Equilibrium constant-presentation

Equilibrium Constant Reaction Quotient

Expression for Equilibrium ConstantConsider the following equilibrium system:

wA + xB ⇄ yC + zD

Kc =

• The numerical value of Kc is calculated using the concentrations of reactants and products that exist at equilibrium.

xw

z

[B][A][D][C]y

Expression for Equilibrium Constant

• Examples:N2(g) + 3H2(g) ⇄ 2NH3(g); Kc =

PCl5(g) ⇄ PCl3(g) + Cl2(g); Kc =

CH4(g) + H2(g) ⇄ CO(g) + 3H2(g);

Kc =

322

23

]][H[N][NH

][PCl]][Cl[PCl

5

23

O]][H[CH][CO][H

24

32

Calculating Equilibrium Constant

• Example-1:1.000 mole of H2 gas and 1.000 mole of I2 vapor are introduced into a 5.00-liter sealed flask. The mixture is heated to a certain temperature and the following reaction occurs until equilibrium is established.

H2(g) + I2(g) ⇄ 2HI(g)

At equilibrium, the mixture is found to contain 1.580 mole of HI. (a) What are the concentrations of H2, I2 and HI at equilibrium? (b) Calculate the equilibrium constant Kc.

Calculating Equilibrium Constantfor reaction: H2(g) + I2(g) ⇄ 2HI(g)• ————————————————————————————• H2(g) + I2(g) ⇄ 2 HI(g)• ————————————————————————————• Initial [ ], M: 0.200 0.200 0.000• Change in [ ], M: -0.158 -0.158 + 0.316• Equilibrium [ ], M 0.042 0.042 0.316• ————————————————————————————

Kc = = = 57]][I[H[HI]

22

2

2)(0.042(0.316)2

Calculating Equilibrium Constant• Example-2:

0.500 mole of HI is introduced into a 1.00 liter sealed flask and heated to a certain temperature. Under this condition HI decomposes to produce H2 and I2 until an equilibrium is established. An analysis of the equilibrium mixture shows that 0.105 mole of HI has decomposed. Calculate the equilibrium concentrations of H2, I2 and HI, and the equilibrium constant Kc for the following reaction:

H2(g) + I2(g) ⇄ 2HI(g),

Calculating Equilibrium Constant• The reaction: H2(g) + I2(g) 2HI⇄ (g), proceeds from right to left.• ————————————————————————————• H2(g) + I2(g) ⇄ 2HI(g)• ————————————————————————————• Initial [ ], M: 0.000 0.000 0.500• Change in [ ], M: +0.0525 +0.0525 -0.105• Equil’m [ ], M 0.0525 0.0525 0.395• ————————————————————————————

Kc = = 56.62

2

(0.0525)(0.395)

Equilibrium Constant Expression for Concentration (Kc)

At equilibrium, the concentration in moles/L of N2, H2, and NH3 at 300°C are 0.25, 0.15, and 0.90, respectively. Find the KC for the reaction

N2 (g) + 3H2 (g) 2NH3 (g)

Equilibrium Constant Expression for Concentration (Kc)Use the value of KC calculated in problem 1 to decide in which direction the reaction will occur given the following concentrations in mol/L.

N2 H2 NH3

0.25 0.25 0.000.15 0.45 0.300.10 0.50 0.20

Predicting the Direction of a ReactionAt any one point during the course of a reaction (nonequilibrium state), it is possible to determine the concentration ratio of the products and reactants using the same form as the equilibrium constant expression. This ratio is called the reaction quotient designated by the symbol Q. The computed values of Q are compared with KC.

Predicting the Direction of a ReactionReaction Quotient (QC)

• The quantity obtained by substituting the initial concentrations into the equilibrium constant expression

Predicting the Direction of a ReactionWhen Q < Kc, the reaction proceeds forward

( )

When Q > Kc, the reaction proceeds backward

( )

When Q = Kc, the system at equilibrium

Predicting the Direction of a ReactionConsider the following reaction and concentration data at 445°C.

H2 (g) + I2 (g) 2HI (g) Kc = 50.2 at 445°C

Predicting the Direction of a ReactionExperiment Initial Concentration

H2 (g) I2 (g) HI (g) Q

1 1.5 x 10 -3 1.5 x 10-3 0 0

2 1.5 x 10 -3 1.5 x 10-3 1.5 x 10-3 1

Predicting the Direction of a ReactionThe composition of PCl5 is shown below.

PCl5(g) PCl3(g) + Cl2(g) Kc = 0.029 at 85°CFind out if the reaction proceeds forward, backward, or if the system is at equilibrium when 1.0 L container is filled with 0.400 mole PCl5, 0.500 mole PCl3, and 0.200 mole Cl2.