Download Thermochem problems

10/02/2016
Enthalpy of reaction
ΔH = Hfinal  Hinitial
so in chemical reactions ΔH = Hproducts  Hreactants
ΔH is called the enthalpy of reaction,
written as ΔHrxn
Thermochemical equation is the balanced chemical reaction and the value of ΔH
provides relationship between amounts of chemicals and the enthalpy change.
Things to note about enthalpy:
1. ΔH depends on amounts of reactants and products
2H2O2(ℓ)
2H2O2(ℓ)  2H2O(ℓ) + O2(g) ΔH = 196 kJ
H2O2(ℓ)  H2O(ℓ) + ½O2(g) ΔH = ½[196 kJ]
2. ΔH forward = - ΔH reverse
2H2O(ℓ) + O2(g)  2H2O2(ℓ) ΔH = +196 kJ
ΔH = 196 kJ
exothermic
ΔH = +196 kJ
endothermic
3. ΔH depends on the physical state of reactants and products
2H2O2(ℓ)  2H2O(g) + O2(g) ΔH = 108 kJ
2H2O(ℓ) + O2(g)
Question:
It is said that James Joule (1818 – 1889) was almost fanatical about measurement.
On his honeymoon he took a very sensitive thermometer and measured the
temperature of water at the top and at the bottom of a scenic waterfall.
You would expect the water at the bottom of the waterfall to be …
A)
B)
C)
D)
E)
a little cooler than the water at the top
a little warmer than the water at the top
exactly the same temperature as the water at the top
harder to get to
Wetter
Not as simple as it might seem! Kinetic energy of falling water converted into heat
will warm the water. Evaporation as the liquid falls will cool it…
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10/02/2016
Question:
Which of these is NOT a state function?
a. internal energy
b. temperature
c. enthalpy
d. work
Question:
Which of these is NOT a state function?
a. internal energy
b. temperature
c. enthalpy
d. work
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10/02/2016
Specific heat can be determined experimentally
specific heat 
amount of heat transferred
(grams of substance) x (change in temperature)
Rearrange to give:
q = CS x mass x ΔT
An example: determining the specific heat of lead:
(a) Heat 150g lead to 100°C
(b) Add the hot lead to 50g water at 22°C
(c) Measure the final temperature
Suppose Tfinal is 28.8 oC
q lost = Cs(Pb) x 150.0 g x (28.8 - 100.0)°C
q gained = Cs(H2O) x 50.0 g x (28.8 – 22.0)°C
heat lost =  heat gained,
hence calculate Cs(Pb)
Hess’s Law
We can use tabulated ΔH values to calculate the enthalpy of reactions
ΔH depends on amounts of reactants and products and their initial and final states
ΔH is a state function, so does not depend upon how we get from reactants to products
Example: N2(g) + 3H2(g)  2 NH3(g) Hrxn = ?
 N2(g) + 3H2(g)  N2H4(g) + H2(g) H = 95.4 kJ
 N2H4(g) + H2(g)  2 NH3(g) ΔH =  187.6 kJ
step 1
ΔH = 95.4 kJ
step 2
ΔH = -187.6 kJ
Hrxn = + 95.4 kJ -187.6 kJ =  92.2 kJ
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10/02/2016
Hess’s Law : The enthalpy change of an overall process is the sum of the
enthalpy changes of its individual steps
Procedure: combine the individual reactions so their sum gives the
desired reaction
e.g. what is ΔH for ½N2(g) + O2(g) → NO2(g)
• Arrange reactions so (final) reactants appear on
given:
the left and products appear on the right
N2(g) + O2(g)  2NO( g) ΔH = +180.50 kJ
NO2(g)  NO(g) + ½ O2(g) ΔH = +57.07 kJ
• Any reaction that is reversed must
have the sign of its ΔH changed
• All intermediates must occur on both
the right and the left so they cancel
NO(g) + ½ O2(g) → NO2(g)
ΔH =- 57.07 kJ
• A reaction can be multiplied by
N2(g) + O2(g) → 2NO( g) ΔH = +180.50 kJ
a coefficient, but ΔH for that reaction
must be multiplied by the same
coefficient.
½N2(g) + ½O2(g) → NO( g) ΔH = (½) 180.50 kJ
½N2(g) + O2(g) → NO2(g) ΔH = -57.07 kJ + 90.25 kJ = 33.18 kJ
Hess’s Law
We can use tabulated ΔH values to calculate the enthalpy of reactions
ΔH depends on amounts and initial and final states of reactants and products
ΔH is a state function, so does not depend upon how we get from reactants to products
Example: N2(g) + 3H2(g)  2 NH3(g) Hrxn = ?
 N2(g) + 2H2(g)  N2H4(g)
H = 95.4 kJ
 NH3(g)  ½ N2H4(g) + ½ H2(g)
ΔH = + 93.8 kJ
Hrxn = -187.6 kJ + 95.4 kJ =  92.2 kJ
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10/02/2016
If X is a state function, then the change in X is given by (Xfinal …. Xinitial)
a.
b.
c.
d.
+
–
×
÷
If X is a state function, then the change in X is given by (Xfinal …. Xinitial)
a.
b.
c.
d.
+
–
×
÷
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10/02/2016
The standard enthalpy of formation of carbon in its graphite form is
_______ kilojoules per mole.
a.
b.
c.
d.
100
1000
1
0
The standard enthalpy of formation of carbon in its graphite form is
_______ kilojoules per mole.
a.
b.
c.
d.
100
1000
1
0
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10/02/2016
The standard enthalpy of formation of carbon in its diamond form is
+1.88 kJ/mole, which means that diamond is _______ graphite.
a.
b.
c.
d.
as stable as
more stable than
less stable than
an isotope of
The standard enthalpy of formation of carbon in its diamond form is
+1.88 kJ/mole, which means that diamond is _______ graphite.
a.
b.
c.
d.
as stable as
more stable than
less stable than
an isotope of
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10/02/2016
2 H2 + O2  2 H2O
If the reaction above releases 483.6 kJ, then the standard enthalpy of
formation of H2O is
a.
b.
c.
d.
–483.6 kJ/mole.
+483.6 kJ/mole.
–241.8 kJ/mole.
+241.8 kJ/mole.
2 H2 + O2  2 H2O
If the reaction above releases 483.6 kJ, then the standard enthalpy of
formation of H2O is
a.
b.
c.
d.
–483.6 kJ/mole.
+483.6 kJ/mole.
–241.8 kJ/mole.
+241.8 kJ/mole.
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10/02/2016
……………………………………………………………………………….
What effect does reversing a reaction have on the value of H?
A.
B.
C.
D.
No change.
Sign of H changes.
Value of H increases.
Value of H decreases.
……………………………………………………………………………….
What effect does reversing a reaction have on the value of H?
A.
B.
C.
D.
No change.
Sign of H changes.
Value of H increases.
Value of H decreases.
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10/02/2016
Ozone, O3(g), is a form of elemental oxygen produced by electric motors, during
electrical storms and in the upper atmosphere. Is Hf for O3(g) necessarily zero?
A.
B.
C.
D.
Yes, because it is just another elemental form of oxygen.
No, because it is not the most stable form of the element oxygen at the given
conditions.
Yes, because changing the subscripts of an elemental formula does not change
standard enthalpy of formation.
No, because there is a temperature change when ozone is formed.
Ozone, O3(g), is a form of elemental oxygen produced by electric motors, during
electrical storms and in the upper atmosphere. Is Hf for O3(g) necessarily zero?
A.
B.
C.
D.
Yes, because it is just another elemental form of oxygen.
No, because it is not the most stable form of the element oxygen at the given
conditions.
Yes, because changing the subscripts of an elemental formula does not change
standard enthalpy of formation.
No, because there is a temperature change when ozone is formed.
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10/02/2016
It’s now possible to buy cars that run on hydrogen gas as a fuel; what advantage(s)
does hydrogen have as a fuel?
A.
B.
C.
D.
It is easily stored as a gas for a long time.
Its product of combustion is only H2O(g).
Safety considerations using hydrogen gas are minimal.
It is easily found in nature as an element.
It’s now possible to buy cars that run on hydrogen gas as a fuel; what advantage(s)
does hydrogen have as a fuel?
A.
B.
C.
D.
It is easily stored as a gas for a long time.
Its product of combustion is only H2O(g).
Safety considerations using hydrogen gas are minimal.
It is easily found in nature as an element.
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10/02/2016
The table at the right lists some specific heat
capacities. Which of the following substances
requires the most heat to increase a 5 gram
sample from 5°C to 12°C?
A) P4
B) Hg
C) Mg
D) Cu
E) Al
The table at the right lists some specific heat
capacities. Which of the following substances
requires the most heat to increase a 5 gram
sample from 5°C to 12°C?
A) P4
B) Hg
C) Mg
D) Cu
E) Al
Would the same apply to molar heat capacity?
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10/02/2016
The heats of formation for each of the components
involved in the dissolution of sodium chloride are given
below.
NaCl(s)  Na + (aq) + Cl  (aq)
 f H o / kJ mol -1
 411
 240
H2O
 167
What is the heat of dissolution of NaCl?
A. 4 kJ mol-1
D. -668 kJ mol-1
B. -4 kJ mol-1
E. none of the above
C. 668 kJ mol-1
The heats of formation for each of the components
involved in the dissolution of sodium chloride are given
below.
NaCl(s)  Na + (aq) + Cl  (aq)
 f H o / kJ mol -1
 411
 240
H2O
 167
What is the heat of dissolution of NaCl?
A. 4 kJ mol-1
D. -668 kJ mol-1
B. -4 kJ mol-1
E. none of the above
C. 668 kJ mol-1
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10/02/2016
Which of the following statements are correct if the surroundings absorb 145 J of heat
from the system while doing 98 J of work on the system?
•
•
•
•
•
I. The system loses 145 J of energy and the surroundings gain 98 J of energy.
II. The surroundings absorb heat, therefore q is positive.
III. The system loses 47 J of energy and the surroundings gain 47 J of energy..
IV. The system gains 47 J of energy and the surroundings lose 47 J of energy.
V. Work is done on the system, therefore w is positive
A. I & II
B. II & III
C. II, III, V
D. III & V
E. IV & V
Which of the following statements are correct if the surroundings absorb 145 J of heat
from the system while doing 98 J of work on the system?
•
•
•
•
•
I. The system loses 145 J of energy and the surroundings gain 98 J of energy.
II. The surroundings absorb heat, therefore q is positive.
III. The system loses 47 J of energy and the surroundings gain 47 J of energy..
IV. The system gains 47 J of energy and the surroundings lose 47 J of energy.
V. Work is done on the system, therefore w is positive
A. I & II
B. II & III
C. II, III, V
D. III & V
E. IV & V
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10/02/2016
The standard enthalpy of formation of NiSO4(s) at 25 °C is -872.9 kJ/mole. The
chemical equation to which this value applies is
•
•
•
•
A. ½ Ni(s) + ½ S(s) + ½ O2(g) → ½ NiSO4(s)
B. Ni(s) + S(s) + 4 O(g) → NiSO4(s)
C. Ni(s) + S(s) + 2 O2(g) → NiSO4(s)
D. Ni(s) + ½ S2(s) + ½ O2(g) → NiSO4 (s)
The standard enthalpy of formation of NiSO4(s) at 25 °C is -872.9 kJ/mole. The
chemical equation to which this value applies is
•
•
•
•
A. ½ Ni(s) + ½ S(s) + ½ O2(g) → ½ NiSO4(s)
B. Ni(s) + S(s) + 4 O(g) → NiSO4(s)
C. Ni(s) + S(s) + 2 O2(g) → NiSO4(s)
D. Ni(s) + ½ S2(s) + ½ O2(g) → NiSO4 (s)
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10/02/2016
Calculate the enthalpy of formation for gaseous HCl from the following data:
•
•
•
N2(g) + 4 H2(g) + Cl2(g) → 2 NH4Cl(s) Hf = - 628.8 kJ
(1)
NH3(g) + HCl(g) → NH4Cl(s)
Hf = - 176.2 kJ
(2)
The enthalpy of formation of ammonia gas is known to be 46.19 kJ mol-1
Solution:
a) Write down the equation for the formation of ammonia; call this (3)
b) Write down the equation for the formation of HCl (g); call this (4)
c) Add and subtract equations (1)-(3) to get equation (4).
d) Combine the H values to get the answer: -628.8 + (2)(176.2) + (2)(46.19)
= -184.0 kJ for two moles of HCl, so f(HCl) = -92.01 kJ/mol
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