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# Week 2 Notes Chem 130

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COURSE
General Chemistry II
PROF.
Dr. Yang
TYPE
Class Notes
PAGES
4
WORDS
KARMA
25 ?

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This 4 page Class Notes was uploaded by n/a on Wednesday January 20, 2016. The Class Notes belongs to Chem 130 at University of Tennessee - Knoxville taught by Dr. Yang in Spring 2016. Since its upload, it has received 29 views. For similar materials see General Chemistry II in Chemistry at University of Tennessee - Knoxville.

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Date Created: 01/20/16
Chem  130  Lecture  Notes  1.19.16  and  1.21.16   Ex.  Tennis  balls  are  usually  filled  with  air  or  N  gas  to  a  pressure  above  atmospher2c  pressure  to     increase  their  "bounce".  If  a  particular  tennis  ball  has  a  volume  of  144cm  and  contains   3   0.33g  of  N  gas,  2hat  is  the  pressure  inside  the  ball  at  24°C?     Given:  V=144  cm =  144  mL     3       PV=nRT       T=  24°C  =  297  K         P(144)=(0.33/28)(0.0821)(297)       n=0.33g  (1  mole/28g)         P=2.0  atm     Molar  Volume  at  STP     -­‐For  1,000  mol  of  ideal  gas  at  1.000  atm  and  0.00°C  (273.15  K),  the  V  is  22.4  L  at  STP     -­‐You  can  calculate  this  using  the  ideal  gas  law  and  STP  values  (T=273,  n=1.00  mol,     P=1.00  atm)     -­‐At  constant  T  and  P,  the  gases  with  the  same  volume  have  the  same  moles  and  number      gas  particles.     Ex.  What  is  th  emass  of  16.2  L  of  SF  at  STP?   6   Method  1:  PV=nRT         Method  2:  At  STP,  1  mole  occupies  22.4  L                  (1)(16.2)=n(0.08206)(273)                    16.2  L*(1  mol/22.4  L)=0.723  mol                  n=0.723  mol  SF  (conv6rt  to  g)                  Covert  to  g,  106  g  SF   6 Density  of  a  Gas     -­‐Density=molar  mass/molar  volume     Unit:  g/L     -­‐At  STP:  d =  4.00He  g/mol  =  0.179  g/L                                22.4  L/mol     -­‐Larger  molar  mass=more  dense     -­‐d=m=PM                V        RT     -­‐One  only  needs  to  know  the  molar  mass,  P  and  T  to  calculate  density  of  gas.     *T  increase,  density  decreases        P  increases,  density  increases          Molar  mass  increases,  density  increases     -­‐At  constant  T  and  P,  the  higher  the  molar  mass,  the  denser  the  gas.     Ex.  Which  of  the  following  is  most  dense  at  1.00  atm  and  20°C?       Cl  because  it  has  the  largest  atomic  mass   2 Molar  Mass  of  a  Gas     -­‐molar  mass=  mass  (g)                              mole  (n)     Ex.  Calculate  the  molar  mass  of  a  liquid  which,  when  vaporized,  at  98°C  and  715  mmHg,         yields  121  mL  of  vapor  with  a  mass  of  0.471  g.       T=98°C         Use  PV=nRT  to  calculate  n:       P=715  mmHg       (715/760)(0.121)=n(0.0821)(273+98)       V=  121  mL       n=0.00373       m=  0.471  g               Molar  Mass=0.471  g/  0.00373  mol=126  g/mol     Chem  130  Lecture  Notes  1.19.16  and  1.21.16   Mixtures  of  Gases     -­‐The  pressure  due  to  any  individual  component  in  a  gas  mixture  is  its  partial  pressure     (P )n     -­‐Ideal  Gas  Law  can  be  used  for  each  individual  gas.  P =n (RT/V)   n n   -­‐Dalton's  Law  of  Partial  Pressure:  P =P +P +P +....   total a b c •   Ideal  Gas  Law  can  be  used  with  a  mixture:  P total =n total (RT/V)   Mole  Fraction:  X a     -­‐The  number  of  moles  of  a  component  in  a  mixture  divided  by  the  total  number  of       moles  in  the  mixture  is  the  mole  fraction.     -­‐X =n /n       P =X (P )   a a total a a total   -­‐The  ration  of  the  partial  pressure  a  single  gas  contributes  and  total  pressure  is  equal  to     mole  fraction.     Ex.  The  air  contains  78%  N  by 2volume.  What  is  the  partial  pressure  of  N  in  a  sample 2of         air  at  1.0  atm?       Given:  composition  of  dry  air       P =N278(1.00  atm)=0.78  atm   Ex.  In  a  2.16  L  container  at  323  K  are  placed  3.00  g  He  and  26.0  g  Ne:     a)  What  is  the  pressure  of  each  gas?       P =He RT)/V=(He00  g/4.0  g)(0.0821)(323  K)/(21.6  L)=1.58  atm     b)  What  is  the  total  pressure?       P total    PHeP = Ne    +  1.58  =  2.50  atm     c)  Mole  fraction  of  Ne?       X =1.58  atm/2.50  atm   He Ex.  Consider  an  apparatus  connecting  two  containers  with  a  closed  valve.  One  container     contains  N  at 2  2.0  L,  1.0  atm,  and  25°C.  The  second  container  contains  O  at  3.0  L,  2.0 2    atm,  and  25°C.  When  the  valve  between  the  two  containers  is  opened,  what  is  the     partial  pressure  of  N ?  O ? 2   2   P V1 1 V  2 2   For  N : 2  5.0  L(P)=1.0  atm  (2.0  L)   P=0.40  atm     For  O : 2  5.0  L(P)=2.0  atm(3.0  L)   P=1.2  atm     P total    PN2  P = O2    +  1.2  =  1.6  atm   Collecting  Gas  over  Water     -­‐Gases  are  often  collected  by  having  them  displace  water  from  a  container.     •   The  problem  is  that  because  water  evaporates,  water  vapor  is  also   collected.   -­‐Partial  pressure  of  water  vapor  (vapor  pressure)  depends  only  on  the  temperature   (Table  11.3)   -­‐Ex.  A  gas  sample  is  collected  over  water.  The  gas  sample  has  a  total  pressure  of  758.2     mmHg  at  25°C.  From  a  vapor  pressure  of  water  table,  the  partial  pressure  of  the     water  vapor  is  23.78  mmHg  at  25°C.     Partial  Pressure=  758.2  -­‐  23.78  =  734.4  mmHg   Gases  in  Chemical  Reactions:  Stoichiometry  Revisited     -­‐mass  A  →  amount  A  (in  moles)  →  amount  B  (in  moles)  →  mass  B   Chem  130  Lecture  Notes  1.19.16  and  1.21.16     OR  use  P,V,T  of  gas  A  to  find  the  amount  in  moles  and  continue  from  there.     -­‐When  gases  are  at  STP,  use  1  mole=22.4  L     -­‐Pressure  here  could  also  be  partial  pressure.     *MOST  IMPORTANT  CONCEPT  IN  CHAPTER  11*   Ex.  Methanol  can  be  synthesized  by  the  reaction:     CO(g)+2H (g)  →CH O2  (g)   3            What  volume  (in  L)  of  2  gas  at  a  T  of  355  K  and  a  P  of  738  mmHg  do  we  need  to  synthesize                 35.7  g  CH OH?   3   Plan  to  Solve:       1.  g  CH OH3  →  mole  CH OH   3     2.  mol  CH OH  3  mol  H   2     3.  mol  H , 2P,  T  →  V H2   Ex.  How  many  grams  of  water  form  when  1.24  L  of  H  gas  at  STP  completely  reacts  with  O ?   2 2   2H (g)2  +  O (g) 2→  2H O  (g)   2   T=273  K;  P=1  atm;  1.24  L  H     2   *STP  Special  Condition:       *Use  Ideal  Gas  Law  if  you  forget  STP        1.24  L  2    1  mole  *  2  mol  *  18.02  g       special  conditions                            22.4  L          2  mol   1  mol                 =0.998  g  H O   2 Kinetic  Molecular  Theory     -­‐The  particles  of  the  gas  (either  atoms  or  molecules)  are  constantly  moving.     -­‐The  attraction  between  particles  is  negligible.     -­‐There  is  a  lot  of  empty  space  between  the  gas  particles  compared  to  the  size  of     particles.     -­‐Average  kinetic  energy  is  directly  proportional  to  Kelvin  T   •   As  temperature  increases,  average  speed  increases   -­‐Collision  of  one  particle  with  another  is  completely  elastic   •   may  exchange  energy,  but  no  overall  loss  of  energy   Temperature  and  Molecular  Velocities     -­‐Average  kinetic  energy  of  one  mole 2 of  gas  particles:       KE =(1avgN mu   a N =Avogadro's  Number       KE=1/2mv   2     m=  mass  of  particle   2           u =  average  square  of  velocity     -­‐Square  Root  of  u =u 2 rms   Root  mean  Square  Velocity  of  a  Particle     -­‐From  Kinetic  Molecular  Theory:  u =  square  root  of  (3RT)/M   rms   -­‐Lighter  particles=faster  average  velocity     -­‐Temperature  increases=more  spread  out   Diffusion  and  Effusion     -­‐Diffusion:gas  molecules  spreading  from  high  to  low  concentration     -­‐Effusion:  gas  molecules  escape  via  small  hole  to  vacuum.     -­‐Graham's  Law  of  Effusion:  rate  =  square  root A of  M /M B A                    B                   -­‐Lighter  particles  effuse  more  quickly  than  heavier  particles.     Chem  130  Lecture  Notes  1.19.16  and  1.21.16

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