Energy, Enthalpy and Thermochemistry. Energy: The capacity to do work or to produce heat

Transcription

1 9 Energy, Enthalpy and Thermochemistry Energy: The capacity to do work or to produce heat The law of conservation of energy Energy can be converted but the total is a constant Two types of energy: Kinetic energy due to motion ½mv 2 Potential energy due to position, composition, attraction, repulsion. Example A B B A Potential energy has changed for A and B: A: decreased B: increased Frictional energy is involved Total energy : no change 1

2 Energy transfer: through work and heat Work: a force acting over a distance In previous example: work is done on B by A The nature of energy A B B A For A: different path different heat and work the change of PE for A the same Energy is a state function independent of path Heat and work are path dependent NOT state function From one state to another state may go through different path 2

3 Chemical energy CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(g) + heat Energy diagram E CH 4 (g) + 2O 2 (g) The PE stored in chemical bonds is released to the surroundings An exothermic reaction PE: released as heat CO 2 (g) + 2H 2 O(g) Ex: N 2 (g) + O 2 (g) + heat 2NO(g) PE 2NO(g) PE: flows into the system N 2 (g) + O 2 (g) An endothermic reaction

4 Ex. CH CH 2 CH=CH 2 + 6O 2 4CO 2 + 4H 2 O H = kcal/mol (CH ) 2 C=CH 2 + 6O 2 4CO 2 + 4H 2 O H = kcal/mol PE 6O 2 + CH CH 2 CH=CH 2 6O 2 + (CH ) 2 CH=CH kcal/mol kcal/mol ~ ~ 4CO 2 + 4H 2 O Thermodynamics The study of energy and its interconversions Questions: How far will a reaction proceed? How much energy will be released or consumed? The first law of thermodynamics The law of conservation of energy 4

5 Internal energy (E) Define internal energy E: the sum of PE and KE in the system Difficult to measure The change is measurable: E = q + w heat work The sign of q and w From the system s point of view: q absorbed by the system is positive Increases E of the system q released by the system is negative Decreases E of the system Ex: endothermic reaction: q (+) exothermic reaction: q (-) 5

6 w done by the system is negative Decreases E of the system w done by the surroundings on the system is positive Increases E of the system A model study ideal gas P = F A expansion against P P h work = force distance = F h Since P = F/A F = PA work = P A h = P V 6

7 Work is done by the system negative V = V final - V initial V final > V initial (for expansion) V positive (for expansion) Therefore we must define w = PV For compression V negative w positive Ex: A balloon heat L L Soln: 1 atm L = 101. Expands against 1.0 atm, E? E = q + w q = w = PV = (1.0 atm)( L) = atm.l = )(101.) = E = +( ) =

8 Enthalpy ( 焓 ) The definition of enthalpy higher T P lower T q P = E + (PV) = E f E i + (PV) f (PV) i = E f + (PV) f E i (PV) i P E = q + w = q PV At constant P: E = q P (PV) Define enthalpy H = E + PV H f = E f + (PV) f H i = E i + (PV) i q p = H f H i = H q P = H at constant P = E + PV at constant P q P E (unless V = 0) What s so important? Usually reactions are performed at constant P q P is obtained experimentally Therefore, H can be measured directly The beauty of H H = E + PV A state function Path independent For a reaction H = H products H reactants Exothermic reaction: q P is negative negative H Endothermic reaction: q P is positive positive H H and E may be different 8

9 Thermodynamics of ideal gases Heating 1 mol of an ideal gas at constant V No change T 1 T 2 V = 0 (KE) avg = / 2 RT (note: total E = KE + PE) KE = / 2 RT 2 / 2 RT 1 = / 2 RT = E V = 0 no PV work (w = 0) E = q v + w = / 2 RT q v = / 2 RT E = q v at constant V Molar heat capacity ( 熱容 ) C heat absorbed increase in T q T (per mol) q C v v C T v = / 2 R Heat required to change T of 1 mol ideal gas by 1 K at constant V 9

10 Heating an ideal gas at constant P T 1 T 2 (KE increases) V 1 V 2 (PV work is done) KE = / 2 RT (1 mol) w = PV = nrt (note: PV 1 = nrt 1 ) For 1 mol of ideal gas E = q p + w q p = E w = / 2 RT + RT = H q C p p C T p = / 2 R + R = C v + R Heat required to change T of 1 mol ideal gas by 1 K at constant P C v C p C p C v (K 1 mol 1 ) He, Ne, Ar H / 2 R = CO > / 2 R 7.27 > 5 / 2 R C 2 H Extra E is required to increase the rotational and vibrational E of polyatomic molecule The difference is due to the expansion work No change 10

11 Heating a gas: energy and enthalpy Translational E for 1 mol of ideal gas: E = / 2 RT T dependent only E = / 2 RT General form: E = / 2 nrt Since C v = / 2 R E = nc v T Recall H = E + PV H = E + (PV) = E + nrt = nc v T + nrt = n(c v + R)T H = nc p T T dependent only E T slope = nc v H T slope = nc p Independent of P or V But q = nct is dependent on condition Summary C v = / 2 R Monoatomic ideal gas C v > / 2 R Polyatomic ideal gas (measured by exp.) C p = C v + R All ideal gases C p = 5 / 2 R = / 2 R + R Monoatomic ideal gas C p > 5 / 2 R Polyatomic ideal gas (measured by exp.) E = nc v T All ideal gases H = nc p T All ideal gases 11

12 Ex. From state A state B P (atm) Q: q, w, E, H? A 1 C Step 1 w 1 = PV = (2.00 atm)(0.0 L 10.0 L) = L atm = D 4 2 B V (L) At A: P 1 V 1 /nr = T 1 At C: P 2 V 2 /nr = T 2 P( V2 V1 ) (2.00)(20.0) 40.0 L atm T T2 T1 nr nr nr nr q1 qp ncpt n( R)( ) nr E1 ncvt n( R)( ) nr H 1 = q p = q 1 = Step 2 w 2 = 0 ( Pf Pi ) V ( )(0.0) 0.0 L atm T nr nr nr nr q2 qv ncvt n( R)( ) nr E 2 = q v = q 2 = H2 ncpt n( R)( ) nr P (atm) A D 1 4 C 2 B V (L) 12

13 Another way to calculate H 2 H 2 = E 2 + (PV) = E 2 + ( )0.0 L atm = Step w = 0 ( Pf Pi ) V ( 1.00)(10.0) 10.0 L atm T nr nr nr nr q qv ncvt n( R)( ) nr E = q v = q = P (atm) H ncpt n( R)( ) nr 1.00 A D 1 4 C 2 B V (L) Step 4 w 4 = PV = (1.00 atm)(0.0 L 10.0 L) = L atm = P( Vf Vi ) (1.00)(20.0) 20.0 L atm T nr nr nr nr q4 qp ncpt n( R)( ) nr E4 ncvt n( R)( ) nr H 4 = q p = q 4 = P (atm) A D 1 4 C 2 B V (L) 1

14 Summary w 1 = q 1 = E 1 = H 1 = w 2 = 0 q 2 = E 2 = H 2 = A C B q = q 1 + q 2 = w = w 1 + w 2 = E = q + w = H = H 1 + H 2 = w = 0 q = E = H = w 4 = q 4 = E 4 = H 4 = A D B q = q + q 4 = w = w + w 4 = E = q + w = H = H + H 4 = Conclusion From A B E, and H are the same for the two pathways q, and w are path dependent P (atm) 2.00 A 1 C D 4 B V (L) 14

15 Calorimetry ( 量熱學 ) The science of measuring heat observing T change C = heat absorbed increase in T Heat capacity ( 熱容 ) Specific heat capacity: per gram sample Unit: o or C g K g Molar heat capacity: per mole sample Unit: or o C mol K mol Constant P calorimetry Ex: Purpose: get q P q P = H Mix 50.0 ml 1.0 M HCl at 25.0 o C 50.0 ml 1.0 M NaOH at 25.0 o C Soln: T = = 6.9 o C mass ml 1.0 g/ml = g specific heat capacity mass Energy released = s m T = (4.18 / o Cg)( g)(6.9 o C) = o C 15

16 Heat is an extensive ( 外延 ) property related to the amount of substance T is an intensive ( 內涵 ) property not related to the amount of substance 50.0 ml 1.0 M HCl mol H mol = /mol H + (aq) + OH (aq) H 2 O(l) H = 58 k/mol At constant P with PV work Example above: reaction in solution assumption: V = 0 w = 0 E = q p + w = q p = H Ex. 2SO 2 (g) + O 2 (g) 2SO (g) There is volume change volume decreased w = PV () (+) E = q p + w = H + w Different values 16

17 2.00 mol SO 2 (g) mol O 2 (g) at 25 o C, 1.00 atm q = 198 k (T kept constant) Q: H, E? Soln: w = PV = P(V 2 V 1 ) = (n 2 RT n 1 RT) = (1.00 mol)(8.145 /K mol)(298 K) = = 2.48 k E = q + w = = 196 k H = q p = 198 k Constant V calorimetry V = 0 w = 0 E = q V + w = q V Therefore, E = q V at constant V Use a bomb calorimeter insulating container water steel bomb 17

18 Ex: Combustion of g octane (C 8 H 18 ) The bomb calorimeter used: C = 11. k/ o C T = 2.25 o C Soln: Heat released = CT = = 25.4 k MW of octane = g/mol g octane g g/mol = mol For 1 mol octane: 25.4 Heat = = k/mol E = kmol 1 Ex. Same bomb calorimeter 1.50 g methane + O 2 (excess) T = 7. o C 1.15 g H 2 + O 2 (excess) T = 14. o C For methane For H 2 CT (11.)(7.) 55 k/g mass 1.50 CT (11.)(14.) 141k/g mass 1.15 Cf. Octane: 48.2 k/g More efficient fuel 18

19 Hess s law Reactants Products H = Σn p H products Σn r H reactants B R P A H is a state function H RP is the same for different pathways Ex: N 2 (g) + 2O 2 (g) 2NO 2 (g) H 1 = 68 k H (k) O 2 (g), 2NO(g) B H = 112 H 2 = 180 C 2NO 2 (g) H 1 A H 1 C A N 2 (g), 2O 2 (g) H 2 B H H 1 = H 2 + H must be 68 k 19

20 Characteristics of H 1. H = H reverse N 2 (g) + O 2 (g) 2NO(g) H 2 = +180 k Reverse reaction: 2NO(g) N 2 (g) + O 2 (g) endothermic H 2 = 180 k 2. H is an extensive property depending on the amount of reactants Xe(g) + 2F 2 (g) XeF 4 (s) 2Xe(g) + 4F 2 (g) 2XeF 4 (s) exothermic H = 251 k H = 2(251 k) = 502 k Ex: 2B(s) + H 2 (g) B 2 H 6 (g) H =? Data available: (a) 2B(s) + / 2 O 2 (g) B 2 O (s) H = 127 k (b) B 2 H 6 (g) + O 2 (g) B 2 O (s) + H 2 O(g) H = 205 k (c) H 2 (g) + 1 / 2 O 2 (g) H 2 O(l) H = 286 k (d) H 2 O(l) H 2 O(g) H = 44 k 20

21 Ex: 2B(s) + H 2 (g) B 2 H 6 (g) H =? Soln: +a (for B) 2B(s) + / 2 O 2 (g) B 2 O (s) H = 127 k +c (for H 2 ) H 2 (g) + / 2 O 2 (g) H 2 O(l) H = 286 k b (for B 2 H 6 ) B 2 O (s) + H 2 O(g) B 2 H 6 (g) + O 2 (g) H = 205 k +d (for H 2 O) H 2 O(l) H 2 O(g) H = 44 k 2B(s) + H 2 (g) B 2 H 6 (g) H = 127 (286 ) (44 ) = +6 k Standard enthalpies of formation Real interest: calculate H for reactions Problem: no way to know the exact value of enthalpy Solution: resolve each compound into the basic elements Define a standard enthalpy of formation (H fo ) 1 mol of a compound from its elements in their standard states 21

22 Standard state: The physical state in which the substance is most stable at 1 atm and the temperature of interest (usually 25 o C) gas: a pressure of 1 atm solution: a concentration of 1 M a pure liquid or solid compound: the pure liquid or solid element: the state at 1 atm (25 o C) Ex: Na(s), Hg(l) *Do not be confused with STP Ex: 1 / 2 N 2 (g) + O 2 (g) NO 2 (g) H fo = 4 k/mol Elements in standard state C(s) + 2H 2 (g) + 1 / 2 O 2 (g) CH OH(l) graphite 1 mol (in standard state) H fo = 29 k/mol The use of H f o Ex: CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(l) H o =? Data available: C(s) + 2H 2 (g) CH 4 (g) C(s) + O 2 (g) CO 2 (g) H 2 (g) + 1 / 2 O 2 (g) H 2 O(l) H fo = 75 k/mol H fo = 94 k/mol H fo = 286 k/mol CH 4 (g) + 2O 2 (g) CO 2 (g) + 2H 2 O(l) C(s) 2H 2 (g) 2O 2 (g) C(s) O 2 (g) 2H 2 (g) O 2 (g) H o = (286 2) = 891 k = H f o (CH4 ) + H f o (CO2 ) + 2(H f o (H2 O)) = H f o (CO2 ) + 2(H f o (H2 O)) H f o (CH4 ) 22

23 Conclusion: H o rxn = Σn p H f o (products) Σn r H f o(reactants) (elements in their standard state are not included) Cf. H = Σn p H products Σn r H reactants Ex: 4NH (g) + 7O 2 (g) 4NO 2 (g) + 6H 2 O(l) H f o : k/mol H o = 6(286) + 4(4) 4(46) = 196 k Present sources of energy Natural gas Mainly CH 4 with some ethane propane butane Petroleum C 510 gasoline distillation C 1018 kerosene cracking 700 o C C 1525 diesel catalyst More gasoline fraction Coal: contains sulfur gives off SO 2, and CO 2 O 2 sun light earth Acid rain Absorbs IR IR: absorbed by CO 2, H 2 O 2

24 New energy sources ~600 o C ~900 o C ~1200 o C Coal gasification ~900 o C C(s) + steam + air: C(s) + 2H 2 (g) CH 4 (g) H o = 75 k C(s) + ½O 2 (g) CO(g) H o = 111 k C(s) + O 2 (g) CO 2 (g) H o = 94 k (+ sulfur compounds) C(s) + H 2 O(g) H 2 (g) + CO(g) H o = 11 k Water gas shift reaction CO(g) + H 2 O(g) CO 2 (g) + H 2 (g) CO(g) + H 2 (g) CH 4 (g) + H 2 O(g) 400 o C Catalyst: Fe 2 O or Ni Remove CO 2, H 2 O & impurities Left with CH 4, CO, and H 2 Used in synthesis o C CO(g) + 2H 2 (g) CH OH(l) 00 atm Ag or Cu catalyst Fischer-Tropsch synthesis nco(g) + (2n+1)H 2 (g) Oxo process o C 1 to several hundred atm catalyst C n H 2n+2 + nh 2 O(g) RCH=CH 2 + H 2 (g) + CO(g) o C catalyst RCH 2 CH 2 CHO 24

25 Hydrogen Problems Costs H 2 (g) + ½O 2 (g) H 2 O(l) H o = 286 k Highly exothermic No pollution Main source: CH 4 (g) + H 2 O(g) H 2 (g) + CO(g) Electrolysis of water? Cheap electricity required Thermolysis of water? Efficient catalytic system required Photo-biological decomposition of water? New biological system required Storage Reacts with metals to give hydrides causing structural change Large volumes are required Ex L gasoline g assume hydrogen is three times more efficient 59200/ = g H L liquid H L gaseous H 2 Better way: H 2 (g) + M(s) MH 2 (s) metal MH 2 (s) H 2 (g) + M(s) Efficient system required 25

26 Other source of energy Ethanol Gasohol (10% ethanol in gasoline) Pure ethanol has too high a bp Methanol Possible Biodiesel From seed oil Exercise Knowing: H f o H2O(l) = k/mol Q: H 2 O(l) H 2 (g) + ½ O 2 (g) At 298 K and 1 atm, E o =? Soln: H 2 (g) + ½ O 2 (g) H 2 O(l) H f o H2O(l) H 2 O(l) H 2 (g) + ½ O 2 (g) H o = k/mol Assuming V H2 + ½ V O2 >> V H2O(l) V = V H2 + ½ V O2 = 1.5(24.6 L) = 6.9 L E o = H o + w = H o PV E o = (1)(6.9)(0.101) = k/mol 26