Chem 2000
Spring 2004
Assignment 6
1)
Friday March 5,
2004
Calculate q when 12.0 g of water is heated from 20 oC to 100 oC.
cs = 4.18 J/g
q = csm∆T = (4.18 J/g)*(12.0 g)*(100 oC – 20 oC) = 4013 J
2)
A 295 g Aluminium engine part at an initial temperature of 3.0 oC absorbs 85.0 kJ
of heat. What is the final temperature of this part if its specific heat cs = 2.42J/gK.
q = csm∆T = csm(Tf -Ti) => q/ csm + Ti = Tf
Tf = (85,000 J)/(2.42 J/gK)(295 K) + (276.0 oC) = 119 + 276 K = 395 K
3)
When 165 ml of water at 22 oC is mixed with 85 ml of water at 82 oC, what is the
final temperature? (assume no heat is lost to the surroundings and that the density
of water is 1.00 g/ml)
q1 = csm1∆T1
q2 = csm1∆T2
q1 = (4.18 J/g)*(165 ml)*(1 g/ml)*(Tf – (273 +22 K))
= (689.7 1/K)(Tf – 295 K)
q2 = (4.18 J/g)*(85 ml)*(1 g/ml)*(Tf – (273 +82 K))
= (355.3 1/K) )*(Tf – 355 K)
System 1 is gaining heat from system 2, ie. q2 = - q1
(355.3 1/K) )*(Tf – 355 K) = -(689.7 1/K)(Tf – 295 K)
(355.3 + 689.7 1/K)*Tf = 355.3*355 +689.7*295
Tf = (355.3*355 +689.7*295)/( (355.3 + 689.7 1/K)
= 315.4 K (42.2 oC)
4)
A 505 g piece of copper tubing is heated to 99.9 oC and placed in an insulated
vessel containing 59.8 g of water at 24.8 oC. Assuming no loss of water and a heat
capacity for the vessel of 10.0 J/K, what is the final temperature of the system. (cs
(Cu) = 0.387 J/gK)
-qtubing = qwater + qcontainer
-qtubing = -cs(tubing) mtubing∆Ttubing
=-(0.387 J/gK)*(505 g)*(Tf – (99.9 +273.15 K))
= (195.4 J/K)*(Tf – 373.05 K)
qwater= cs(water) mwater∆Twater
=(4.18 J/gK)*(59.8 g)*(Tf – (24.8 + 273.15 K))
=(249.96 J/K)*(Tf – 297.95 K)
qcontainer= Ccontainer∆Tcontainer
= (10.0 J/K)*(Tf –(24.8 + 273.15K))
=(10.0 J/K)*(Tf –297.95K)
-(195.4 J/K)*(Tf – 373.05 K) =
(249.96 J/K)*(Tf – 297.95 K) +(10.0 J/K)*(Tf –297.95K)
(195.4 J/K)*(373.05 - Tf K) = (249.96 + 10.0 J/K) *(Tf –297.95K)
= (259.96 J/K) *(Tf –297.95K)
373.05 - Tf K = (259.96 J/K) *(Tf –297.95K)/ (195.4 J/K)
373.05 - Tf K = 1.33*(Tf – 297.95 K)
373.05 + 1.33*297.95 K = 1.33*Tf + Tf
Tf = (373.05 + 1.33*297.95 K)/2.33 = 330.23 K (57.08 oC)
5)
Consider the following balanced thermochemical equation for the reaction
sometimes used to produce H2S:
1/8 S8(s) + H2(g) R H2S(g) ∆Hrxn = -20.2 kJ
a)
Is this an endothermic or exothermic reaction?
This reaction is exothermic.
b)
What is the enthalpy of the reverse reaction?
∆Hrxn = 20.2 kJ
c)
What is the enthalpy when 3.2 mol of S8 reacts?
(3.2 mol)/(1/8) = 25.6 times the reaction as stated above
therefore ∆Hrxn = 25.6*20.2 kJ = -520 kJ
d)
What is the enthalpy when 20.0 g of S8 reacts?
MWt(S8) = 8*32.07 g/mol = 256.56 g/mol
nS8 = (20.0 g)/( 256.56 g/mol) = 0.078 mol
0.078/(1/8) = 0.623 times the reaction as stated above
therefore ∆Hrxn = 0.623*20.2 kJ = -12.6 kJ
6)
Calculate ∆Hrxn for
Ca(s) + ½ O2(g) + CO2(g) R CaCO3 (s)
Given that
7)
Ca(s) + ½ O2(g) R CaO(s)
∆Hrxn= -635.1 kJ
CaCO3(s) R CaO(s) + CO2(g)
∆Hrxn= 178.3 kJ
Ca(s) + ½ O2(g) R CaO(s)
∆Hrxn= -635.1 kJ
CaO(s) + CO2(g) R CaCO3(s)
∆Hrxn= -178.3 kJ
Ca(s) + ½ O2(g) + CO2(g) R CaCO3 (s)
∆Hrxn= -813.4 kJ
At a specific set of conditions, 241.8 kJ is given off when 1 mol of H2O(g) forms
from its elements. Under the same conditions, 285.8 kJ is given off when 1 mol of
H2O(l) form from its elements. Calculate the heat of vaporization of water at these
conditions.
H2(g) + ½ O2 (g) R H2O(g)
∆Hf = -241.8 kJ
H2(g) + ½ O2 (g) R H2O(l)
∆Hf = -285.8 kJ
H2O(l) R H2(g) + ½ O2 (g) ∆Hf = 285.8 kJ
8)
H2(g) + ½ O2 (g) R H2O(g)
∆Hf = -241.8 kJ
H2O(l) R H2O(g)
∆Hvap = 44.0 kJ
Diamond and graphite are two crystalline forms of carbon. At 1 atm and 25 oC,
diamond changes to graphite so slowly that the enthalpy of the process must be
obtained indirectly. Determine ∆Hrxn for:
C(diamond) R C(graphite)
with the equations from the following list:
(1)
C(diamond) + O2 R CO2(g)
∆Hrxn= -395.4 kJ
(2)
2CO2(g) R 2CO(g) + O2(g)
∆Hrxn= 566.0 kJ
(3)
C(graphite) + O2(g) R CO2(g)
∆Hrxn = -393.5 kJ
(4)
2CO(g) R C(graphite) + CO2(g)
∆Hrxn = -172.5 kJ
C(diamond) + O2 R CO2(g)
∆Hrxn= -395.4 kJ
2CO2(g) R 2CO(g) + O2(g)
∆Hrxn= 566.0 kJ
2CO(g) R C(graphite) + CO2(g)
∆Hrxn = -172.5 kJ
C(diamond) R C(graphite)
∆Hrxn= -1.9 kJ
or
9)
C(diamond) + O2 R CO2(g)
∆Hrxn= -395.4 kJ
CO2(g)R C(graphite) + O2(g)
∆Hrxn = 393.5 kJ
C(diamond) R C(graphite)
∆Hrxn= -1.9 kJ
Calculate ∆Horxn for each of the following using molar enthalpies of formation in
appendix D. (hint be sure the equations are balanced before you proceed)
a)
∆Horxn
2H2S(g) + 3O2(g) R 2SO2(g) + 2H2O(g)
= 2∆Hof(H2O(g)) + 2∆Hof(SO2(g))- 2∆Hof(H2S(g))- 3∆Hof(O2(g))
=
2*(-241.82) + 2*(-296.83) – 2*(-20.63) –3*(0) kJ/mol
= -1036.04 kJ/mol
b)
CH4(g) + Cl2(g) R CCl4(l) + HCl(g)
balance equation first!!!!
CH4(g) + 4Cl2(g) R CCl4(l) + 4HCl(g)
∆Horxn = ∆Hof(CCl4(l)) + 4∆Hof(HCl(g))- ∆Hof(CH4(g))- 4∆Hof(Cl2(g))
= 1*(-135.44) + 4*(-92.31) – (-74.81)- 4*(0) kJ
= -429.87 kJ
10)
Electric generating plants transport large amounts of hot water through metal
pipes, and oxygen dissolved in the water can cause a major corrosion problem.
Hydrazine (N2H4) added to the water avoids the problem by reacting with oxygen:
N2H4(aq) + O2(q) R N2(g) + 2H2O(l)
About 4*107 kg of hydrazine is produced every year by reacting ammonia with
sodium hypochlorite in the Raching process:
2NH3(aq) + NaOCl(aq) R N2H4(aq) + NaCl(aq) + H2O(l) ∆Horxn = -151 kJ
a) If ∆Hof of NaOCl(aq) = -346 kJ/mol, calculate ∆Hof of N2H4(aq).
2NH3(aq) + NaOCl(aq) R N2H4(aq) + NaCl(aq) + H2O(l) ∆Horxn = -151 kJ
∆Horxn = ∆Hof(N2H4(aq)) + ∆Hof(Na+(aq))+ ∆Hof (Cl-(aq)) + ∆Hof(H2O(l))
- 2∆Hof(NH3(aq)) - ∆Hof(Na+(aq)) - ∆Hof (OCl(aq))
-151 kJ = ∆Hof(N2H4(aq)) + ∆Hof(Na+(aq)) + ∆Hof (Cl-(aq)) + ∆Hof(H2O(l))
- 2∆Hof(NH3(aq)) - ∆Hof(NaOCl+(aq))
-151 kJ = ∆Hof(N2H4(aq)) +(-240.12) + (–167.16) + (-285.83) –2*(-80.29)
– (-346) kJ
-151 kJ = ∆Hof(N2H4(aq)) –186.53 kJ
36 kJ = ∆Hof(N2H4(aq))
b) What is the heat released when aqueous N2H4 is added to 5.00*103 L of plant
water that is 2.50*10-4 M in O2.
N2H4(aq) + O2(q) R N2(g) + 2H2O(l)
∆Horxn = ∆Hof(N2(g)) + 2*∆Hof(H2O(l)) - ∆Hof(O2(g))- ∆Hof(N2H4(aq))
= 0 + 2*(-285.83) – 0 – 36 kJ = -608 kJ
nO2 = (5.00*103 L)( 2.50*10-4 M) = 1.25 mol O2
q = nO2*∆Horxn = (-608 kJ/mol O2) *(1.25 mol O2) = -760 kJ
11)
Using Table 10-3 of bond enthalpies on page 462 in the text determine the ∆Hrxn
for the combustion of propane (C3H8).
C3H8 + 5O2 R 3CO2 + 4H2O
CH3CH2CH3 + 5 O=O R 3 O=C=O + 4 H-O-H
bonds broken:
bonds made:
8 C-H 8*(412 kJ)
2 C-C 2*(348 kJ)
5 O=O 5*(497 kJ)
∆Hbroken = 6477 kJ
6 C=O 6*(-743)
8 O-H 8*(-463)
∆Hmade = - 8162 kJ
∆Hrxn = ∆Hbroken + ∆Hmade = 6477 + -8162 kJ
= - 1685 kJ/mol