Download Practice Problems: Thermodynamics

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Practice Problems: Thermodynamics
CHEM 1A
1. Answer the questions below for each of the following reaction coordinate diagrams:
*
*
reactants
H
products
potential energy 
potential energy 
Ea
Ea
products
H
reactants
reaction coordinate 
reaction coordinate 
a) Is the reaction exothermic or endothermic?
exothermic
endothermic
b) What is the sign of H?
(–)
(+)
c) Is heat absorbed or released?
absorbed
released
d) What happens to the temperature of the surroundings?
increases (gets warmer)
decreases (gets colder)
e) What does the asterisk (*) at the top of the hill represent?
activated complex or high energy transition state
2. Based on the following equation:
2 C3H7OH (l) + 9 O2 (g)  6 CO2 (g) + 8 H2O (g) + heat
a) Is this reaction exothermic or endothermic ? (circle one) Heat is released as a product
more
b) Is the entropy (randomness) of the system increasing or decreasing ? (circle one) liquid  gas
c) Label the sign (+ or –) of each variable in the equation below. If G depends on the
temperature, enter a question mark (?).
G = H – T S
( – ) = ( – ) – ( + )( + )
G is negative (spontaneous) at any Kelvin temperature (which is always positive in sign)
d) At what temperatures is the reaction above spontaneous? (circle one)
at all temperatures
only spontaneous at
only spontaneous at
at no temperatures
(always spontaneous)
high temperatures
low temperatures
(never spontaneous)
3. Based on the following equation:
NH4Cl (s) + heat  NH4+ (aq) + Cl– (aq)
a) Is this reaction exothermic or endothermic ? (circle one) Heat is absorbed as a reactant
b) Is the entropy (randomness) of the system increasing or decreasing ? (circle one) solid  aqueous
c) Label the sign (+ or –) of each variable in the equation below. If G depends on the
temperature, enter a question mark (?).
G = H – T S
( ? ) = ( + ) – ( + )( + )
d) At what temperatures is the reaction above spontaneous? (circle one)
at all temperatures
only spontaneous at
only spontaneous at
at no temperatures
(always spontaneous)
high temperatures
low temperatures
(never spontaneous)
high temperatures allow the favorable S term to dominate in magnitude over the unfavorable H term
4. The 150.0 g of water in a cup goes from 20.1ºC to 48.5ºC. How much heat was absorbed by the
water?
q = mCT
qw = mw Cw (Tw final – Tw initial)
qw = (150.0 g) (4.184 J/gºC) (48.5ºC – 20.1ºC)
qw = (150.0 g) (4.184 J/gºC) (28.4 ºC)
qw = 17823.84 J
J or 1.78 x 104 J
Answer: 17,800
___________________
5. A 28.5 g piece of metal at 99.8ºC is placed in a calorimeter containing 50.0 g of water at 22.3ºC.
The calorimeter itself has a heat capacity of 3.40 J/°C, and the temperature rises to 26.1ºC. Calculate
the specific heat of the metal (J/gºC).
subscripts:
q w + qc = – qx
q = mCT
w = water
c = calorimeter
x = unknown metal
Note: the brackets and labeling of
equation parts, the subscript key, and
the intermediate math steps are not
part of the required work, but are
shown here for clarity
absorbing releasing
heat
heat
heat capacity
mw Cw (Tfinal – Tw initial) + mc Cc (Tfinal – Tc initial) = – mx Cx (Tfinal – Tx initial)
water
calorimeter
metal
(50.0 g) (4.184 J/gºC) (26.1ºC – 22.3ºC) + (3.40 J/ºC) (26.1ºC – 22.3ºC) = – (28.5 g) Cx (26.1ºC – 99.8ºC)
(50.0 g) (4.184 J/gºC) (3.8ºC) + (3.40 J/ºC) (3.8ºC) = – (28.5 g) Cx (–73.7ºC)
(794.96 J) + (12.92 J) = – (–2100.45 gºC) Cx
(80 7.88 J) = (2100.45 gºC) Cx
0.38 J/gºC
Answer: ___________________
Cx = 0.3846223428 J/gºC
6. 38.1 g of zinc and 25.9 g of aluminum are heated to 100.2ºC, and dropped into a container with
200.0 g of water at 22.4ºC. Assuming that no heat is lost, what is the final temperature of the water?
Note: CAl = 0.897 J/gºC and CZn = 0.388 J/gºC
qw = – ( qZn + qAl )
mw Cw (Tf – Tw i ) = – (mZn CZn (Tf – TZn i) + mAl CAl (Tf – TAl i))
(200.0 g)(4.184 J/g°C)(Tf – 22.4°C) = – ((38.1 g)(0.388 J/g°C)(Tf – 100.2°C) + (25.9 g)(0.897 J/g°C)(Tf – 100.2°C))
836.8 J/°C Tf – 18744.32 J = – (14.7828 J/°C Tf – 1481.23656 J + 23.2323 J/°C Tf – 2327.87646 J)
874.8151 J/°C Tf = 22553.43302 J
Tf = 25.78079987°C
25.8°C
Answer: ___________________
7. Given the equation below, determine how many grams of methane (CH4) must be combusted in a
Bunsen burner to bring a beaker with 325.7 g of water at 24.3ºC up to the normal boiling point?
Assume that all of the heat generated by the burner is absorbed by the water.
CH4(g) + 2 O2(g)  CO2(g) + 2 H2O(g) + 891 kJ
qw = mw Cw (Tw f – Tw i)
75.7°C
qw = (325.7 g) (4.184 J/g°C) (100.0°C – 24.3°C) = 103158.5702 J
103158.5702 J
1 kJ
1 mol CH4
16.0426 g CH4
1000 J
891 kJ
1 mol CH4
= 1.857386844 g CH4
1.86 g CH4
Answer: ___________________
8. A 79.3 L sample of a gas, at a constant pressure of 2.05 atm, absorbs 0.695 kJ of heat and expands to
82.1 L. What is the change in internal energy of the gas in joules?
E = q + w = q + – pV
2.8 L
E = 0.695 kJ
1000 J
+ – (2.05 atm) (82.1 L – 79.3L)
1kJ
E = 695 J – 581.462 J = 11 3.538 J
110 J or 1.1 x 102 J
Answer: ___________________
101.3 J
1 L·atm
9. Given the following data, use Hess’s Law to calculate the H for the overall reaction:
H = – 191.2 kJ
H2(g) + O2(g)  H2O2(aq)
C6H4O2(aq) + H2(g)  C6H4(OH)2(aq)
H = – 177.4 kJ
H2(g) + ½ O2(g)  H2O(g)
H = – 241.8 kJ
H2O(g)  H2O(l)
H = – 43.8 kJ
Overall reaction:
C6H4(OH)2(aq) + H2O2(aq)  C6H4O2(aq) + 2 H2O(l)
Required work:
 rewrite the equations above with the correct orientation, multipliers, and H values
 show that after cancelling, the sum of your reactions gives the desired overall reaction
C6H4(OH)2(aq)  C6H4O2(aq) + H2(g)
H = –1 (– 177.4 kJ)
H2O2(aq)  H2(g) + O2(g)
H = –1 (– 191.2 kJ)
H = 2 (– 241.8 kJ)
2 H2(g) + O2(g)  2 H2O(g)
H = 2 (– 43.8 kJ)
2 H2O(g)  2 H2O(l)
C6H4(OH)2(aq) + H2O2(aq)  C6H4O2(aq) + 2 H2O(l)
H = – 202.6 kJ
– 202.6 kJ
Answer: ___________________
10. Using the provided H values, calculate H
Note: H
NH3(g)
= – 46 kJ/mol, H
NO(g)
for the following reaction:
= 90 kJ/mol, and H H2O(g) = – 242 kJ/mol
4 NH3(g) + 5 O2(g)  4 NO(g) + 6 H2O(g)
H
H
H
= [n NOH
=  [nH (products)] −  [nH (reactants)]
NO(g)
+ n H2OH
H2O(g)]
− [n NH3H
NH3(g)
+ n O2(g)H
O2(g)]
= [4 mol (90 kJ/mol) + 6 mol(−242 kJ/mol)] − [4 mol(−46 kJ/mol) + 5 mol(0 kJ/mol)]
H = [−1092 kJ] − [−184 kJ]
H
– 908 kJ
Answer: ___________________
= −908 kJ
11. Ethane, C2H6 (H3C–CH3), is burned in air. Write the balanced equation, and calculate H
the average bond energies provided.
2 C2H6 + 7 O2  4 CO2 + 6 H2O
Check:
C 4
H 12
O 14
charge 0




Bond
C≡C
C=C
C–C
C–H
C=O
O=O
O–H
using
Bond Energy
(kJ/mol)
839
614
347
413
799 (in CO2)
498
467
H
=  [ n reactants)] −  [ n products)]
H
= [ 2(C–C) + 12(C–H) + 7(O=O) ] – [ 8(C=O) + 12(O–H) ]
H
= [2 mol(347 kJ/mol) + 12 mol(413 kJ/mol) + 7 mol(498 kJ/mol)] – [8 mol(799 kJ/mol) + 12 mol(467 kJ/mol)]
H = [ 9136 kJ ] – [ 11996 kJ ]

H
− 2860. kJ
Answer: ____________________
= – 2860. kJ