The following reaction is endothermic: C(s) + H!O(g) CO (g) + H!(ūĚĎĒ)

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Chemistry ¬†122 ¬†(General ¬†College ¬†Chemistry ¬†I)-¬≠‚Äź ¬†Fall ¬†2018 ¬† Discussion ¬†Worksheet ¬†Week ¬†10 ¬†(Coordinated ¬†by ¬†Dr. ¬†R. ¬†Hatfield)

Topics: ¬†equilibrium ¬†state ¬†and ¬†the ¬†equilibrium ¬†constant ¬†(K), ¬†reaction ¬†quotient ¬†(Q), ¬†expressing ¬† equilibrium ¬†with ¬†pressure ¬†terms-¬≠‚Äź ¬†relationship ¬†between ¬†Kc ¬†and ¬†Kp, ¬†determining ¬†reaction ¬†direction, ¬† solving ¬†a ¬†variety ¬†of ¬†equilibrium ¬†problems, ¬†Le ¬†Chatelier‚Äôs ¬†principle ¬† ¬† Relevant ¬†Chapters ¬†in ¬†Silberberg, ¬†8e: ¬†Chapter ¬†17 ¬†Sections ¬†1-¬≠‚Äź6, ¬†pp. ¬†747-¬≠‚Äź780. ¬† ¬† ¬† Student ¬†Directions: ¬† ¬†Try ¬†all ¬†problems ¬†with ¬†your ¬†best ¬†effort ¬†and ¬†bring ¬†your ¬†work ¬†to ¬†discussion. ¬† ¬†Use ¬† your ¬†text ¬†as ¬†a ¬†resource ¬†for ¬†any ¬†constants ¬†or ¬†relationships ¬†needed. ¬†You ¬†will ¬†have ¬†to ¬†work ¬†ahead. ¬† When ¬†topics ¬†are ¬†completely ¬†covered ¬†in ¬†lecture, ¬†go ¬†back ¬†and ¬†redo ¬†any ¬†problems ¬†you ¬†did ¬†not ¬† understand. ¬† ¬†See ¬†pages ¬†782-¬≠‚Äź783 ¬†of ¬†your ¬†text ¬†for ¬†a ¬†summary ¬†list ¬†of ¬†key ¬†equations ¬†and ¬†relationships. ¬† ¬† Basic ¬†Math: ¬†Review ¬†the ¬†general ¬†form ¬†for ¬†quadratic ¬†equations ¬†and ¬†solving ¬†quadratic ¬†equations. ¬† Mon., ¬†Tues., ¬†and ¬†Wed. ¬†Discussions ¬†Only: ¬†Complete ¬†up ¬†through ¬†#5 ¬†for ¬†credit ¬†this ¬†week.

1. The  reaction  for  the  formation  of  ammonia  is  shown  as:

N!(g)    +    3  H!(g)    2NH!(g)     K!=  ?     a) Write  the  equilibrium  constant  expression  (K!)  for  this  reaction.

 

b) One  equation  for  the  breakdown  of  ammonia  is  shown  as:       NH!(g)     !!  N!(g)    +

! !  H!(g)        K! ’  =  ?

Manipulate  the  reaction  for  ammonia  formation,  as  necessary,  to  express  K! ’  in   terms  of  K! .    (Hint:  A  summary  table  for  ways  of  writing  and  calculating  K  is  found   on  page  755  of  your  text).

 

c) Calculate ¬†the ¬†value ¬†of ¬†K! ¬†at ¬†500. ¬†K ¬†for ¬†the ¬†formation ¬†of ¬†ammonia ¬†in ¬†part ¬†a) ¬†using ¬†the ¬† following ¬†measured ¬†concentrations ¬†for ¬†the ¬†equilibrium ¬†mixture: ¬†[N!] ¬†= ¬†3.0 ¬†x ¬†10-¬≠‚Äź2 ¬†M; ¬† [H!] ¬†= ¬†3.7 ¬†x ¬†10-¬≠‚Äź2 ¬†M; ¬† ¬†[NH!] ¬†= ¬†1.6 ¬†x ¬†10-¬≠‚Äź2 ¬†M. ¬† ¬† ¬† ¬† [1.7 ¬†x ¬†102]

 

d) Next,  calculate  the  value  of  K!  for  the  formation  of  ammonia  at  500.  K.       [0.10]

e) Using  the  calculated  value  of    K!  at  500.  K  from  part  a  as  well  as  the  expression  you

derived ¬†for ¬†K! ‚Äô ¬†in ¬†part ¬†b, ¬†calculate ¬†the ¬†value ¬†of ¬†K! ‚Äô. ¬† ¬† ¬† [7.7 ¬†x ¬†10-¬≠‚Äź2]

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2. Equilibrium  terminology.    When  solving  equilibrium  problems,  K  and  Q  are  used.    Define  Q.     When  does  K  =  Q?

3. Balance  each  of  the  following  examples  of  heterogeneous  equilibria  and  write  each  reaction   quotient,  Q!:

a) KNO!(s)        KNO!(s)  +    O!(g)

b) HCl(g)    +    O!(g)      H!O(l)    +        Cl!(g)

 

4. Calculate ¬†the ¬†value ¬†of ¬†the ¬†equilibrium ¬†constant ¬†at ¬†427 ¬†¬įC ¬†for ¬†the ¬†reaction:

Na!O ¬†(s) ¬†+!! ¬†O! ¬†(g) ¬† ¬† ¬†Na!O!(s) ¬† Kc ¬†= ¬†? ¬† ¬† ¬† ¬† ¬† [4000] ¬† ¬† Equilibrium ¬†constants ¬†for ¬†the ¬†following ¬†reactions ¬†are ¬†given ¬†at ¬†427 ¬†¬įC: ¬† ¬† Na!O(s) ¬† ¬† ¬† ¬† ¬†2 ¬†Na(l) ¬† ¬†+ ¬† ¬†!! ¬†O!(g) ¬† ¬† ¬† ¬† Kc ¬†= ¬†2 ¬†x ¬†10-¬≠‚Äź25 ¬† ¬† Na!O!(s) ¬† ¬† ¬†2 ¬†Na(l) ¬† ¬†+ ¬† ¬† ¬†O!(g) ¬† ¬† ¬† ¬† Kc ¬†= ¬†5 ¬†x ¬†10-¬≠‚Äź29

 

 

5. Phosphine  (PH!)  decomposes  at  higher  temperatures  as  follows:

2 ¬†PH! ¬†(ūĚĎĒ) ¬† ¬† ¬† ¬†P! ¬†(g) ¬† ¬†+ ¬† ¬†3 ¬†H! ¬†(g) ¬† ¬† ¬† Kp ¬†= ¬†398 ¬†at ¬†873 ¬†K

a) If  the  initial  partial  pressures  are  PPH3  =  0.0260  atm,  PP2  =  0.871  atm,  PH2  =  0.517  atm,   calculate  Qp  and  determine  which  direction  the  reaction  proceeds  in  to  reach   equilibrium.                 [178]

 

b) When  a  mixture  of  PH!,  P!,  and  H!  reaches  equilibrium  at  873  K,  PP2  =  0.412  atm  and   PH2  =  0.822  atm.    Find  PPH3.             [0.0240  atm]

 

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6. Iodine  gas  and  hydrogen  gas  are  sealed  in  a  flask  and  heated  to  600.  K.    The  initial  partial   pressures  of  iodine  and  hydrogen  gas  at  600.  K  before  they  react  are:    PI2  =3.00  atm  and  PH2   =2.00  atm.    No  other  reactions  take  place.    Determine  the  equilibrium  partial  pressures  of   I!,H!,  and  HI  at  600.  K.               [I! =  1.13  atm]

[H!  =  0.134  atm]   I!(g)  +  H!(g)   2  HI(g)   K!  =  92          at  600.  K       [HI  =  3.73  atm]

 

7. In ¬†the ¬†decomposition ¬†of ¬†hydrogen ¬†sulfide: ¬† 2H!S(g) ¬† ¬†2H!(g) ¬† ¬†+ ¬† ¬†S!(g) ¬† ¬† ¬† ¬† Kc ¬†= ¬†9.30 ¬†x ¬†10-¬≠‚Äź8 ¬† ¬†at ¬†700 ¬†¬įC

If ¬†0.51 ¬†mol ¬†H!S ¬†is ¬†placed ¬†in ¬†a ¬†3.0 ¬†liter ¬†container, ¬†what ¬†is ¬†the ¬†equilibrium ¬†concentration ¬†of ¬† H!(g) ¬†at ¬†700 ¬†¬įC? ¬† ¬†Can ¬†the ¬†small ¬†x ¬†approximation ¬†be ¬†used ¬†here ¬†when ¬†solving ¬†for ¬†x? ¬† ¬† ¬† ¬† ¬† ¬† ¬† ¬† ¬† ¬† ¬† ¬† ¬† [1.8 ¬†x ¬†10-¬≠‚Äź3 ¬†M]

 

 

 

8. The ¬†following ¬†reaction ¬†is ¬†endothermic: ¬† ¬† ¬† ¬†C(s) ¬† ¬† ¬†+ ¬† ¬†H!O(g) ¬† ¬† ¬† ¬† ¬†CO ¬†(g) ¬† ¬†+ ¬† ¬†H!(ūĚĎĒ)

For  a  reaction  mixture  at  equilibrium,  how  would  the  following  changes  affect  the  equilibrium   (shift  right,  shift  left,  no  change)  and  the  equilibrium  concentration  of    CO  (g)  (increase,   decrease,  no  change)?   Change   Effect  of  equilibrium   Effect  on  concentration  of  CO   Decreasing  the  temperature       Adding  C  to  the  mixture       Adding  H!O  to  the  mixture       Adding  a  catalyst       Adding  H!  to  the  mixture       Lowering  the  volume         Adding  an  inert  gas