GENERAL CHEMISTRY I
QUIZ 5 and QUIZ 6
November 14, 2002
INSTRUCTIONS:
PRINT OUR NAME ————>
NAME
1. Determine the reaction enthalpy for the hydrogenation of ethyne to ethene,
C2H2(g) + H2(g)
C2H4(g)
From the following data:
.
First circle the reactants and box the products for the target reaction in Rxn 1-3
2 C2H2(g) + 5 O2(g)
4 CO2(g) +
2 H2O(liq)
-2600
C2H4(g) + 3 O2(g)
2 CO2(g) +
2 H2O(liq)
-1408
2 H2(g) + O2(g)
2 H2O(l)
-572
Rxn 1 and 3 are written forward to keep the reactants C2H2 and H2 on the products side. Rxn 2
much be reversed to get C2H4 onto the products side. Also Rxn 1 and 3 must be multiplied by ½ to
get the correct number of moles of the two reactants. The same mathematical operations must be
applied to the ∆H values.
(Rxn 1 × ½)
C2H2(g) + 5/2 O2(g)
2 CO2(g) +
2 CO2(g) +
C2H4(g) + 3 O2(g)
-1408 × (-1) = +1408
H2O(l)
-572 × ½
2 H2O(liq)
H2(g) + ½ O2(g)
H2O(liq)
-2600 × ½
=
=
-1300
-286
-178 kJ
2. The following reaction for the combustion of methane has a ∆Hrxn of –890 kJ/mol.
CH4(g) + 2 O2(g)
CO2(g) +
2H2O(liq)
If 2.98 g of methane are completely burned. The heat released is transferred to a sample
of liquid ethanol, C2H5OH, which is at the boiling point of 351.5 K. How many g of
ethanol can be converted to the gaseous phase if none of the heat is lost to the
surroundings? The enthalpy of vaporization , ∆Hvap, of ethanol is 838 J/g
g CH4
mol CH4
heat(kJ)
heat (J)
g ethanol
1 _ mol _ CH 4
2.98 _ g _ CH 4 ×
= 0.186 _ mol _ CH 4
16.04 _ g _ CH 4
890 _ kJ
0.186 _ mol _ CH 4 ×
= 166 _ kj _ heat = 166,000 J heat
1 _ mol _ CH 4
1 _ g _ ethanol
166000 _ kJ _ heat ×
= 198 _ g _ ethanol
838 _ J _ ethanol
3. A sample of water (24.8 g) at 42.7 °C is mixed with 186.0 g of ethanol at 25.0 °C in
an insulated container. When the mixture reaches thermal equilibrium, the final
temperature is 30.0 °C. Assuming no heat is lost to the surroundings, what is the specific
heat capacity of ethanol in J/(g °C)? [The specific heat of water is 4.184 J/(g °C)]
No heat is lost to the surroundings then we only worry about the heat transferred from the hot to
the cold object. We can write
qwater + qethanol = 0
mwater×Cwater×∆Twater + methanol×Cethanol×∆Tethanol = 0 (remember ∆T = Tf - Ti)
(24.8 g)×( 4.184 J/(g °C)×( 30.0 - 42.7 °C) + (186.0 g)×(Cethanol)×(30.0 - 25.7 °C) = 0
(24.8 g)×( 4.184 J/(g °C)×( -12.7 °C) = -[(186.0 g)×(Cethanol)×(+4.30 °C)]
(-1318 _J)
= C ethanol
- 800 g o C
1.65 J/(g °C) = Cethanol (Note: I set up the problem incorrectly. The actual C = 2.44 J/(g °C) )
4. Calculate the values of ∆Hºrxn using the data in the table.
H2O(liq) + 3NO2(g)
2HNO3(aq)
Compound
H2O(liq)
NO2(g)
∆Hºf(kJ/mol)
-285.9
+33.1
HNO3(g)
-207.4
NO(g)
+90.3
+
NO(g)
∆Hrxn = Σnp(∆H°f(products)) - Σnr(∆H°f(reactants))
∆Hrxn = 2∆H°f(HNO3) + ∆H°f(NO) -[(∆H°f(H2O)) + 3(∆H°f(NO2)]
∆Hrxn = 2 mol(-207.4 kJ/mol) + 1 mol(+90.3 kJ/mol) - [1mol(-285.9kJ/mol) + 3 mol(+33.1 kJ/mol)]
∆Hrxn = -324.5 kJ - [186.6 kJ]
∆Hrxn = -137.9 kJ
5. For an H atom the energy of a transition between energy levels is given by the formula:
1
-18
∆E = -2.178 X 10
J
2
nf
-
1
2
ni
For the transition for n = 2 to n = 4,
(a) Calculate the energy of this transition.
1
-18
∆E = -2.178 X 10
J
2
2f
-
1
2
4i
∆E = -2.178 × 10-18 J (+0.1875) = -4.08 × 10-19 J (negative means energy released)
(b) Calculate the wavelength of the photon associated with this transition.
∆E = hν and (ν λ = c or
∆E =
hc
λ
⇒ λ =
hc
∆Ε
λ = 4.87 _× 10 − 7 _ m ×
ν =
c
)
λ
⇒ ∆Ε =
10 9 _ nm
1_ m
(6.626 _× _ 10 −34 _ Js ) 2.997 _× _ 10 +8 _ ms 
4.08 _× 10 −19 _ J
= 487 _ nm
6. Each of the following sets of quantum numbers is incorrect. Explain which quantum
number makes the set incorrect. In each case, indicate a correct value for the quantum
number that is incorrect.
Incorrect
Quantum
Why is the value
A correct
number
incorrect?
value
(a) n = 5,
l = 0, ml = -1
____
ml ______
(b) n = 2.6,
l = 1, ml = 0
_____
n _____
(c) n = 4,
l = 4, ml = -1
_____
l _____
(d) n = -2,
l = 2, ml = -1
(e) n = 20,
l = 19, ml = -20
_____
_____
n _____
ml _____
__
ml = -l, …0 ….l_
n = positive integer__
___
l =0,1,2,3 …(n-1)_
____
__
____0________
___2,3,4
etc._
____0,1,2,3________
n = positive integer ____3,4, 5 etc
ml = -l, …0 ….l -19…+19____
7. Given the energy level diagram in the hydrogen atom below. Using vertical lines
indicate the following transitions on the energy level diagram:
(1) n= 2 ————> n= 6
(2) n =1 ————> n = 3
(3) n = 4 ————> n = 1
Circle the correct answer(s) for the following (some of (a)-(e) may require more than one answer
to be circled.)
A transition with an upward arrow requires energy and is endothermic. A transition with a
downward arrow releases energy (as a photon of light) and is exothermic.
(a) which transition(s) will a photon be emitted?
1
2
3
none of 1-3
all of 1-3
(b) which transition(s) will a photon be absorbed?
1
2
3
none of 1-3
all of 1-3
(c) which transition(s) are endothermic?
1
2
3
none of 1-3
all of 1-3
(d) which transition(s) are endothermic? (should be exothermic)
1
2
3
none of 1-3
all of 1-3
(e) which transition(s) is the electron in its most stable state in the final state of
the transition
1
2
3
none of 1-3
all of 1-3
n =∞
n = 6
n = 5
n = 4
E
n
e
r
g
y
n = 3
n = 2
n = 1
1
2
3
8. The most prominent line in the spectrum of neon is at 865.4 nm. Other lines in the
spectrum are found at 837.8 nm, 878.1 nm, 878.4 nm, and 1885.4 nm. (1 m = 109 nm)
(a) Which of these lines represents the most energetic light?
hc
Energy is proportional to frequency and inversely
λ
proportional to wavelength, so the line with the shortest wavelength has the greatest
energy or 837.8 nm represents the most energetic light
(b) What is the frequency of the light for the most prominent line?
∆E = hν
and
865.4 _ nm ×
νλ = c
∆E =
1_ m
9
10 _ nm
ν=
c
/λ
= 8.654 _× _ 10 − 7 _ m
ν
=
m
s
−7
8.654 _× _ 10 _ m
2.997 _× _ 10 + 8 _
⇒ 3.463 _× _ 10 +14 _
(c) What is the energy of one photon of light from the most prominent line?
∆E = hν
∆Ε
=
(6.626 _× _ 10 −34 _ Js ) 3.463 _× _ 10 +14 _ 1s  ⇒
2.29 _× _ 10 −19 _ J
(d) What is the energy of one mole of photons from the most prominent line?
(Avagadro’s number = 6.02 x 1023)
J
(2.29 × 10-19 /photon ) × (6.02 × 1023
photon
J
/mol) = 137,858 /mol or 138
kJ
/mol
1
s