Numerical Applications of Equilibrium Constants

1
Numerical Applications of Equilibrium Constants

• Predicting the direction in which a reaction will
  proceed.
• Ex For the reaction H2(g) I2(g) ? 2 HI(g),
  the Kc value is 50.7 at 448C.
• Suppose that a flask contains the following
  concentrations of the three species
• H2 0.0250 M, I2 0.0180 M, HI
  0.0500M.
• In which direction will the reaction proceed?

2

• The mass-action expression is designated Q, the
  reaction quotient, if we dont know whether or
  not the system is at equilibrium.
• Since Q lt Kc, the reaction must make more
  products and use up reactants in order to reach
  equilibrium (increase the numerator and decrease
  the denominator).
• Therefore, the reaction will spontaneously
  proceed to the right to reach equilibrium.

3

• Summary given an arbitrary set of reactant and
  product concentrations, if
• Q lt Kc , the reaction will proceed to the right
• Q gt Kc , the reaction will proceed to the left
• Q Kc , the reaction is at equilibrium

4

• Relating Initial and Equilibrium Concentrations
  Using Kc
• A general setup for equilibrium problems
• 1) Need balanced chemical equation
• 2) Set up I.C.E. table
• a) Initial concentrations of all species
• b) Algebraic expression for changes in
  concentration of all species (in terms of x)
• c) Equilibrium concentrations (a b)
• 3) Substitute from line c) into Kc expression.

5

• Ex H2(g) I2(g) ? 2 HI(g), Kc 29.1 at 1000
  K
• If 10.0 mol of HI is initially placed in a
  1.00-L flask at 1000K, what is the concentration
  of I2 after equilibrium is established?
• The I.C.E. table
• H2 I2
  ? 2 HI
• initial 0 0
  10.0 M
• change x x
  2x
• equil x x
  (10.0 2x)

6

• Now set up the mass-action expression
• Note that this is a quadratic equation.
• 29.1 x2 100.0 40.0 x 4 x2
• 25.1 x2 40.0 x 100.0 0
• in the form a x2 b x c 0

7

• Recall the quadratic formula
• Substituting
• Only the root 1.35 mol/L makes sense.
• H2 I2 1.35 mol/L
• HI 10.0 2(1.35) 7.30 mol/L

8

• But note that the mass-action expression is a
  perfect square. Take the square root
• 5.39 x 10.0 2 x
• 7.39 x 10.0
• x 1.35 mol/L

9

• Ex. 2 N2O4(g) ? 2 NO2(g) at 380C
• Initially, there is 0.625 mol of N2O4 in a 5.00-L
  flask. After equilibrium is reached, there is
  0.0750 mol/L of N2O4 along with some NO2.
  Determine Kc for this reaction.
• I.C.E. table N2O4 ? 2
  NO2
• initial (0.625 mol/5.00 L) 0
• change x
  2x
• equil. (0.125 M x)
  2x

10

• We also know that the equilibrium concentration
  of N2O4 is 0.0750 M, which we have set equal to
  (0.125 x).
• Thus, (0.125 x) 0.0750 M, and
• x 0.125 0.0750 0.0500 mol/L
• Now, NO2 2x 2 (0.0500 mol/L) 0.100 M
• Kc 0.133 mol/L

11

• Now find the value of Kp for this reaction.
• ?n 2 mol (prods) 1 mol (reacts.) 1 mol
• Kp Kc (RT)?n
• (0.133 mol/L) (0.08206 L atm/mol K)(653 K)1
• 7.13 atm