
= {[C]c[D]d/[A]a[B]b}x = Kx
aA <==> cC
(1)
K(1) = [C]c/[A]a
bB <==> dD (2)
K(2) = [D]d /[B]b
When add equation (1) and equation (2) together, one obtains the resultant
equation as
Application: Given
aA + bB <==> cC + dD
(3)
the equilibrium constant, K(3), for the resultant equation, (3), is
K(3) = [C]c[D]d/[A]a[B]b = K(1) x K(2) = K
18. What is the equilibrium constant for
2NO(g)  N2O(g) + ½ O2(g)?
(A) 2.40 x 10-18
(B) 4.1 x 10-31
(C) 4.17 x 1017
12
35
(D) 5.85 x 10
(E) 1.74 x 10
Hint: For 9th ed., See End-of-Chapter Exercises: 71 & 72 and the above question.
2NO(g)  N2(g) + O2(g) K = 1/Kc = 1/(4.1 x 10-31) = 2.439 x 10+30
+) N2(g) + ½ O2(g)  N2O(g)
Kc = 2.40 x 10-18
2NO(g)  N2O(g) + ½ O2(g) K = (2.439 x 10+30) x (2.40 x 10-18) = 5.853 x 10+12
6
Q 19. What is the equilibrium law (Kc) for
CaCO3(s) + 2HCl(aq)  Ca2+(aq) + 2Cl-(aq) + H2O(l) + CO2(g)
(A) [Ca2+(aq)][Cl-(aq)]2[CO2(g)]/[HCl(aq)]2
(B) [Ca2+(aq)][Cl-(aq)]2[CO2(g)] [H2O(l)]/[HCl(aq)]2
(C) [Ca2+(aq)][Cl-(aq)]2[CO2(g)] [H2O(l)]/[CaCO3(s)][HCl(aq)]2
(D) [H2O(l)]/[CaCO3(s)]
(E) [H2O(g)]/[HCl(g)]
Hint: For 9th ed., see Section 13.4: p.p. 617-618 or Definition in Q 17-18.
Q 20. Which of the following equilibrium will have Kp ≠ Kc?
(A) SO3(g) + NO(g)  NO2(g) + SO2(g)
(B) N2(g) + O2(g)  2NO(g)
(C) 2NO(g)  N2(g) + O2(g)
(D) 3H2(g) + N2(g)  2NH3(g)
(E) 6CO2(g) + 6H2O(l)  C6H12O6(s) + 6O2(g)
Hint: For 9th ed., see Section 13.3: p.p. 614-616:
Relationship between Kp and Kc:
For a hypothetical reaction: aA(g) + bB(g) <==> cC(g) + dD(g)
Kc = [C]c[D]d/[A]a[B]b
Since the reactants and products are all gases, one can write the equilibrium constant as
Kp = (PC)c(PD)d/(PA)a(PB)b and from PV = nRT, one can write P = (n/V)RT = MRT where M is
molarity. Because all gases are in the same container and thus they have same
temperature and volume. Therefore,
PA = (nA/V)RT = [A]RT, PB = (nB/V)RT = [B]RT
PC = (nC/V)RT = [C]RT, and PD = (nD/V)RT = [D]RT
By substituting all into Kp = {([C]RT)c([D]RT)d}/{([A]RT)a([B]RT)b}
= ([C]c[D]d/[A]a[B]b)(RT)c+d-a-b = Kc(RT)Δn(g)
where Δn(g) = (c+d) – (a+b)
Note: When Δn(g) > 0, then Kp > Kc. When Δn(g) = 0, then Kp = Kc. When Δn(g) < 0,
then Kp < Kc.
Q 21-22. A mixture of 5.00 x 10-3 mole of H2 and 1.00 x 10-2 mole of I2 is placed in a 5.00
L container at 448 oC and allowed to come to equilibrium. At equilibrium, the
concentration of HI is 1.87 x 10-3 M.
21. What is the Kc at 448 oC for the reaction H2(g) + I2(g)  2HI(g)?
(A) 50.5
(B) 2.5
(C) 1.0
(D) 0.67
(E) 0.44
7
Hint: For 9th ed., See Interactive Example 13.9: p. 623. Initial concentration for H2(g) =
5.00 x 10-3 mole/5.00 L = 0.001 M; initial concentration for I2(g) = 1.00 x 10-2 mole/5.00 =
0.002 M
H2(g) +
I2(g) 
2HI(g)
[in]
0.001
0.002
0
[react] -x
-x
+2x
[eq] 0.001-x
0.002-x
+2x
Here, it says “at equilibrium, the concentration of HI is 1.87 x 10-3 M” and thus
[HI] = +2x = 1.87 x 10-3 M  x = 9.35 x 10-4 M;
[H2(g)]eq = 0.001 - x = 0.001 - 9.35 x 10-4 = 6.5 x 10-5 M
[I2(g)]eq = 0.002 - x = 0.002 - 9.35 x 10-4 = 1.065 x 10-3 M
Kc = [HI(g)]2/[H2(g)][I2(g)] = (1.87 x 10-3)2/[(6.5 x 10-5)(1.065 x 10-3)]
= 3.4969 x 10-6 /(6.9225 x 10-8) = 50.5
Kp = Kc(RT)Δn(g) ; Δn(g) = 2 – (1+1) = 0; Kp = 50.5
22. What is the equilibrium concentration of H2(g)?
(A) 5.00 x 10-3
(B) 5.00 x 10-4
(D) 2.50 x 10-4
(E) 6.5 x 10-5
Hint: See Hint in the above question.
(C) 1.00 x 10-3
Another practice:
At 25oC, Kc = 4.500 for the reaction SO3(g) + NO(g)  NO2(g) + SO2(g). If 0.50 mole of
SO3 and 0.60 mole of NO are placed in a 2.00 L container and allowed to react,
What will be the equilibrium concentration of NO(g)?
(A) 0.032
(B) 0.068
(C) 0.21
(D) 0.12
(E) 3.12
Hint: For 9th ed., see Section 13.5, Interactive Example 13.9, p. 623.
Hint: Initial concentration for [SO3] = 0.50/2.00 =0.25 M; for [NO] = 0.60/2.00 = 0.30;
Need to solve quadric equation. 0 < x < 0.25 M and thus x = 0.1845 M, NOT 0.523 M.
At equilibrium, [NO2]=[SO2] = 0.1845 M, [SO3] = 0.0655 M, [NO] = 0.1155 M.
SO3(g) + NO(g)  NO2(g) + SO2(g)
[in]
0.25
0.30
0
0
[react] -x
-x
+x
+x
[eq] 0.25-x 0.30-x
+x
+x
Kc = [NO2(g)][SO2(g)]/[SO3(g)][NO(g)] = (0.25-x)(0.30-x)/(x)(x) = 4.500 
(0.25-x)(0.30-x) = 4.5x2  0.075 – 0.55x + x2 = 4.5x2 3.5x2 + 0.55x -0.075 = 0 
use the formula of quadric equation to solve x. So x = 0.1845 or 0.523. Be aware that
ALL EQUILIBRIUM CONCENTRATIONS, [eq], MUST BE POSITIVE. Thus, x = 0.1845, not
0.523.
8
What will be the equilibrium concentration of SO2(g)?
(A) 0.032
(B) 0.068
(C) 0.19
(D) 2.12
Hint: See above calculation procedure.
(E) 3.12
What is the Kp at 25oC?
(A) 1.5
(B) 3.0
(C) 4.5
(D) 6.0
(E) 7.5
Hint: 25oC = 273.15 + 25 = 298.15 Kelvin; Δn(g) = (1+1) – (1+1) = 0
Kp = Kc(RT)Δn(g) = 4.500 (0.082 x 298.15)0 = 4.500 x 1 = 4.500
Which of the following is not true after adding H2O(g) to react with SO3(g)?
(A) The equilibrium shifts to the left.
(B) The concentration of NO increases.
(C) The partial pressure of NO2(g) decreases.
(D) The partial pressure of SO2(g) decreases.
(E) H2O(g) has no effect on the equilibrium.
Hint: For 9th ed., see Section 13.7: Le Chatelier’s Principle. Interactive Examples 13.13,
13.14, & 13.15; End-of-Chapter Exercises 63, 65, 66, 67 & 68. “Adding H2O(g) to react
with SO3(g)” has the same physical meaning of “removing the SO3(g) from the
reaction, SO3(g) + NO(g)  NO2(g) + SO2(g), and thus the reaction will shift to the left
side, that is, the reactant side.”
Q 23-24. Consider the equilibrium:
Consider the equilibrium: 2SO2(g) + O2(g)  2SO3(g) Ho < 0
23. Which of the following action will cause the equilibrium shifts to the left?
(A) Adding Ne(g)
(B) Adding O2(g)
(C) Adding SO3(g)
(D) Decreasing temperature
(E) Adding catalyst
Hint: For 9th ed., see Section 13.7: Le Chatelier’s Principle. Interactive Examples 13.13,
13.14, & 13.15; End-of-Chapter Exercises 63, 65, 66, 67 & 68. (a) Adding non-reactive
gasNo effect in the equilibrium; (b) shifts to the right; (c) equilibrium shifts to the
left; (d) shifts to the right; (e) Catalyst has no effect on equilibrium.
24. Which of the following action will cause the equilibrium shifts to the left?
(A) The volume of the reaction vessel is doubled
(B) The pressure of the reaction is doubled
(C) A catalyst is added to the mixture
(D) Adding C6H12O6(s) to the mixture
(E) Decreasing temperature
Hint: For 9th ed., see Section 13.7: Le Chatelier’s Principle. Practice more from the Endof-Chapter Exercises.
Q 25. Given 6CO2(g) + 6H2O(l)  C6H12O6(s) + 6O2(g)
Ho = 2816 kJ
Which of the following action will affect the equilibrium?
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(A) Removing part of the C6H12O6(s)
(B) Removing part of the H2O(l)
(C) Removing all of the H2O(l)
(D) Adding C6H12O6(s)
(E) Adding H2O(l)
Hint: For 9th ed., see Section 13.7: Le Chatelier’s Principle. Practice more from the Endof-Chapter Exercises.
Additional information:
Density = mass/volume
Percent by mass = {mass of solute/mass of solution}  100
Molarity = mole of solute / Liter of solution
Molality = mole of solute / Kg of solvent
R = 8.314 J/mol.K = 0.082 atm/mol.K
Tb = Tb, solution – Tb, solvent = Kb i m
Tf = Tf, solvent – Tf, solution = Kf i m
π = M i  R  T
t1/2 = 0.693/k
ln([A]0/[A]t) = kt.
t1/2 = 1/(k  [A]initial)
1/[A]t –1/[A]o = kt
ln(k2/k1) = (-Ea/R)  (1/T2 –1/T1)
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