Gaseous Chemical Equilibrium Chapter 12

2HI(g) H2(g) + I2(g) If pure HI(g) is placed in a sealed container at 52 C, H2(g) and I2(g) are formed. Then the reverse reaction can occur forming HI(g) The partial pressure of HI(g) drops, slowing the forward reaction, all the while, the partial pressures of H2(g) and I2(g) increase; increasing the rate of the reverse reaction

Soon the forward and reverse rates become equal and a dynamic equilibrium is established. At equilibrium appreciable amounts of all gases are present. At equilibrium the partial pressures of all gases remain constant.

Note: P HI = -2 P H2 = -2 P I2 2HI(g) H2(g) + I2(g) Po(atm) 0.200 0.000 P(atm) - 0.040 + 0.020 Peq(atm) 0.160 0.020 Note: P HI = -2 P H2 = -2 P I2

Equilibrium constant Kp Depends only on temperature Is independent of: Original pressures Total pressure Volume

Equilibrium expression Only gaseous and aqueous species allowed Products numerator Reactants denominator Gases enter as pressures (atm) Aqueous enter as concentrations [M] Adding or removing a solid or liquid does NOT affect the position of the equilibrium

2 HI H2 + I2 (g) (g) (g) ( ) ( ) = Kp Equilibrium expression ( )

Try: Remember: solids and liquids do not appear in equilibrium expressions The pressure of CO2 will be the same regardless of how much CaO or CaCO3 is present

Try:

2HI(g) H2(g) + I2(g) Po(atm) 0.200 0.000 P(atm) - 0.040 + 0.020 Peq(atm) 0.160 0.020

For the reverse rxn for

If Eqn 3 = Eqn 1+ Eqn 2 then K3 = K1 x K2

Applications of K If K is large If K is small Equilibrium moves Product formation is favored If K is small Reactant formation is favored

Determining the direction of a reaction Compare the actual pressure quotient, Q, to K. If Q < K the forward reaction occurs to make Q larger If Q > K the reverse reaction occurs to make Q smaller If Q = K system is at equilibrium nothing happens

K = 0.016 Predict the direction of the reaction if HI = I2 = H2 = 0.10 atm Q > K reaction shifts

What is the pressure of all the gases at equilibrium when 0 What is the pressure of all the gases at equilibrium when 0.100 atm of pure HI is allowed to react? 2HI(g) H2(g) + I2(g) K = 0.016 Po 0.100 atm 0.000 atm P -2X +X Peq 0.100 - 2X X

What are the partial pressure of N2O4 and NO2 when 0 What are the partial pressure of N2O4 and NO2 when 0.100 atm of N2O4 is allowed to come to equilibrium? N2O4(g) 2NO2(g) K = 11 Po 0.100 atm 0.000 atm P - X + 2X Peq 0.100 - X 2X Quadratic equation X = 0.097; NO2 = 0.194atm; N2O4 = 0.003atm

Le Chatelier’s Principle If a system at equilibrium is subjected to stress, the reaction occurs in a direction so as to partially relieve the stress.

2HI(g) H2(g) + I2(g) If H2 is added Reaction shifts P HI is greater than before disturbance P I2 is less P H2 is intermediate between original equilibrium value and that immediately after disturbance

2HI(g) H2(g) + I2(g) If H2 is removed reaction shifts Some HI decomposes to bring P H2 back part way to its equilibrium value P I2 is increased P HI decrease

Increasing the volume Immediate effect is to lower the concentration of molecules To counteract this, reaction occurs in which increases the number of molecules of gas

Expansion ( V; P) Contraction ( V; P) N2O4(g) 2NO2(g) N2(g) + 3H2(g) 2NH3(g) H2(g) + I2(g) 2HI(g) No effect No effect

Changes in temperature Endothermic process is favored by increased Temperature Temperature change is the only change that affects the value of K If forward reaction is endothermic: K as T If forward reaction is exothermic: K as T van’t Hoff equation

H T N2O4(g) 2NO2(g) +58.2 kJ N2(g) + 3H2(g) 2NH3(g) - 924 kJ