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What is Energy

What is Energy
Lecture 11 thermodynamics We’ll be dealing with the energy of chemical reactions How do you keep track of it Where does it come from Thermochemistry Energy   • Can  come  from  a  variety  of  sources:   Ø Light  (photochemistry)   Ø Electricity  (electrochemistry)   Ø Heat  (thermochemistry)   Thermochemistry What  is  Energy   •  The  ability  to:   •   do  work     •  transfer  heat.   Ø Work:  Energy  used  to  cause  an  object  that  has  mass  to   move.   Ø Heat:  Energy  used  to  cause  the  temperature  of  an   object  to  rise.   Thermochemistry Work •  Energy  used  to  move  an  object   over  some  distance.   •  w  =  F  ž d,    w  =  work,     F  =  force     d  =  distance  over  which  the  force   is  exerted.     Note  units:   2 F  =  ma,  mass(distance/s )   2 2 W  =  F(d)  =  mass(distance /s )   Thermochemistry 2 =  mv  Heat   •  Energy  can  also  be   transferred  as  heat.   •  Heat  flows  from   warmer  objects  to   cooler  objects.   Thermochemistry Kinetic Energy  Energy  an  object  possesses  by  virtue  of  its   moMon.   1 2 KE = ⎯ mv 2 Thermochemistry PotenMal  Energy    Energy  an  object  possesses  by  virtue  of  its   posiMon  or  chemical  composiMon.   More potential E Less P.E. as bike goes down. Thermochemistry Transferal  of  Energy   a)  Add  P.E.  to  a  ball  by  liPing  it  to  the  top  of   the  wall   Thermochemistry Transferal  of  Energy   a)  Add  P.E.  to  a  ball  by  liPing  it  to  the  top  of   the  wall   b)  As  the  ball  falls,     2 P.E  ­‐­‐­‐­‐­‐­‐  K.  E.  (1/2mv )   Thermochemistry Transferal  of  Energy   a)  Add  P.E.  to  a  ball  by  liPing  it  to  the  top  of   the  wall   b)  As  the  ball  falls,     2 P.E  ­‐­‐­‐­‐­‐­‐  K.  E.  (1/2mv )   Ball  hits  ground,  K.E.  =0,  but  E  has  to  go   somewhere.  So   1.  Ball  gets  squashed   2.  Heat  comes  out.   Thermochemistry Energy accounting • We  must  idenMfy  where  different  types  of   energy  go.   • Therefore,  we  must  idenMfy  the  places.   Thermochemistry System and Surroundings •  The  system  includes  the   molecules  we  want  to   study  (here,  the   hydrogen  and  oxygen   molecules).   •  The  surroundings  are   everything  else  (here,   the  cylinder  and  piston).   Thermochemistry First Law of Thermodynamics •  Energy  is  conserved.   •  In  other  words,  the  total  energy  of  the  universe  is  a   constant;                    ΔE  =  ­‐ΛE   System surroundings Thermochemistry Internal Energy  The  internal  energy  of  a  system  is  the  sum  of  all   kineMc  and  potenMal  energies  of  all  components  of   the  system;  we  call  it  E.     E = E + E + E + E +…… internal,total KE PE electrons nuclei Almost impossible to calculate total internal energy Instead we always look at the change in energy (ΔE). Thermochemistry Internal  Energy    By  definiMon,  the  change  in  internal  energy,Δ   E,  is  the   final  energy  of  the  system  minus  the  iniMal  energy  of   the  system:        ΔE  =  E  −  E   final iniMal Thermochemistry Changes  in  Internal  Energy   • If  ΔE    0,  E    E     final iniMal Ø Therefore,  the  system   absorbed  energy  from   the  surroundings.   Thermochemistry Changes  in  Internal  Energy   • If  ΔE    0,  E    E     final iniMal Ø Therefore,  the  system   released  energy  to  the   surroundings.   Thermochemistry Changes in Internal Energy •  When  energy  is   exchanged  between   the  system  and  the   surroundings,  it  is   exchanged  as  either   heat  (q)  or  work  (w).   • That  is,  ΔE  =  q  +  w.   Thermochemistry ΔE,  q,  w,  and  Their  Signs   q +q Surroundings hot plate adds suck heat out of heat to water Thermochemistry water. Sign  of  work   block pushes truck down does work on truck w block w + Truck pushes block up. truck Does work on block w truck w + block Thermochemistry Exchange of Heat between System and Surroundings •  When  heat  is  absorbed  by  the  system  from  the   surroundings,  the  process  is  endothermic.   Thermochemistry Exchange of Heat between System and Surroundings •  When  heat  is  absorbed  by  the  system  from  the   surroundings,  the  process  is  endothermic.   •  When  heat  is  released  by  the  system  to  the   surroundings,  the  process  is  exothermic.   Thermochemistry Units of Energy • The  SI  unit  of  energy  is  the  joule  (J).     2 kg m 1 J = 1 ⎯⎯   2 s • An  older,  non­‐SI  unit  is  sMll  in  widespread   use:  The  calorie  (cal).      1  cal  =  4.184  J   2 Energy  has  units  of  (mass)(velocity)   2 Remember  kineMc  energy  was  1/2mv   Thermochemistry Energy, specific heat and temperature chang e Thermochemistry Calorimetry   How  is  heat  measured  Calorimetry     Color  =  heat   metry  =  measure   A  calorimeter  is  a  device  for   measuring  heat  transfer     How  does  this  work     Thermochemistry Heat Capacity and Specific Heat • Heat  capacity   • The  amount  of  energy  required  to  raise  the   temperature  of  an  object  by  1  K  (1°C)  is  its   heat  capacity.   • We  define  specific  heat  capacity  (or  simply   specific  heat)  as  the  amount  of  energy   required  to  raise  the  temperature  of  1  g  of  a   specific  substance  by  1  K.   Thermochemistry Heat Capacity and Specific Heat heat transferred Specific heat = mass × temperature change q s = m ΔT Heat change(q) = specific heat(s) mass(m) temp. change(T) 1 1 J JK g g K smΔT = q Thermochemistry Exampl es • How  much  heat  is  required  to  raise  the   temperature  of  50  g  of  water  from  20°  C  to   ­‐1 ­‐1 45  °C    Sp.  Heat  =  4.184  JK g   • Q  =  sm(T ­‐T )   final init ­‐1 ­‐1 • Q    =    4.184  JK g 50  g(45  –  20  °C)  =  5230  J   • How  much  will  the  temperature  rise  if  1000  J   of  energy  are  used  to  heat  a  10g  block  of   ­‐1 ­‐1 copper  (s =0.385  JK g )   Cu ­‐1 ­‐1 • 1000  J  =  0.385  JK g 10gΔT   Thermochemistry • ΔT=260  °C  Lecture 12 Energy and changes of sta te • Change  of  state:   • Solid  à  liquid    liquid  à  solid   • Liquid  à  gas        gas    à    liquid   • A  change  of  state  will  occur  at  a  constant  T.    melMng  ice  will  stay  at  0°C  unMl  all  the   ice  is  melted.    You  can’t  heat  ice  to  higher  T   than  0  °C  at  1  atm.   Thermochemistry HeaMng  curve:   110 Heating gas 100 Water Heating T Boiling water °C (heat of vaporization) Heat 0 Ice melting ice 10 (heat of fusion) Heat added to system Example  calculate  heat  used  in  above  (10g).    Five  stages:   ­‐1 ­‐1 ­‐1 ­‐1 Heat  ice:    2.1  JK g  10g10°C=210  J  (smΔΤ, s =2.1  JK g  )   ice ­‐1 Melt  ice:    333Jg 10g  =  3330  J    (g(heat  of  fusion )     wat Heat  water:    4.18510g100°C=4184  J   ­‐1 Vaporize  water:    2260  Jg 10g=22600J  (g(heat  of  vap )   wat ­‐1 ­‐1 ­‐1 ­‐1 Thermochemistry Heat  steam:    2.0  JK g 10g(10°C)=200  J  (s =2.0  JK g )   steamWork and heat • 2  possibiliMes:   Ø 1  The  system  does  not  expand.    Then,  there   . can  be  no  force  over  a  distance  (FD)  and  no   work.   • ΔE  =  q  +  w  =  q  constant  volume.   V Ø 2.    The  system  expands  and  work  is  done.   Thermochemistry Work   process  in  an  open  container  (chemical  reacMon  in  a   beaker)   w    (can  there  be  any  work)       Thermochemistry Yes, evolving gases could push on the surroundings. Catch the work, do the same process in a cylinder. Note, P is constant. Process evolves gas, pushes on piston, work done Thermochemistry on piston Catch the work, do the same process in a cylinder Note: P is constant. w = Fd, F = PA, d=Δh w = PAΔh= PΔV Negative because an increase in Volume means that the system is doing work on the surroundings. ΔE  =  q  +  w    =    q  ­‐  PΔV q = ΔE + PΔV P Thermochemistry Catch the work, do the same process in a cylinder Note: P is constant. w = Fd, F = PA, d=Δh w = PAΔh= PΔV Negative because an increase in Volume means that the system is doing work on the surroundings. ΔE  =  q  +  w    =    q  ­‐  PΔV q = ΔE + PΔV P Thermochemistry Enthalp y • ΔE  =  q  +  w    =    q  ­‐  PΔV   • q  =  ΔE  +    PΔV   P • A  new  funcMon  is  defined,  Enthalpy  (H)   Enthalpy  means  “heat  inside”  or  “heat  content”   Change  in  enthalpy  is  the  heat  gained  or  lost  by  a  system   at  constant  pressure.    The  difference  between  ΔH  and  ΔE   is  small  for  chemical  reacMons  that  do  not  involve  a   change  in  gas  volume  (solids  and  liquids  don’t  have  much   Thermochemistry change  in  volume).  The case of chemical reaction s Endothermic Exothermic Heat in products reactants Heat out H H products reactants reaction reaction 90 kJ + 1/2N + 1/2O à NO H + 1/2O à H O + 242 kJ 2 2 2 2 2 1/2N + 1/2O à NO ΔH =90 kJ H + 1/2O à H O ΔH  =242 kJ 2 2 2 2 2 Thermochemistry Hess’s la w Thermochemistry Hess s Law • ΔH  is    known  for  many  reacMons.   • measuring  ΔH  can  be  a  pain   • Can  we  esMmateΔ   H  using  ΔH  values  for   other  reacMons   Thermochemistry Hess s  Law   Yes  Hess s  law:  states   that:    ΔH  for  the  overall   reacMon  will  be   equal  to  the  sum   of  the  enthalpy   changes  for  the   individual  steps.   Thermochemistry Hess s Law Why     Because  ΔH  is  a  state  funcMon,   and  is  pathway  independent.   Only  depends  on  iniMal  state  of   the  reactants  and  the  final   state  of  the  products.   Thermochemistry Hess’s law, an examp le • Given  the  enthalpy  changes  for  the  following   reacMons:   • C(s)    +    O    à    CO(g)    +    1/2O (g)    ΔH  =  ­‐110  kJ   2 2 • C(s)    +    O    à    CO (g)                                ΔH  =  ­‐394  kJ   2 2 • What  is  ΔH  for  the  following  reacMon:   • CO(g)    +    1/2O (g)  à  CO (g)     2 2 Thermochemistry Hess’s law, an examp le • Given  the  enthalpy  changes  for  the  following   reacMons:   • C(s)    +    O    à    CO(g)    +    1/2O (g)    ΔH  =  ­‐110  kJ   2 2 • C(s)    +    O    à    CO (g)                                ΔH  =  ­‐394  kJ   2 2 • What  is  ΔH  for  the  following  reacMon:   • CO(g)    +    1/2O (g)  à  CO (g)     2 2 st • 1.    reverse  1  reacMon  to  get  reactants:   • CO(g)    +    1/2O (g)    à    C(s)    +    O    ΔH  =  110  kJ   2 2 Thermochemistry Hess’s law, an examp le C(s)    +    O    à    CO(g)    +    1/2O (g)    ΔH  =  ­‐110  kJ   2 2 C(s)    +    O    à    CO (g)                                ΔH  =  ­‐394  kJ   2 2 What  is  ΔH  for  the  following  reacMon:   CO(g)    +    1/2O (g)  à  CO (g)     2 2 st 1.    reverse  1  reacMon  to  get  reactants:   CO(g)    +    1/2O (g)    à    C(s)    +    O    ΔH  =  110  kJ   2 2 nd 2.  Add  2  reacMon:   C(s)    +                O                      à    CO (g)                ΔH  =  ­‐394  kJ   2 2 Thermochemistry Hess’s law, an examp le C(s)    +    O    à    CO(g)    +    1/2O (g)    ΔH  =  ­‐110  kJ   2 2 C(s)    +    O    à    CO (g)                                ΔH  =  ­‐394  kJ   2 2 What  is  ΔH  for  the  following  reacMon:   CO(g)    +    1/2O (g)  à  CO (g)     2 2 st 1.    reverse  1  reacMon  to  get  reactants:   CO(g)    +    1/2O (g)    à    C(s)    +    O    ΔH  =  110  kJ   2 2 nd 2.  Add  2  reacMon:   C(s)    +                O                      à    CO (g)                ΔH  =  ­‐394  kJ   2 2 CO(g)    +    1/2O (g)  à  CO (g)                      ΔH  =  ­‐284  kJ   2 2 Thermochemistry    Hess s  law,  example:   •  Given:   •  N (g)  +  O (g)    ­‐­‐­‐­‐    2NO(g)  ΔH  =      180.7  kJ   2 2 •  2NO(g)  +  O (g)  ­‐­‐­‐­‐    2NO (g)    ΔH  =    ­‐113.1  kJ   2 2 •  2N O(g)  ­‐­‐­‐­‐    2N (g)    +    O (g)  ΔH  =    ­‐163.2  kJ   2 2 2 •  use  Hess s  law  to  calculate  ΔH  for  the  reacMon:   •  N O(g)  +  NO (g)­‐­‐­‐­‐    3NO(g)   2 2   Thermochemistry Hess s  law,  example:   •  Given:   •  N (g)  +  O (g)    ­‐­‐­‐­‐    2NO(g)  ΔH  =      180.7  kJ   2 2 •  2NO(g)  +  O (g)  ­‐­‐­‐­‐    2NO (g)      ΔH  =    ­‐113.1  kJ   2 2 •  2N O(g)  ­‐­‐­‐­‐    2N (g)    +    O (g)    ΔH  =    ­‐163.2  kJ   2 2 2 •  use  Hess s  law  to  calculate  ΔH  for  the  reacMon:   •  N O(g)  +  NO (g)­‐­‐­‐­‐    3NO(g)   2 2   • N O(g)  ­‐­‐­‐­‐    N (g)    +  1/2O (g)    ΔH  =­‐163.2/2  =    ­‐81.6  kJ   2 2 2 • NO (g)  ­‐­‐­‐­‐  NO(g)  +  1/2O (g)    ΔH  =  113.1  kJ/2  =  56.6  kJ   2 2 • N (g)  +  O (g)        ­‐­‐­‐­‐    2NO(g)        ΔH  =                                              180.7  kJ   2 2       • N O(g)  +  NO (g)­‐­‐­‐­‐    3NO(g)                    ΔH  =155.7  kJ     Thermochemistry 2 2    Enthalpies  of  FormaMon    An  enthalpy  of  formaMon,Δ   H,  is  defined  as   f the  ΔH  for  the  reacMon  in  which  a   compound  is  made  from  its  consMtuent   elements  in  their  elemental  forms.   That s  what  we  did  for  the  Thermite   reacMon:   •2   Al + Fe O Al O + 2Fe 2 3 2 3 •W   hat is the heat of reaction given: • 2Fe + 3/2O Fe O ΔH = 825.5 KJ 2 2 3 Thermochemistry •2   Al + 3/2O Al O ΔH = 1675.7 KJ 2 2 3CalculaMon  ofΔ   H   C H (g) + 5 O (g)⎯→ 3 CO (g) + 4 H O (l) 3 8 2 2 2 • Imagine  this  as  occurring    in  3  steps:     C H (g) ⎯→ 3 C + 4 H (g) 3 8 2 (graphite) 3 C + 3 O (g) ⎯→ 3 CO (g) 2 2 (graphite) 4 H (g) + 2 O (g) ⎯→ 4 H O (l) 2 2 2 Thermochemistry CalculaMon  ofΔ   H   C H (g) + 5 O (g)⎯→ 3 CO (g) + 4 H O (l) 3 8 2 2 2 • Imagine  this  as  occurring    in  3  steps:     C H (g) ⎯→ 3 C + 4 H (g) 3 8 2 (graphite) 3 C + 3 O (g) ⎯→ 3 CO (g) 2 2 (graphite) 4 H (g) + 2 O (g) ⎯→ 4 H O (l) 2 2 2 Thermochemistry CalculaMon  ofΔ   H   C H (g) + 5 O (g)⎯→ 3 CO (g) + 4 H O (l) 3 8 2 2 2 • Imagine  this  as  occurring    in  3  steps:     C H (g) ⎯→ 3 C + 4 H (g) 3 8 2 (graphite) 3 C + 3 O (g) ⎯→ 3 CO (g) 2 2 (graphite) 4 H (g) + 2 O (g) ⎯→ 4 H O (l) 2 2 2 Thermochemistry CalculaMon  ofΔ   H   C H (g) + 5 O (g)⎯→ 3 CO (g) + 4 H O (l) 3 8 2 2 2 •  The  sum  of  these   equaMons  is:   C H (g) ⎯→ 3 C + 4 H (g) 3 8 2 (graphite) 3 C + 3 O (g) ⎯→ 3 CO (g) 2 2 (graphite) 4 H (g) + 2 O (g) ⎯→ 4 H O (l) 2 2 2 C H (g) + 5 O (g) ⎯→ 3 CO (g) + 4 H O (l) 3 8 2 2 2 Make each reactant or product from its elements Thermochemistry This is called the heat of formation of a compound CalculaMon  ofΔ   H    We  can  use  Hess s  law  in  this  way:   ° °  ΔH  =  Σ n ΔH  ­‐  Σ m ΔH     f(products) f(reactants)    where  n  and  m  are  the  stoichiometric   coefficients.     Thermochemistry Standard  Enthalpies  of  FormaMon   °  Standard  enthalpies  of  formaMon,Δ   H,  are   f measured  under  standard  condiMons  (25°C  and   1.00  atm  pressure).   Thermochemistry CalculaMon  ofΔ   H   •  Calculate  ΔH  using  the  table:     •  C H    +    5  O      ­‐­‐­‐­‐­‐    3CO    +    4H O   3 8 2 2 2 Thermochemistry CalculaMon  ofΔ   H   • C H    +    5  O      ­‐­‐­‐­‐­‐    3CO    +    4H O   3 8 2 2 2 ΔH = 3(ΔH CO ) + 4(ΔH H O) (ΔH C H ) + (5ΔH O ) f 2 f 2 f 3 8 f 2 = 3(393.5 kJ) + 4(285.8 kJ) (103.85 kJ) + 5(0) = 1180.5 kJ + (1143.2 kJ) (103.85 kJ)+ 0 kJ = 2323.7 kJ 103.85 kJ) = 2219.9 kJ Thermochemistry Making and Breaking bonds i.e. a chemical reaction • When  a  bond  is  formed,  energy  is  released   Ø An  exothermic  process   • When  a  bond  is  broken,  energy  is  required   Ø An  endothermic  proces   Thermochemistry Example   Example: Calculate the heat of formation ΔH of hydrazine N H f 2 4 Using the bond energies given. 1 436 kJmol N N 159 946 N + 2H à N H N N 2 2 2 4 H H N H 389 H H N N H H N N 436 H H H H Need to be broken bonds made N—N 159 946+   2(436) (159 + 4(389)) = 103 kJ Thermochemistry Examp le Given: H + H à H (g) ΔH = 436 kJ 2 C(s) à C(g) ΔH = 717 kJ (sublimation) 1 Heat of formation of methane is: ΔH = 75 kJmol f What is the bond energy of a CH bond Sketch an energy cycle: 75 C(s) + 2H (g) à CH (g) 75 kJ = 717kJ+2(436kJ)4(CH) 2 4 4(CH) = 1664 kJ 2(436) 717 (CH) = 416 kJ C(g) + 4H 4(CH) Thermochemistry Thermochemistry