Chemical Reaction Equilibria

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Chemical Reaction Equilibria

• Part III

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Equilibrium constant K
For gas-phase reactions, fio Po 1 bar
At a given temperature, if P changes, the
compositions at equilibrium will change in such a
way that K remains constant
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K for gas phase reactions
this is the equation that we used in the previous
example
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analysis of K for ideal gases
At constant P, if the reaction is endothermic,
DH0 gt0, K increases when T increases, therefore
the LHS term will increase, the reaction shifts
to the right, eeq increases if the reaction is
exothermic, DH0 lt0, K decreases when T
increases, therefore the LHS term will decrease,
the reaction shifts to the left, eeq decreases
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effect of pressure (at constant T)

• it depends on n, which is the change in the total
  number of moles of the reaction
• if n is negative gt the total number of moles
  decreases
• if P increases, the LHS must increase to keep K
  constant, gt the equilibrium shifts to the right,
  eeq increases
• if n is positive gt the total number of moles
  increases
• if P increases, the LHS must decrease to keep K
  constant, gt the equilibrium shifts to the left,
  eeq decreases

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example

• the production of 1,3-butadiene can be carried
  out by dehydrogenation of n-butane
• C4H10(g)?CH2CHCHCH2 (g) 2H2(g)
• Side reactions are suppressed by introduction of
  steam. If equilibrium is attained at 925 K and 1
  bar and if the reactor product contains 12 mol
  of 1,3 butadiene, find
• the mole fractions of the other species in the
  product gas
• the mole fraction of steam required in the feed

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solution

• C4H10(g)?CH2CHCHCH2 (g) 2H2(g)
• n 2
• no 1x
• y1 (1-e)/(1x2e)
• y2 e/(1x2e)0.12
• y3 2 y20.24
• In order to calculate K, we need DG at 925 K
• First calculate DGo and DHo at 298 K, from
• tables appendix C
• DGo 235030 J/mol
• DHo 166365 J/mol

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• Get A, B, C, D for each component and calculate
  DA, DB, DC, and DD.
• Calculate DG at 925K using equation 13. 18 DG
  9.242 x103 J/mol
• Calculate K exp (- DG/RT) 0.30
• K (y3)2(y2)/y1(0.24)2(0.12)(1x2e)/(1-e)0.3
• here there are two unknowns, x, and e
• However since we know y2 we have another equation
  y2 e/(1x2e)0.12
• Therefore, solve for e0.84 and x4.31mol steam
• Get y1 (1-e)/(1x2e) 0.023
• ysteam 4.31/5.31 0.812 (in the feed)
• yH2O at equilibrium 1-0.24-0.12-0.023 0.617

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Ammonia synthesis reaction

• ½ N2(g) 3/2 H2(g) ? NH3 (g)
• the equilibrium conversion of ammonia is large at
  300K but decreases rapidly with increasing T.
  However, reaction rates become appreciable only
  at higher temperatures. For a feed mixture of
  hydrogen and nitrogen in the stoichiometric
  proportions,
• what is the equilibrium mole fraction of ammonia
  at 1 bar and 300 K?

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solution
½ N2(g) 3/2 H2(g) ? NH3 (g)

• n -1
• no 2
• In order to calculate K, we need DG at 300 K
• First calculate DGo and DHo at 298 K, from
• tables appendix C
• DGo -16450 J/mol
• DHo - 46110J/mol
• Get A, B, C, D for each component and calculate
  DA, DB, DC, and DD.
• Calculate DG at 300K using equation 13. 18 DG
  -16270 J/mol
• Calculate K exp (- DG/RT) 679.57

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• K (y3)/(y2)3/2(y1)1/2
• you can show (see problem 13.9) that
• eeq 1-(11.299KP/Po)-1/20.9664
• yNH3 e/(2-e)0.935
• (b) At what T does the equilibrium mole fraction
  of ammonia equal 0.5 for a pressure of 1 bar?
• if yNH3 0.5, eeq 2/3 1-(11.299KP/Po)-1/2
• K 6.16? at what T, K has this value?
• K exp (- DG (T)/RT) 6.16? solve for T
  iterative T399.5 K

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• (c) At what temperature does the equilibrium mole
  fraction of ammonia equal 0.5 at a pressure of
  100 bar, assuming the equilibrium mixture is an
  ideal gas?
• For P 100 bar,
• eeq 1-(11.299KP/Po)-1/2 2/3
• K 0.06159? at what T, K has this value?
• K exp (- DG (T)/RT) 0.06159
• ? solve for T577.6 K

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• (d) at what temperature does the equilibrium mole
  fraction of ammonia equal 0.5 for a pressure of
  100 bar, assuming the equilibrium mixture is an
  ideal solution of gases?
• for an ideal solution model,
• (I)
• i) Define a guess T. Start with Tguess 578 K
  (obtained in part a) and 100 bar. Use virial
  equation of state, for fis that are functions of
  temperature
• ii) Obtain new expression for K using the
  calculated fis in equation (I)
• 1-(11.299K/1.184xP/Po)-1/2 2/3
• iii) solve for K 0.0729 iv) since K (Tguess)
  0.0729 must beexp (- DG (T)/RT), evaluate exp (-
  DG (Tguess)/RTguess). Is it equal to 0.0729? If
  not, change Tguess and go to (i)
• Solution after convergence T 568.6 K