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FOCUS 4 Online Cumulative Example
Your first experience with chemical transformations in a classroom might have been the
reaction of baking soda (NaHCO3) with vinegar (CH3COOH(aq)) to produce carbon
dioxide, water, and sodium acetate (NaCH3CO2). You wish to examine the thermodynamics of this reaction that you enjoyed as a child. A solution of baking soda is prepared
by dissolving 15.9 g of NaHCO3 in 50. mL of water. To this you add 200. mL of household vinegar, which is 0.833 m CH3COOH(aq). The reaction is carried out in an open
container in a room in which the temperature is 22.1 8C and the atmospheric pressure
is 1.00 atm.
(a) Use the data in Appendix 2A to write the thermochemical equation.
(b) Calculate the following quantities for the reaction: DH8, DU8, DS8, and DG8.
(c) What volume of CO2(g) is produced?
(d) Predict the values of w and q for this experiment and the corresponding changes
in the internal energy and enthalpy (DUexpt and DHexpt).
(a) PLAN The chemical equation for the reaction is
HCO32 (aq) 1CH3COOH(aq) ¡ CO2 (g) 1 H2O(l) 1 CH3CO2
2 (aq)
Use Appendix 2A to find DH8 for this reaction.
SOLVE
From Eq. 2 in Topic 4D,
¢H° 5 a n¢Hf°(products) 2 a n¢Hf°(reactants)
HO
2
3
2
6
474
8
3 (2285.83 kJ?mol21) 1 (1 mol) 3 (2486.0 kJ?mol21 ) 4
HCO32
CH3COOH
6
474
8
6
474
8
23(1 mol) 3 (2692.0 kJ?mol21 ) 1 (1 mol)
–1177.8 kJ
¢H° 5 3 (1 mol) 3 (2393.51 kJ?mol ) 1 (1 mol)
CH CO
Elements
2
6
474
8
21
–1165.3 kJ
CO
2
6
474
8
DHº = +12.4 kJ
Products
Reactants
3 (2485.76 kJ?mol21 )4 5 112.4 kJ
The thermochemical equation is therefore
HCO32 (aq) 1 CH3COOH(aq) ¡ CO2 (g) 1 H2O(l) 1 CH3CO22 (aq)
¢H° 5 112.4 kJ
(b) ANTICIPATE Because a gas is being formed, you should expect the value for DS8 to
be positive.
PLAN Use Eq. 1 of Topic 4D, to find DU8, Eq. 2 from Topic 4H to find DS8 and Eq. 3
from Topic 4J to find DG8.
SOLVE
Find DU8, Eq. 1 from Topic 4D in the form, ¢U ° 5 ¢H ° 2 ¢ngas RT
From the chemical equation in part (a), ¢ngas 5 11 mol
Dn
gas
6
474
8
¢U° 5 12 500 J 2 (1 mol) 3 (8.3145 J?K21?mol21 ) 3 (295.2 K) 5 110.0 kJ
1
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Online Cumulative Example
Find DS8 from Eq. 2 in Topic 4H,
¢S° 5 a nSm°(products) 2 a nSm°(reactants)
CO
Products
HO
2
6
474
8
2
6
474
8
¢S° 5 3(1 mol) 3 (213.74 J?K ?mol ) 1 (1 mol)
21
21
CH CO
Reactants
2
3
2
6
474
8
3 (69.91 J?K21?mol21) 1 (1 mol) 3 (86.60 J?K21?mol21 )4
HCO
2
CH3COOH
3
6
474
8
6
474
8
2 3 (1 mol) 3 (91.2 J?K ?mol ) 1 (1 mol)
21
21
DSº =
+100.3 J·K–1
+370.2 J·K–1
Focus 4
+269.9 J·K–1
2
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Elements
3 (178.7 J?K21?mol21 ) 4 5 1100.3 J?K21
For the next step, note that this value can be written as 10.1003 kJ?mol21.
From Eq. 3 in Topic 4J, ¢G° 5 ¢H ° 2 T¢S°
DH°
6
474
8
T
64748
DS°
644474448
¢G° 5 12.4 kJ 2 (295.2 K) 3 (0.1003 kJ?K21 ) 5 217.2 kJ
This negative value implies that the reaction is spontaneous under standard conditions
(Topic 4J).
EVALUATE The value for DS8 is positive as predicted.
(c) PLAN Write the balanced equation, determine the limiting reactant, then use the
ideal gas law to find the volume of CO2(g).
SOLVE
First write the complete equation
NaHCO3 (aq) 1 CH3COOH(aq) ¡ CO2 (g) 1 H2O(l) 1 NaCH3CO2 (aq)
Find the limiting reactant
15.9 g
Amount of CO2 produced from 15.9 g of NaHCO3
15.9 g NaHCO3 3
1 mol NaHCO3
1 mol CO2
3
84.01 g
1 mol NaHCO3
1 mol
3
1 mol
84.01 g
= 0.189 mol
5 0.189 mol CO2
Amount of CO2 produced from 0.200 L of CH3COOH(aq)
0.200 L 3
0.833 mol CH3COOH
1 mol CO2
3
L
1 mol CH3COOH
1 mol
3
0.833 mol
1 mol
3
3
1 mol
0.200 L 1 L
5 0.167 mol CO2
= 0.167 mol
Therefore, CH3COOH(aq) is limiting and 0.167 mol of CO2 is produced.
Use the ideal gas law in the form of V 5 nRT/P, with the value of R chosen to match the units
supplied and required (atm and L, respectively), to convert this amount of CO2 to a volume of
gas at 22.1 8C and 1.00 atm:
T 5 22.1 1 273.15 K 5 295.2 K
V5
(0.167 mol) 3 (0.08206 L?atm?K21?mol21 ) 3 (295.2 K)
5 4.04 L
1.00 atm
The change in volume of gas in the experiment is therefore DV 5 14.04 L.
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Focus 4
(d) ANTICIPATE Because a gas is being formed, the system is losing energy as expansion work and you should expect w to be negative; because the reaction is endothermic,
you should expect q to be positive.
PLAN Find the expansion work from w 5 2PexDV. Find q from the amount of products
formed and the thermochemical equation; since the reaction is performed at constant
pressure, find the enthalpy change from DHexpt 5 qP. The change in internal energy is
found by using the first law of thermodynamics in the form DUexpt 5 w + q.
What should you assume? Assume that the volume of solution remains unchanged.
SOLVE
From w 5 2PexDV
From part (a) you know that DV 5 14.04 L
w 5 21.00 atm 3 4.04 L 5 24.04 L?atm
DV = +4.04 L
Convert this quantity to joules
w 5 24.04 L?atm 3
101.325 J
5 2409 J
1 L?atm
corresponding to 20.409 kJ (for later use).
w = –PexDV = –409 J
Find q from the amount of CO2 produced and the thermochemical equation from part (b),
noting that the equation refers to the formation of 1 mol CO2 but in the experiment only
0.167 mol CO2 is produced.
q 5 12.5 kJ 3
0.167 mol CO2
5 12.09 kJ
1 mol CO2
Find DHexpt from DHexpt 5 qP
Because the reaction is occurring at constant pressure
¢Hexpt 5 qP 5 12.09 kJ
Find DUexpt from the first law, DUexpt 5 w + q
w
64748
q
+2.09 kJ
+1.68 kJ
64748
¢Uexpt 5 20.409 kJ 1 2.09 kJ 5 11.68 kJ
w
q
–0.409 kJ
ΔUexpt
EVALUATE The value for the transfer of energy as work is negative (work is done) and
that for heat is positive (heat is supplied), as anticipated.
Online Cumulative Example
3