Winter 2009
Chemistry 452
First Examination February 4, 2009 First Page: Useful information and equations: Law IA: ∆U = q + w U = U (T ,V ) Law IB:
Equations of State (EoS) I.G.: PVm = RT
∂U ∂U dT + dV ∂ ∂ T V V T
dU =
H = U + PV Heat Capacity ∂U ∂H CV = and C P = ∂T V ∂T P
Work: w = − Pext ∆V or w = − Pext d V w = mgh and w = IQt
Kinetic Energy: ε =
mv
2
2 Isothermal Reversible work: V f w = −nRT ln V i PV = Const Adiabatic Reversible work: w = − PdV C P , m γ P (V ) = Const ; where γ = C V , m
Integral Identity:
∂ Z ∆ Z = ∫ dx ∂ x y x = z
dx dx dx dy dx =
dz
dz dy
+y
P=
VdW EoS:
RT Vm − b
−
a V m
2
Thermodyanamic EoS ∂U ∂P = T ∂V ∂T − P T V Cyclic rule: dx dz dy = −1 dy z dx y dz x Reaction Enthalpies o o ∆ H rxn = ν i ∆H f ,i
∑ i
o
dni = ni − ni = ν i dx ∆C P , rxn =
∑ν C i
P,i
i
∆ H
o
o
= ∆H rxn dx
R = 0.082 L − atm / mol − K
i
d ( y)
∆U = CV ∆T and ∆H = C P ∆T
Constants: R = 8.314 J / mol − K
x f
d ( yz )
CP , m = CV , m + R
d ( z) dx
R ⋅ 298.13 = 2.48kJ / mol
1Cal = 4.18kJ 1cal = 4.18 J 1bar = 105 Pa T ( K ) = T (C ) + 273.15 g = 9.8 m
sec
2
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Winter 2009
Chemistry 452 Name ______________________________ ID____________
Show your work throughout, and always show units for computed quantities Q1) Given the heats of formation and Heat Capacities of the following substances, o o Substance C P J/mol-K ∆ H f kJ/mol
O2 ( g )
0.0
29.4
CO2 ( g )
-393.5
37.1
H 2O ( l )
-285.8
75.3
C6 H12O6 ( s )
-1273.0
219.2
The most basic form of energy consumption people do is burn sugar for energy. The 0 atmospheric pressure is 1 Bar and body temperature is 37 C. The central chemical equation for this process is: C6 H12O6 ( s ) + 6O2 ( g ) → 6CO2 ( g ) + 6H 2O (l ) o a) What is the standard enthalpy of this reaction as written at 298K, ∆ H rxn ( 298K ) ?
∆ H rxn = − {6 ⋅ 393.5+6 ⋅ 285.8 − 1⋅ 1273.0 − 6 ⋅ 0} = − 2.8 MJ o
mol − rxn
b) The standard reference temperature is different from body temperature. Do we get more (or less) energy by burning the sugar in our bodies as opposed to standard temperature? [Be sure to show the computation that bolsters your conclusion.] CP ,Rxn = {6 ⋅ 37.1 + 6 ⋅ 75.3 −1 ⋅ 219.2 − 6 ⋅29.4} = 280 J / mol − K > 0 o
∆ H rxn (T ) = ∆H rxn (To ) + CP , Rxn ∆T
Because the net heat capacity is positive, and DT>0, the heat of reaction will be positive or less exothermic. So we will get less energy from the reaction, as a result of doing it at a higher temperature. c) The molecular weight of glucose is 180g/mol, how many Calories (kcal) do we get from this reaction per gram of glucose? 1 mol 1 cal o 6 ∆ H rxn = −2.8 ⋅10 J ⋅ 1 mol − rxn ⋅ ⋅ mol − rxn mol − glucose 180 gram 4.18 J =−
2.8 ⋅103 180 ⋅ 4.18
3 ⋅10 cal
= −3.7 ⋅10 = −3.7 Cal 3
g
g
The minus sign can be omitted because we know it is energy we get from the reaction, so the reaction is exothermic.
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Winter 2009
Chemistry 452
Q2) A typical soft drink (12 ounce, Oz.) contains 40 grams of glucose.
a) How much energy do we obtain from a single soft drink? [You can assume the burning is done under standard conditions.] 40 ⋅ 3.7 = 148Cal / soft − drink b) Many times we add ice to cool a soda. Two ice cubes weights about 1.5 ounces. How cool would the (12 ounce) soda be after adding the ice, assuming the soda started at o 25 C? The heat capacity of water is 1 cal/gram/degree (that’s how a calorie was defined) and the latent heat of ice is 80 cal/gram. [16 ozs is one pound, and 1kg is 2.2 pounds.] The soda drops in temperature from Tw=25C to the final temperature T, the ice melts with the heat of melting (which is positive, as melting is endothermic) and the water of the ice cubes warm to the final temperature from the Tice which is 0C. All heats are kept in the container (it is adiabatic) so the total heat is zero. ∆ H = 0 = mSoda CP , g (T − Tw ) + mIce ∆H f , g + mIceC P , g (T − T Ice ) CP , g = 1 cal mSoda =
12
gK
mIce =
1.5 16
⋅ lb ⋅
1 kg 3 g ⋅1⋅ 10 = 42.6 g lb kg 2.2
1 kg 3 g ⋅1⋅ 10 = 341g lb kg 16 2.2 ∆T ≡ T − T Ice T − Tw = T − TIce + TIce − Tw = ∆T − (Tw − TIce ) = ∆T − 25 ⋅ lb ⋅
∆ H = 0 = mSoda CP , g (T − Tw ) + mIce ∆H f , g + m IceC P , g (T − T Ice )
0 = mSoda CP , g ( ∆T − 25 ) + mIce ∆H f , g + mIceC P , g ( ∆T ) 25 ⋅ mSoda = ( mSoda + mIce ) ∆T + mIce ∆H f ,g ∆T =
25 ⋅ mSoda − mIce ∆H f , g
( mSoda + mIce )
=
25 ⋅ 341 − 42.6 ⋅ 80
( 341 + 42.6 )
= 13K
The final temperature then is 13 degrees C, because the rise is 13K, and the start point is 0C, the freezing point of ice. A mass of ice, a bit over 12% of the mass of the soda, cooled the soda half way to freezing.
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Winter 2009
Chemistry 452
c) Your legs can easily push 100 pounds (60 kg) up and down. Assuming the leg extension is about 2 feet (0.7 meters), how many times of pushing 100 pounds would you need to burn off the energy from the soft drink (in part a of this question)? ∆ ( PV ) = RT ∆ngas = 0 3
∆ H = 148 ⋅ 4.18 ⋅10 J = ∆U = w = Nmg ∆h = N 60 ⋅ 9.8 ⋅ 0.7
N =
148 ⋅ 4.18 ⋅103 60 ⋅ 9.8 ⋅ 0.7
= 1,500reps
d) Let’s consider the basal level of energy consumption by just breathing. In breathing we consume about 0.01 moles of oxygen per minute. Assuming that the oxygen is used to burn glucose, how much energy do we consume per hour, just breathing? [Use results from Q1) a] 0 ∆ H rxn 2.8MJ ∆U = ⋅ nO = ⋅ 0.6 = 280 kJ 2 hr ν 6 O2
nO2 = 0.01 moles
min
⋅ 60 min
hr
= 0.6 moles
hr
e) If you sit around and drink the can of soda (above) your body will convert the sugar to fat. Neglecting the energy of conversion, but noting that fats have energy of 9 Cal/gram as opposed to sugars (which have the amount from above) how much weight would you gain if the soda were converted to body fat? 3.7 Cal m = 40 g − sugar *
9 Cal
g − sugar
= 16. g − fat
g − fat So you don’t gain weight directly by the sugar, but get about ½ the weight because it is converted to fat. Of course some is lost in just converting sugar to fat (and burning the fat to get energy).
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Winter 2009
Chemistry 452
Q3) One mole of an ideal gas, for which CV , m = 52 R undergoes two successive changes
in state. 1) From 25C at 2 Atm, the gas expands isothermally against a constant pressure of .5 atm to twice its initial volume. 2) At the end of the previous step the gas is cooled at constant volume from 25C to -40C. a) Calculate q, w, ∆U and ∆H for the first step Isothermal ∆U = q + w = CV ∆T = 0
∆H = C P ∆T = 0
q = − w = Pext ∆V = 0.5 ⋅12 = 6 − Atm = 600 J V2 = 2V1 ⇒ ∆V = 2V1 − V1 = V1 =
nRT 1 P1
=
0.08 ⋅ 298 2
= 12
w = −600 J
b) Calculate q, w, ∆U and ∆H for the second step Cool at constant Volume ∆V = 0 w=0 ∆U = CV ∆T = 1 ⋅ 5 R ∆T = − 5 ⋅ 8.3 ⋅ 65 = −1.35kJ 2 2 ∆ H = C P ∆T = 1⋅ 7 R∆T = −1.9kJ 2
q = ∆U = −1.35kJ
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Winter 2009
Chemistry 452
∂U for the van der ∂V T
Q4) a) Using the Thermodynamic equation of state, evaluate
Waals gas. P=
RT Vm − b
∂U ∂V T
−
a Vm 2
R ∂P ⇒ = ∂T V Vm − b
RT RT a a ∂P =T − = − − = P 2 Vm − b Vm − b V m Vm 2 ∂T V
=
an
2
V2
b) Assume 5 moles of a van der Waals’ gas is allowed to expand isothermally, initially at 2 Atm, from ½ liter to 5 liters. This van der Waals gas (similar to oxygen) is described 2 5 by b = 0.003 and a = 2 atm 2 , and CV , m = 2 R . What is the change in the mol mol internal energy of the gas for this process? V2 =5
∂U ∆U = ∫ dV ∂ V T V =.5 1
2 a = 2 atm 2
V 2 = 5
= an
2
∫
V 1 =.5
1 V
2
2
1
dV = − an
V2
−
1
= an {2 − .2} 1 V1 2
mol 2
∆U = 2 ⋅ 5 ⋅1.8 = 90 − Atm = 9kJ
c) Compare this result with that of an ideal gas. The energy increased as the volume went up. For an ideal gas there would be no change.
d) Explain why the change in energy has the sign that it does. The internal energy increased because the mean distance between molecules increased. Therefore, the stable interactions among the gas molecules were broken or disrupted. This is the equivalent of bond breaking, which is endothermic, but it is a small effect, 9 kJ.
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