1 What is ∆U?
• A monatomic perfect gas at 0 oC is expanded
from 22.4 dm3 to 44.8 dm3.
0
Paul Franklyn – School of Chemistry – Wits - 1
J
1 What is ∆U?
• A monatomic perfect gas at 0 oC is expanded
from 22.4 dm3 to 44.8 dm3.
• ANSWER:
– ∆U=0 internal energy is a function of T only
Paul Franklyn – School of Chemistry – Wits - 2
2 What is ∆H?
• A monatomic perfect gas at 0 oC is expanded
from 22.4 dm3 to 44.8 dm3.
0
Paul Franklyn – School of Chemistry – Wits - 3
J
2 What is ∆H?
• A monatomic perfect gas at 0 oC is expanded
from 22.4 dm3 to 44.8 dm3.
• ANSWER:
– ∆H = 0J because it is isothermal
Paul Franklyn – School of Chemistry – Wits - 4
3 Calculate w:
• A monatomic perfect gas at 0 oC is expanded
from 22.4 dm3 to 44.8 dm3 when this is done
reversibly
-15700
Paul Franklyn – School of Chemistry – Wits - 5
J
3 Calculate w:
• A monatomic perfect gas at 0 oC is expanded
from 22.4 dm3 to 44.8 dm3 when this is done
reversibly
• ANSWER:
Vf
dV
w = −nRT ∫
= - nRT In[Vf/Vi]
V
Vi
w = -1.57 kJ
Paul Franklyn – School of Chemistry – Wits - 6
4 Calculate w:
• A monatomic perfect gas at 0 oC is expanded
from 22.4 dm3 to 44.8 dm3 when this is done
under constant external pressure.
-1130
Paul Franklyn – School of Chemistry – Wits - 7
J
4 Calculate w:
• A monatomic perfect gas at 0 oC is expanded from 22.4 dm3 to 44.8 dm3
when this is done under constant external pressure.
• ANSWER:
•
w = −p.∆V
∆V = (44.8 -22.4) dm3 = 22.4 dm3
pV = nRT
so pext = pf = nRT/ Vf = (1.00 mol)(0.08206 dm3 atm K-1 mol-1)(273K)
44.8 dm
= 0.500 atm = 5.065 x 104 Pa
w = −p.∆V = −5.065 x 104 Pa x 22.4 dm3 = -1.13 kJ
Paul Franklyn – School of Chemistry – Wits - 8
5 What is w?
• A monatomic perfect gas at 0 oC is expanded
from 22.4 dm3 to 44.8 dm3 when this is done
freely.
0
Paul Franklyn – School of Chemistry – Wits - 9
J
5 What is w?
• A monatomic perfect gas at 0 oC is expanded
from 22.4 dm3 to 44.8 dm3 when this is done
freely.
• ANSWER:
– No restoring pressure
– So nothing to do work against
Paul Franklyn – School of Chemistry – Wits - 10
6 Calculate ∆S for:
• the following changes in state of 2.50 mol of a
perfect monoatomic gas with CV.m = 1.5 R for
all temperatures.
• (1.50 atm, 400 K) → (3.00 atm, 600 K)
6.66
Paul Franklyn – School of Chemistry – Wits - 11
J/K
6 Calculate ∆S for:
• the following changes in state of 2.50 mol of a perfect
monoatomic gas with CV.m = 1.5 R for all temperatures.
• (1.50 atm, 400 K) → (3.00 atm, 600 K)
• ANSWER:
– For an ideal gas with CV.m = 1.5 R at all T's, Eq (Eq 1 on
page 26 of pink book) gives:
∆S = CV ln (T2/T1) + nR ln (V2/V1)
= nR[1.5 ln(T2/T1) + ln(V2/V1)]
= (2.50 mol)(8.314 J mol-1 K-1)(1.5 ln 1.5 + ln 0.75)
= 6.66 J/K-1
Paul Franklyn – School of Chemistry – Wits - 12
7 How long (in seconds): to 3 sig figs?
• Given the following second order
reaction studied at 383oC:
NO2(g) →NO(g) + O2(g)
• Given that the rate constant for
this reaction is 10 M-1s-1 and the
experimental data on the right.
• How long for the concentration
of NO2(g) to drop down to 2.35%
of what it originally was?
Do not enter any units!!!
41.6
Paul Franklyn – School of Chemistry – Wits - 13
Time / [NO2(g)] /
s
M
0.0
0.100
5.0
0.017
10.0
15.0
0.0090
0.0062
20.0
0.0047
8 What is the activation energy?
• Calculate the activation energy for a reaction
whose rate constant at room temperature is
doubled by an increase in temperature (T) by
10oC.
53
Paul Franklyn – School of Chemistry – Wits - 14
kJ/mol