Chemistry 1011

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Chemistry 1011

• TOPIC
• Gaseous Chemical Equilibrium
• TEXT REFERENCE
• Masterton and Hurley Chapter 12

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12.4 Applications of the Equilibrium Constant

• YOU ARE EXPECTED TO BE ABLE TO
• Determine the potential for a reaction to occur
  from the magnitude of the equilibrium constant.
• Distinguish between a reaction quotient and the
  equilibrium constant.
• Predict the direction in which a reaction is
  proceeding given the value of the reaction
  quotient.
• Calculate the concentration (partial pressure) of
  a component of an equilibrium system, given the
  equation, the value of the equilibrium constant,
  and the partial pressures of other components.

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Application 1 Interpreting the Magnitude of K

• The magnitude of the equilibrium constant can
  give an indication of whether a reaction is
  likely to occur
• If K is very small, then the concentration of
  products at equilibrium would be very low
• Eg N2(g) O2(g) 2NO(g)
• Kp 1.0 x 10-30 at 25oC
• Alternatively, if K is gt1 then the equilibrium
  mixture will contain product
• Eg N2(g) 3H2(g) 2NH3(g)
• Kp (PNH3)2 6.0 x 105 at 25oC
• PN2 x (PH2)3

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Application 2 The Reaction Quotient

• For aA(g) bB (g) cC (g) dD (g)
• K (PC)c x (PD)d
• (PA)a x (PB)b
• If the system is NOT at equilibrium, the actual
  pressure ratio, known as the reaction quotient,
  can have any value at all
• Q (PC)c x (PD)d
• (PA)a x (PB)b

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Using the Reaction Quotient

• Comparison of a reaction quotient to the
  equilibrium constant will indicate whether the
  reaction is at equilibrium
• If not, it will indicate which direction the
  reaction will proceed in

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Q and the Direction of Reaction

• If Q is LESS than K, then the reaction will
  proceed from left to right. The value of Q will
  increase until it becomes equal to K
• If Q is GREATER than K, then the reaction will
  proceed from right to left. The value of Q will
  decrease until it becomes equal to K
• If you start with pure reactants, the value of Q
  will initially be zero
• If you start with pure products, the value of Q
  will initially be infinity

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The NOBr Equilibrium

• Given that Kp for the NOBr equilibrium at 350oC
  is 2.8 x 10-2, will any net reaction occur if
  1.00atm of NOBr, 0.80atm of NO and 0.40atm of Br2
  are mixed?
• If so, will NO be formed or consumed?
• 2 NOBr(g) 2NO(g) Br2(g)
• Q (PNO)2 x PBr2 (0.80)2 x (0.40)
  2.6 x 10-1
• (PNOBr)2
  (1.00)2
• Q gt Kp
• The reaction will move towards the reactants.
• NO will be consumed

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Application 3 Determining Equilibrium Partial
Pressures

• Given a balanced equation, initial partial
  pressures or concentrations and the value for the
  equilibrium constant, it is possible to determine
  the equilibrium partial pressures of reactants
  and products
• Sometimes a quadratic equation will have to be
  solved

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Determining Equilibrium Partial Pressures

1. Write a balanced equation for the equilibrium
2. Write an expression for the equilibrium constant
3. Distinguish equilibrium from initial partial
   pressures
4. Use x for the unknown partial pressures.
   Express the equilibrium partial pressures of all
   species in terms of x
5. Input equilibrium partial pressures into
   expression for Kp
6. Calculate x

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Determining Equilibrium Partial Pressures

• Consider the system
• PCl5(g) PCl3(g) Cl2(g)
• Initially, a system contains PCl5 only, at a
  pressure of 3.00atm at 300oC.
• The value of the equilibrium constant Kp at this
  temperature is 11.2
• Find
• PPCl5 at equilibrium
• PPCl3 at equilibrium
• PCl2 at equilibrium

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Determining Equilibrium Partial Pressures

• PCl5(g)
  PCl3(g) Cl2(g
• Initially 3.00atm
  0.00atm 0.00atm
• At equilibrium ?
  ? ?
• Let the equilibrium partial pressure of PCl3 be
  x atm
• DP - x atm
  x atm x atm
• At equilibrium 3.00 - x atm
  x atm x atm
• Kp
  PPCl3 x PCl2 x x x 11.2
• PPCl5
  (3.00 - x)
• x 2.46atm
• PPCl3 2.46atm PCl2 2.46atm PPCl5 0.54atm

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Determining Equilibrium Partial Pressures

• Hydrogen cyanide can be made by the reaction
• C2N2(g) H2(g) 2HCN(g)
• At a certain temperature, Kp 64
• Calculate the partial pressures of all species at
  equilibrium at this temperature if the initial
  partial pressures of the reactants are 0.50atm

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Determining Equilibrium Partial Pressures

• C2N2(g) H2(g)
  2HCN(g)
• Initially 0.50atm
  0.50atm 0.00atm
• At equilibrium ? ?
  ?
• Let the equilibrium partial pressure of HCN be
  2x atm
• DP - x atm
  -x atm 2x atm
• At equilibrium (0.50 x)atm (0.50 x)atm
  2x atm
• Kp
  (PHCN)2 (2x)2 64
• PC2N2 x PH2
  (0.50 x)2
• x 0.40atm
• PHCN 0.80atm PC2N2 0.10atm PH2 0.10atm