Document 6519418

6/7/2011
1 What is ∆U?
• 2.00 g of He(g) with CV.m = 1.5R essentially
independent of temperature undergoes a
reversible constant-pressure expansion from
20.0 to 40.0 dm3 at 0.800 bar.
2400
Paul Franklyn – School of Chemistry – Wits - 1
J
1 What is ∆U?
•
2.00 g of He(g) with CV.m = 1.5R essentially independent of temperature undergoes a
reversible constant-pressure expansion from 20.0 to 40.0 dm3 at 0.800 bar.
• ANSWER:
For a perfect gas
Cv =
dU
dT
∴ dU = CvdT.
For an appreciable change
f
Tf
∆U = ∫ dU = ∫ CV dT
i
Ti
Since Cv is given as independent of temperature,
∆U = Cv.∆T = nCv.m∆T = (0.500 mol)(1.5R)(384 K) = 2.40 kJ
Paul Franklyn – School of Chemistry – Wits - 2
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6/7/2011
2 What is ∆H?
• 2.00 g of He(g) with CV.m = 1.5R essentially
independent of temperature undergoes a
reversible constant-pressure expansion from
20.0 to 40.0 dm3 at 0.800 bar.
4000
Paul Franklyn – School of Chemistry – Wits - 3
J
2 What is ∆H?
•
•
2.00 g of He(g) with CV.m = 1.5R essentially independent of temperature undergoes
a reversible constant-pressure expansion from 20.0 to 40.0 dm3 at 0.800 bar.
ANSWER:
Similarly, for a perfect gas
∴
Cp = dH
dT
dH = Cp dT ,
which, on integration, gives
∆H = Cp.∆T = nCp.m∆T = (0.500 mol)(2.5R)(384 K)
= 4.00 kJ
Paul Franklyn – School of Chemistry – Wits - 4
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6/7/2011
3 Calculate w:
• 2.00 g of He(g) with CV.m = 1.5R essentially
independent of temperature undergoes a
reversible constant-pressure expansion from
20.0 to 40.0 dm3 at 0.800 bar.
-1600
Paul Franklyn – School of Chemistry – Wits - 5
J
3 Calculate w:
• 2.00 g of He(g) with CV.m = 1.5R essentially
independent of temperature undergoes a
reversible constant-pressure expansion from
20.0 to 40.0 dm3 at 0.800 bar.
• ANSWER:
Vf
w = − ∫ p.dV = -P(V - V ) = -(0.789 atm)(2000cm3)
2
1
Vi
= 1578 atm cm3
Hence, w = 1578 [(8.315 J)/82.06] = -1.60 kJ.
Paul Franklyn – School of Chemistry – Wits - 6
3
6/7/2011
4 Calculate q:
• 2.00 g of He(g) with CV.m = 1.5R essentially
independent of temperature undergoes a
reversible constant-pressure expansion from
20.0 to 40.0 dm3 at 0.800 bar.
4000
Paul Franklyn – School of Chemistry – Wits - 7
J
4 Calculate q:
• 2.00 g of He(g) with CV.m = 1.5R essentially
independent of temperature undergoes a
reversible constant-pressure expansion from
20.0 to 40.0 dm3 at 0.800 bar.
• ANSWER:
– q = qp = ∆H = 4.00 kJ.
Paul Franklyn – School of Chemistry – Wits - 8
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6/7/2011
5 Calculate ∆Ssys?
• When Argon at 25 oC and 1 bar in 0.5 dm3
contained is allowed to compress to 0.0500
dm3 and is simultaneously cooled to -25 oC.
-0.43
Paul Franklyn – School of Chemistry – Wits - 9
J/K
6 Calculate ∆H°370 :
• Given ∆H°298 = -956.5 kJ mol-1
2HN3(g) + 2NO(g) → H2O2(l) + 4N2(g)
Use data in the Appendix and the approximation
of neglecting the temperature dependence of
C°P,m to estimate ∆H°370 for the reaction
above. (to 4 sig figs)
952400
Paul Franklyn – School of Chemistry – Wits - 10
J/mol
5
6/7/2011
6 Calculate ∆H°370 :
• Given ∆H°298 = 956.5 kJ mol-1
2HN3(g) + 2NO(g) → H2O2(l) + 4N2(g)
• ANSWER:
Using Kirchhoff's Law:
∆H°(Tf) = ∆H°(Ti) +
Tf
∫ ∆C dT
o
p
------------------------(1)
Ti
∆C°P = C°P(H2O2) + 4C°P(N2) - 2C°P(HN3) - 2C°P (NO)
= 89.1 + 4(29.125) - 2(43.68) - 2(29.844)
= 58.55 J K−1 mol−1
Tf = 370 K
Ti = 298 K
Substituting numerical values into Kirchhoff's Law := −956 500 J mol−1 + 58.88 J K−1 mol−1 x (70 K)
−1
= −952.4 kJ Paul
mol
Franklyn – School of Chemistry – Wits - 11
7 How old (in years)?
• A Mayan mummy was radiocarbon dated by
three different laboratories. The three 14C/12C
ratios were found to be 63.10%, 62.91% and
64.15% of that found in living organisms. Use
the half-life for 14C as 5715 years to determine
the average age of the Mayan mummy.
Do not enter any units!!!
3759
Paul Franklyn – School of Chemistry – Wits - 12
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6/7/2011
7 Answers
Correct answer is: 3759
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8 Calculate EA
• For a reaction whose rate constant at room
temperature is doubled by an increase in
temperature (T) by 10oC.
53000
Paul Franklyn – School of Chemistry – Wits - 14
J/mol
7
6/7/2011
8 Answers
Correct answer is: 53000J/mol
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8