Solutions to the Chem 203 Term Test 1 Booklet Practice Problems
Thermochemistry
Problem 1.
In a coffee cup calorimeter, 50.0 mL of 0.100 M AgNO3 and 50.0 mL of 0.100 M HCl are mixed
to yield the following reaction:
−
Ag +
(aq) + Cl(aq) → AgCl(s)
The two solutions were initially at 22.60 ˚C, and the final temperature is 23.40 ˚C. Calculate the heat that
accompanies this reaction in
kJ
mol
of AgCl formed.
Solution:
Moles of AgNO3 = 0.050 L x 0.100 M = 0.05 moles AgNO3 = 0.05 moles
Moles of HCl = 0.050 L x 0.100 M = 0.05 moles HCl = 0.05 moles Cl−
(aq)
Since the reaction runs to completion, and all species are in a 1:1 ratio, 0.05 moles of AgCl are produced
(you can prove this to yourself using an ICE table)
Heat lost by reaction = heat gained by solution
Heat gain = m  s  T
= 100.0 g H2 O x 4.18
J
− 1 × (23.40 °C– 22.60 °C)
g °C
= 330 J = Heat Loss, this is the heat evolved from the formation of 0.050 moles of AgCl
Therefore, the heat produced per mol of AgCl =
330 J
1kJ

 6.60 kJ mol 1
0.050 mol AgCl 1000 J
Problem 2.
When a bomb calorimeter is heated electrically for exactly 6 minutes using a 4.00 kilowatt
heater (1 watt = 1 J s-1), its temperature increases from 25.0 C to 50.3 C. When 0.100 mol C2 H5 OH(l) is
burned completely in oxygen in the same calorimeter, the temperature increases from 25.0C to 27.4C.
What is the constant volume heat of combustion of C2 H5 OH(l) ?
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Solution
C2 H5 OH(l) + 3 O2(g) ®2 CO2(g) + 3 H2 O(l)
Answer: -1370 kJ mol-1
Heat capacity of calorimeter = C
q=(360 s)  (4000 J/s) = 1440 kJ
q = C ∆T = C (50.3 - 25.0) rearrange and solve  C = 56.92 kJC-1
Q rxn  
(56.92 kJ C 1 )(27.4 25.0 C)
 1370 kJ / mol
(0.100 mol)
Problem 3.
A gas absorbs 1.0 J of heat and performs 15.2 J of work. What is the change in the internal
energy of the gas?
Solution: -14.2 J
Absorbing heat is an endothermic process (positive)
Work done by the system on the surroundings is negative by convention, therefore
E = q + w = 1.0 J - 15.2 J= -14.2 J
Problem 4.
A balloon filled with 39.1 mol helium has a volume of 876 L at 0.0 °C and 1.00 atm pressure.
The temperature of the balloon is 38.0 °C as it expands to a volume of 998 L, the pressure remaining
constant. Calculate q, w, and ΔE for the helium in the balloon.
J
The molar heat capacity for helium gas is 20.8 mol °C
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Solution: q = n × smol × T
= (20.8 J
J
) × 39.1 mol × (38.0 – 0.0)°C x
mol °C
1kJ
= 30.9 kJ
1000 J
Pressure-Volume work can be calculated by:
w = -PΔV = -1.00 atm x (998 – 876 L) = -122 L atm x
101.3 J
1kJ
x
= -12.4 kJ
1000 J
L atm
So, the total internal energy, ΔE = q + w = 30.9 kJ + (−12.4 kJ) = 18.5 kJ
Problem 5.
Given the following data:
2 O3(g) → 3 O2(g)
O2(g) → 2 O(g)
ΔH = −427 kJ
ΔH = +495 kJ
NO(g) + O3(g) → NO2(g) + O2(g) ΔH = −199 kJ
Calculate ΔH for the reaction:
NO(g) + O(g) → NO2(g)
Solution:
NO(g) + O3(g) → NO2(g) + O2(g)
3
O
→ O3(g)
2 2(g)
O(g) → ½ O2(g)
ΔH = −½ (− 427 kJ)
ΔH = −½ (+ 495 kJ)
NO(g) + O(g) → NO2(g)
Problem 6.
ΔH = − 199 kJ
ΔH = − 233 kJ
Using equations below, find H at 25 C for the oxidation of C2 H5 OH(l) :
C2 H5 OH(l) + 3 O2(g) → 3 H2 O(l) + 2 CO2(g)
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C2 H4(g) + 3 O2(g) → 2 CO2(g) + 2 H2 O(l) ΔH = -1411 kJ
C(s) + 3 H2(g) + ½ O2(g) → C2 H5 OH(l) ΔH = -278 kJ
C2 H4(g) + H2 O(l) → C2 H5 OH(l)
ΔH = -44 kJ
Solution: 1367 kJ
C2 H5 OH(l) → C2 H4(g) + H2 O(l) ΔH = +44 kJ
C2 H4(g) + 3 O2(g) → 2 CO2(g) + 2 H2 O(l) ΔH = −1411 kJ
C2 H5 OH(l) + 3 O2(g) → 3 H2 O(l) + 2 CO2(g) ΔH = −1367 kJ
Problem 7.
Based on the following the data at 25°C, and assuming ΔH° and ΔS° are independent of T, :
J
S° (
)
mol K
kJ
ΔHf0 (
)
mol
Compound
NO(g)
90.29
210.4
NO2(g)
233.82
240.2
Calculate ΔH° at 25°C for 2 NO(g) + O2(g) → 2 NO2(g)
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kJ
Answer: ΔH° = 2(33.82) − 2(90.29) − 0 = −112.94 mol
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Equilibrium
Problem 8.
What is the equilibrium constant expression for the reaction below?
+
3 Sn(s) + 4 H(aq)
+ 4 NO−
3(aq) + H2 O(l) ⇌ 3 H2 SnO3(s) + 4 NO(g)
Answer: K c 
[ NO ]4
[ H  ]4 [ NO3 ]4
Solution: Pure solids and pure liquids and solvents are not included in the expression. Products go over
Reactants, and coefficients in the equation are written as superscripts.
Problem 9.
Give an expression for Kc for the reaction
CaCO3(s) ⇌ CaO(s) + CO2(g) ?
Solution:
K c = [CO2(g) ]
Remember: Solids and liquids are not included in the expression.
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Problem 10.
Given the equilibrium constant for equation 1, what is the equilibrium constant for equation 2?
Equation 1: NH3(g) ⇌
1
3
N2(g) + H2(g) K p = 0.157
2
2
Equation 2: N2(g) + 3 H2(g) ⇌ 2 NH3(g) K p =?
Solution: Equation 2 is equal the double and reverse of equation1 therefore
K p2 = K −2
p1 = (
Problem 11.
1
) 2 = 40.6
0.157
Given:
P4(s) + 6 Cl2(g) ⇌ 4 PCl3(l) Equilibrium Constant = K1
What is the equilibrium constant for the following reaction?
2 PCl3(l) ⇌ 3 Cl2(g) + ½ P4(s)
Answer:
−1/2
Equilibrium Constant = K1
=
1
√K1
Solution: The second equation if reversed (1/K) and halved (K1/2). Combining these two gives 1/K1/2.
Problem 12.
At 25 ˚C Kc for the reaction of carbon monoxide with oxygen to produce carbon dioxide is
3.3x1091. What is Kp for this reaction under the same conditions?
Answer:
So the reaction equations is: 2 CO(g) + O2(g) ⇌ 2 CO2(g) K c = 3.3 × 1091
The relationship between K c and K p is: K p = K c (RT)∆n gas In this case there are 3 moles of gas in the
reactants and 2 moles of gas in the products, so ∆n = -1
So solving for K p :
K p = K c (RT)∆n gas = (3.3 × 1091 ) [(0.0821
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−1
L atm
) (298 K)] = 1.35 × 1090
mol K
The equilibrium constant for the reaction below is K c = 2.5 × 10−4 at 25C. If the initial
Problem 13.
concentrations were [N2 ] = [C2 H2 ] = [HCN] = 1.00
mol
L
, then what is [HCN] at equilibrium?
N2(g) + C2 H2(g) ⇌ 2 HCN(g)
Solution:
N2(g) +
C2 H2(g) ⇌
2 HCN(g)
i:
1.00
1.00
1.00
c:
+x
+x
-2x
e:
1.00+ x
1.00 + x
1.00-2x

Q=1>K so then rxn goes to the left
(1.00  2 x )2
(1.00  2 x)
 Kc 
 Kc
2
(1.00  x )
(1.00  x)
x
1  Kc
2  Kc
Problem 14.
 0.488
Phosgene, COCl2(g) , is a toxic gas that dissociates according to the chemical equation:
COCl2(g) ⇌ CO(g) + Cl2(g)
For the reaction above, K c = 8.05 × 10−4 at 673 K. Exactly 1.00 mol COCl2(g) is placed
in an empty 25.0-L flask. The flask is capped and then warmed to 673 K. What is the equilibrium concentration
of CO at this temperature?
Solution:
COCl2(g)
⇌ CO(g)
+Cl2(g)
initial:
0.04
0
0
change:
-x
+x
+x
equil:
0.04-x
x
x
x2
Kc =
0.04  x
x2 + Kcx – 0.04Kc = 0
x
 Kc  Kc 2  (4)(0.04)( Kc)
 5.29  103
2
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Problem 15.
The equilibrium constant for the gas-phase reaction
CO + H2 O ⇌ CO2 + H2
is K c = 5.10 at 700 K. One mole each of CO, H2 O, CO2 , and H2 is placed in a 10.0-L flask and the
temperature is raised to 700 K. What is the concentration of H2 when the reaction reaches equilibrium?
Solution: Kc 
[CO2 ][ H 2 ]
 5.10
[CO][ H 2O]
CO +
H2 O ⇌
CO2
+H2
I
0.1M
0.1
0.1
0.1
C
-x
-x
+x
+x
E
0.1-x
0.1-x
0.1+x
0.1+x
(0.1  x )2 0.01  0.2 x  x 2
Kc 

 5.10
(0.1  x )2 0.01  0.2 x  x 2
0.01  0.2 x  x 2  0.051  1.02 x  x 2
0  0.041  1.22 x
1.22 x  0.041
x  0.034 M
Therefore at equilibrium [H2] = 0.1 + x = 0.1 + 0.034M = 0.134M
Problem 16.
For the following equilibrium, which is endothermic in the forward direction, are the following
statements true or false?
CH3 CHO(g) ⇌ CH4(g) + CO(g)
a) A decrease in temperature will increase the pressure in the system.
b) Addition of methane will increase the heat absorbed by the system.
c) An increase in the volume of the system will increase the total moles of CO(g) at equilibrium.
d) At constant total pressure, an increase in temperature will increase the amount of CH3 CHO(g) at
equilibrium.
e) As temperature increases, the equilibrium constant for the reaction decreases.
Solution:
Only C is true
An increase in volume will shift the equilibrium to the right thus causing an increase in the total moles of CO
at equilibrium.
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Problem 17.
The following reaction has reached a state of dynamic equilibrium, in the presence of a catalyst.
2 SO2(g) + O2(g) ⇌ 2 SO3(g) ,
∆H = −198 kJ
Which direction will the equilibrium shift with the following changes?
a) increasing the temperature
b) removing the catalyst
c) decreasing the volume of the reaction vessel
d) adding some more O2(g)
Solution:
a) shift to the left
b) no effect
c) shift to the right
d) shift to the right
Problem 18.
Consider the following reaction:
Ni(CO)4(g) ⇌ Ni(s) + 4 CO(g)
A mixture of Ni(CO)4(g) and CO(g) , each with a concentration of 0.800M, and an excess of Ni(s) were
confined in a 1.0 L container at 300K. In which direction will the reaction shift to reach equilibrium?
Solution:
The reaction will shift left forming Ni(CO)4(g) to reach equilibrium
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Problem 19.
The equilibrium constant for the reaction
2 NOCl(g) ⇌ 2 NO(g) + Cl2(g)
is 0.51 at a certain temperature. A mixture of NOCl(g) , NO(g) and Cl2(g) with concentrations 1.3, 1.2, and
0.60 M, respectively, was introduced into a container at this temperature. Which of the following are
true?
a) The concentration of Cl2(g) increases until equilibrium is reached.
b) [NOCl(g) ] = [NO(g) ] = [Cl2(g) ] at equilibrium
c) We are at equilibrium and thus no net change takes place.
d) The concentration of NOCl(g) increases until equilibrium is reached.
Answer: C
Solution: Q = [NO]2[Cl2]/[NOCl]2 = (1.2)2(0.56)/(1.3)2 = 0.51 = K, therefore we are already at equilibrium.
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