CHEM 101DL Study Guide - Midterm Guide: Tetrafluoromethane, Methanol, Oxygen Difluoride
Chemistry 10D1L
Dr. T. Woerner
Summer I, 2018
Thermodynamics Problems - Solutions
1. The combustion of 1.000 g of ethyl alcohol, C2H5OH(
!
), in a bomb calorimeter (at
constant volume) evolves 29.62 kJ of heat at The products of the combustion are
CO2(g) and H2O(
!
). The molecular weight of ethyl alcohol is 46.06 g/mole. a) What is
for the combustion of one mole of ethyl alcohol? b) Write the balanced equation for
the reaction. What is the value of for the combustion? c) Use values from the
attached table and your answer to part (b) above to calculate for ethyl alcohol.
a) a bomb calorimeter measures directly as qv = . 1.000 g C2H5OH(!) gives off
29.62 kJ of heat. There are 46.06 g/m for C2H5OH(!).
Thus =
Note: the (–) sign indicated that heat has been given off.
b) C2H5OH(!) + 3O2(g) 2CO2(g) + 3H2O(!)
c)
25oC
Δ
Eo
Δ
Ho
Δ
Ho
f
ΔEo
ΔEo
ΔEo
!
"– 29.62 kJ
g #
$ !
"46.06 g
mole #
$ = – 1364.3 kJ
mole
ΔHo = ΔEo + ΔnRT
here Δn = (2 moles of gaseous CO2 product) − (3 moles of gaseous O2 reactant)
∴ Δn = – 1
So, ΔHo = $
%– 1364.30 kJ
mole &
' + (– 1) $
%8.314X10−3 kJ
K . mole &
'(298 K)
∴ ΔHo = – 1366.78 kJ
mole
ΔHo = ∑ ΔHo
f (products) − ∑ ΔHo
f (reactants)
ΔHo = [3(ΔHo
f H2O(l)) + 2(ΔHo
f CO2(g))] − [3(ΔHo
f O2(g)) + ΔHo
f C2H5OH(l)]
– 1366.78 = $
%
&3 '
(– 286.10 kJ
mole )
* + 2 '
(– 393.51 kJ
mole )
*
+
,
- − $
%
&3 '
(0 kJ
mole )
* + (ΔHo
f C2H5OH(l)) +
,
-
ΔHo
f C2H5OH(l) = – 278.54 kJ
mole
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2. Calculate for the combustion of octane, C8H18(
!
). The products of the
combustion reaction are CO2(g) and H2O(
!
) at . The value of for this reaction
at is –5470.71 kJ/mole of octane.
C8H18(!) + 12.5O2(g) 8CO2(g) + 9H2O(!)
Note: if this balancing result is bothersome for you due to the fraction, you could double
all the coefficients, but be aware that the reaction would then be written for 2 moles of
octane.
3. Calculate and for the reaction of Ca3P2(s) with H2O(
!
). The reaction
produces Ca(OH)2(s) and PH3(g). The enthalpy of formation values needed for the
calculation may be found in the attached table.
Ca3P2(s) + 6H2O(!) 3Ca(OH)2(s) + 2PH3(g)
4. For the reaction
Δ
Eo
25oC
Δ
Ho
25oC
ΔEo = ΔHo – ΔnRT
ΔEo = – 5470 kJ
mole – (– 4.5) "
#8.314X 10–3
kJ
K mole $
%(298 K)
ΔEo = – 5470 kJ
mole + 11.15 kJ
mole = – 5458. 85 kJ
mole
Δ
Eo
Δ
Ho
ΔHo=ΔHf
o(products) −
∑ΔHf
o(reactants)
∑
ΔHo=2 5.47 kJ
mole
%
&
'
( +3−968.59 kJ
mole
%
&
'
(
)
*
+
,
-
. −6−286.10 kJ
mole
%
&
'
( +−504.17 kJ
mole
%
&
'
(
)
*
+
,
-
.
=−674.06 kJ
mole
ΔEo=ΔHo− ΔnRT
ΔEo=−674.06 kJ
mole −2
( )
8.314 ×10−3kJ
K⋅mole
&
'
(
) 298K
( )
=−679.02 kJ
mole
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find more resources at oneclass.com
Document Summary
Thermodynamics problems - solutions: the combustion of 1. 000 g of ethyl alcohol, c2h5oh(! The products of the combustion are constant volume) evolves 29. 62 kj of heat at. The molecular weight of ethyl alcohol is 46. 06 g/mole. a) what is. E for the combustion of one mole of ethyl alcohol? b) write the balanced equation for the reaction. H o for the combustion? c) use values from the attached table and your answer to part (b) above to calculate. Ho f for ethyl alcohol: a bomb calorimeter measures. $ = 1364. 3 directly as qv = " 29. 62 mole # kj g kj mole o. Note: the ( ) sign indicated that heat has been given off: c2h5oh(!) Ho = eo + nrt here n = (2 moles of gaseous co2 product) (3 moles of gaseous o2 reactant) K . mole & kj mole & kj mole. Ho = 1366. 78 f (reactants) f (products) ho.