30 Zn(s) 45 Rh. Pd(s) Ag(s) Cd(s) In(s) Sn(s) white. 77 Ir. Pt(s) Au. Hg(l) Tl. 109 Mt. 111 Uuu. 112 Uub. 110 Uun. 65 Tb. 62 Sm. 64 Gd. 63 Eu.
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1 Enthalpy changes: experimentally it is much easier to measure heat flow at const pressure - this is enthalpy q p = )H : also nearly all chemical reactions are done at constant pressure. Enthalpy (heat) of formation. )H is a change in enthalpy (like the difference in height of two mountains) 1. )H f0 is defined as the heat change when 1 mole of a substance if formed from its composite elements in their normal (standard) states usually at 25 O C and 1 atm. examples C (s) + O 2 (g) CO 2 (g) C (s) + 1/2 O 2 (g) CO (g) H 2 (g) + O 2 (g) H 2 O 2 (l) H 2 (g) + 1/2 O 2 (g) H 2 O (l) C (s) + 2 H 2 (g) + 1/2 O 2 (g) CH 3 OH (l) 4-1
2 2. Any element in its standard state has )H f0 = 0 you must know/remember the standard states of the elements Mark Baker July Periodic Table of Elements: Standard States 18 1 H 2 (g) 3 4 Li(s) Be(s) Na(s) Mg(s) 19 K(s) B(s) N 2 (g) O 2 (g) F 2 (g) 20 Ca(s) Ti(s) V(s) Cr(s) Mn(s) Fe(s) Co(s) Ni(s) Cu(s) 21 Sc(s) 30 Zn(s) 6 C(s) graphite 2 He(g) 10 Ne(g) Al(s) Si(s) P(s) white S 8 (s) Cl 2 (g) Ar(g) Ga(s) Ge(s) As(s) Se(s) Br 2 (l) Kr(g) Rb(s) Sr(s) Y Cs(s) Ba 87 Fr 88 Ra 57 La 89 Ac 40 Zr 72 Hf 104 Rf 41 Nb 73 Ta 105 Db 42 Mo 74 W 106 Sg 43 Tc 75 Re 107 Bh 44 Ru 76 Os 108 Hs 45 Rh 77 Ir 109 Mt Pd(s) Ag(s) Cd(s) In(s) Sn(s) white Pt(s) Au 110 Uun 111 Uuu Hg(l) Tl 112 Uub ) 82 Pb 114 Uuq Ununq uadium ) 51 Sb 83 Bi 52 Te 84 Po 53 I 2 (s) 85 At 54 Xe 86 Rn ) 58 Ce 59 Pr 60 Nd 61 Pm 62 Sm 63 Eu 64 Gd 65 Tb 66 Dy 67 Ho 68 Er 69 Tm 70 Yb 71 Lu 90 Th 91 Pa 92 U 93 Np 94 Pu 95 Am 96 Cm 97 Bk 98 Cf 99 Es 100 Fm 101 Md 102 No 103 Lr 4-2
3 Thermochemical Equations: you must learn how to write these down Note: when the thermochemical equation is written for a formation reaction all elements are on the LHS in their standard states to form 1 mole of product Some formation reactions proceed readily when the elements Are brought together. The formation reactions shown are: Fe(s) + ½ O 2 (g) FeO(s) (left) K (s) + ½ Cl 2 (g) KCl(s) (right). 4-3
4 Enthalpy changes for any reactions can be determined by using standard enthalpies of formation lets see how. Recall that enthalpy is a state function so the path taken does not matter Energy used/gained in this arm of reaction must equal that in upper arm Any chemical reaction can be imagined to proceed in two stages. First, reactants decompose into their constituent elements. Second, these elements recombine to form products. So if we know energy for two of the steps we can determine the other (unknown). 4-4
5 examples 1. )H fo for CO 2 (g) C(s) + O 2 (g) CO 2 (g) HBrO 4 (l) 1/2 H 2 (g) + 1/2 Br 2 (l) + 2O 2 (g) HBrO 4 (l) Note: you must put in the states 1 mole of product 4-5
6 )H can be determined for any process using the strategy on page 4-4 a: phase changes: let s list the enthalpy of formation for the three phases of water H 2 O(g) : )H fo = kjmol -1 H 2 O(l) : )H fo = kjmol -1 H 2 O(s) : )H fo = kjmol -1 so )Hvap = +44 kjmol -1 : WHY? If we go round the cycle we spend no energy: so starting at bottom left: H 2 O(l) )Hvap H 2 O(g) )Hvap -(-241.8) = 0 )H fo = kjmol -1 )H fo = kjmol -1 H 2 (g) + 1/2O 2 (g) so )Hvap = +44 kjmol
7 All this is really based on: 1st Law of Thermodynamics U universe = U system + Usurroundings = 0 We can transfer energy between System and Surroundings as HEAT and WORK we cannot manufacture energy -- that is why when we go round the cycle there is no energy change 4-7
8 More H 2 O(g) : )H fo = kjmol -1 H 2 O(l) : )H fo = kjmol -1 H 2 O(s) : )H fo = kjmol -1 If you do not want to set up that cycle every time then there is a simple equation that you can use provided you are using enthalpies of formation (there are other types of enthalpy changes that we meet later) this is: )H o reaction = E )H fo (products) -E)H fo (reactants) So: )Hvap = (-285.8) = +44 kjmol -1 (as before) and : )Hsub (ice to vapor) = +50 kjmol -1 )Hcond (vapor to liquid) -44kJmol - )Hfusion (ice to liquid) +6kJmol -1 Make sure you can get these numbers 4-8
9 b. Heats of solution:example: HCl(g) H + (aq) + Cl - (aq) )H O f Zero by definition: see later )H o reaction = E)H fo (products) -E)H fo (reactants) = (-92.3) = kj (solution heats up) Standard enthalpies of formation are given in Appendix C of the text (Ebbing) 4-9
10 Hot and cold packs: MgSO 4 (s) NH 4 NO 3 (s) Mg 2+ (aq) + SO 4 2- (aq) )H o = kj NH 4+ (aq) + NO 3- (aq) )H o = kj 4-10
11 Bond dissociation H 2 (g) 2H(g) )H = +436kJmol -1 H H Strong bond Heats of reaction Fe 2 O 3 (s) + 3CO (g) Fe(s) + 3CO 2 (g) )H O = -6.8 kj (note no per mole ) 4-11
12 Summary: Enthalpies or reaction and formation enthalpies 1. )H o reaction = E)H fo (products) -E)H fo (reactants) 2. Physical state is important 3. Coeffficients in eqn must be used : see example below 4. Units : kj Example: Determine )H o reaction from the given )H fo values. Fe 2 O 3 (s) + 3CO (g) Fe(s) + 3CO 2 (g) )H o reaction = 3 x ()H f O CO 2 (g) - ()H f O Fe 2 O 3 (s) - 3 x ()H f O CO (g) = 3 x (-395) -(-842) - 3x (-110.5) = -6.8kJ 4-12
13 Example heats of combustion: in a combustion reaction the reactants are burnt in oxygen to produce (usually) carbon dioxide and water. If the products are in their standard states (liquid water and gaseous carbon dioxide) then we call the heat of reaction the standard heat of combustion. C 10 H 22 (l) + 31/2 O 2 (g) 10CO 2 (g) + 11H 2 O(l) )H o comb. = 10 x ()H f O CO 2 (g) ) + 11 x ()H f O H 2 O(l)) -)H O f C 10 H 22 (l) = 10 (-393.5) + 11 (-285.8) - ( ) = kj Note the minus sign Combustion reactions are actually often used to get )H f O for a compound Note: enthalpies of combustion are ALWAYS negative 4-13
14 Example: formation enthalpies from combustion reactions C 6 H 12 (l) + 9O 2 (g) Combustion enthalpy is CO 2 (g) + 6H 2 O(l) )H o comb = (6 x )H fo H 2 O(l)) + 6 x()h fo CO 2 (g)) -)H O f C 6 H 12 (l) = 6 (-393.5) + 6 (-285.5) -)H O f C 6 H 12 (l) so : )H O f C 6 H 12 (l) = kj mol -1 This is typical of the kind of question you will get: a load of enthalpies and one is unknown 4-14
15 Example: Calculate )H for the reaction CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(l ) )H = [) H f (CO 2 )(g) + 2) H f (H 2 O)(l)] - [) H f CH 4 (g)] [From Tables] = [ (-286)] - [-75] = -891 kj What about the O 2? Example The heat of formation of liquid ethanol C 2 H 5 OH(l ) is -278 kj mol -1 Write out the relevant thermochemical equation. ) H f C 2 H 5 OH(l) C(s) + H 2 (g) + O 2 (g) C 2 H 5 OH(l ) 2C(s) + 3H 2 (g) + ½O 2 (g) C 2 H 5 OH(l ), )H f = -278 kj Any comments? Can you make whisky by shaking lumps of charcoal with H 2 and O 2 Answer: Not directly - but by cunning chemistry
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