Chemical equilibrium By Dr. Hisham Ezzat Abdellatef Professor of Pharmaceutical Analytical Chemistry First Year

Classification of reaction Homogeneous only one phase Heterogeneous mixture is not uniform e.g: In the gas phase: H 2(g) + I 2 → 2 HI (g) In the liquid phase: CH 3 COCl + CH 3 OH → CH 3 COOCH 3 + HCI Acetyl Chloride methyl alcohol methylacetate

irreversible reaction or complete reaction reversible reaction →  A + B → AB N 2(g) + 3 H 2(g)  2 NH 3(g) at equilibrium ----the rate of forward reaction equal the rate of backward reaction

Factors influencing equilibrium: 1.forward and reverse rates 2.partial pressures 3.concentrations 4.temperature

Law of mass action A 2(g) + B 2(g)  2 AB (g) rate of the forward reaction = K f. [A 2 ] [B 2 ] rate of the reverse reaction is = K r [AB] 2 At equilibrium rate f = rate r K f [A 2 ] [B 2 ] = K r [AB] 2 K which is called the equilibrium constant

aA + bB  eE + fF eE +fF  aA + bB In general, for any reversible reaction

Mechanisms of more than one step 1 2 NO 2 CI (g)  2 NO 2 (g) + CI 2 (g) mechanism consisting of two steps: 1- NO 2 CI  NO 2 + Cl 2. NO 2 CI  NO 2 + CI 2

For reactions involving gases P α C using partial pressures instead of concentration. N 2(g) + 3 H 2(g)  2 NH 3(g) (K c ≠ K p )

The Relationship Between K p and K C : aA + bB  eE + fF Assuming ideal gas PV= nRT concentration of a gas XP x is its partial pressure  P x = [X] RT K p = K c. (RT)  n (g)

Example 1: For the reaction N 2 O 4(g)  2 NO 2(g) The concentrations of the substances present in an equilibrium mixture at 25°C are [N 2 O 4 ] = 4.27 x mol/L [NO 2 ] = 1.41 x mol/L what is the value of K c for this temperature.

Example 2: At 500 K. 1.0 mol of ONCI (g) is introduced into a one - liter container. At equilibrium the ONCI (g) is 9.0% dissociated: 2 ONCI (g)  2 NO (g) + CI 2(g) Calculate the value of K c for equilibrium at 500 K. Solution: [ONCI (g) ] = 1 mol/L since 9.0% dissociated, Number of moles dissociated = at equilibrium, [ONCI] = 1.0 mol/L mol/L = 0.91 mol/L amounts of CI 2 : 2 ONCI  2NO + CI2 2 x x 0.09 mol mol

2ONCI  2NO + CI 2 at start 1.0 mol/L Change mol/L at equilibrium

Example 3: For the reaction 2 SO 3(g)  2 SO 2(g) + O 2(g) at 1100 K K c is mol/L. what is K p at same temperature. Solution:  n = 3-2 =1 K p = K c (RT) +1 = mol/L x ( L. atm / K. mol) (1100 K) = 2.45 atm

Example 4: What is K c for the reaction? N 2(g) + 3 H 2(g)  2 NH 3(g) At 500°C if K p is 1.5 x / atm -2 at this temperature. Solution: K p = K c (RT)  n K p = K c (RT) -2 K c = K p (RT) 2 = x [ atm/K.mol x 773K] 2 = (1.5 x I0 -5 / atm 2 ) (4.03 x 10 3 L 2. atm 2 / mol 2 ) = 6.04 x L 2 / mol 2  n = = -2 T = = 773 K

Try At 127 o C, K = 2.6 x mol 2 /L 2 for the reaction 2NH 3 (g)  N 2 (g) + 2H 2 (g) Calculate Kp at this temperature

Reaction quotient (Q).

Predicting the Direction of a Relation: For the reaction PCl 2(g)  PCl 3(g) + Cl 2(g) at 250 o C Suppose that a mixture of 1.00 mol of PCI 5(g) / 0.05 mol of PCI 3(g) / and 0.03 mol of CI 2(g) is placed in 1.0 L container. Is this an equilibrium system, or will a net reaction occur in one direction or the other?

1- Q < K c from left to right (the forward direction) to approach equilibrium. 2- Q = K c The system is in equilibrium. 3- Q > K c from right to left (the reversible direction) to approach equilibrium.

Q (0.015 mol/L) < k c ( mol/L). The system is not at equilibrium. The reaction will proceed from left to right.

Example 4: For the reaction 2 SO 2(g) + O 2(g)  2 SO 3(g) at 827°C, k c is 36.9 L / mol. If 0.05 mol of SO 2(g), 0.03 mol of O 2(g), and mol of SO 3(g) are mixed in a 1.0 L container, in what direction will the reaction proceed? Solution: Since Q (208 L/mol) > k c (36.9 L/mol), the reaction will proceed from right to left (SO 3 will dissociate).

Heterogeneous Equilibria: The concentration of a pure solid or a pure liquid is constant and do not appear in the expression for the equilibrium constant. For example CaCO 3(S)  CaO (S) + CO 2(g) K c = [CO 2 ] 3 Fe (s) + 4 H 2 O (g)  Fe 3 O 4(s) + 4 H 2(g)

Example 5: K c for the HI equilibrium at 425°C is 54.5: H 2(g) + I 2(g)  2 HI (g) A quantity of HI (g) is placed in a 1.01 container and allowed to come to equilibrium at 425°C What are the concentrations of H 2(g) and I 2(g) in equilibrium with 0.5 mol/L of HI (g)

Solution: at equilibrium [H 2 ] = [I 2 ]  [H 2 ] = [I 2 ] = x [HI] = 0.5 mol/L  x = mol/L The equilibrium concentration is: [HI] = 0.5 mol/ L [H 2 ] = [I 2 ] = mol/ L

Example 6: For the reaction H 2(g) + CO 2(g)  H 2 O (g) + CO (g) k c is at 750°C. If 0.01 mol of H 2 and 0.01 mol of CO 2 are mixed in 1 liter container at 750°C, what are the concentrations of all substances present at equilibrium?

Solution: If x mol of H 2 reacts with x mol of CO 2 out of the total amount supplied, x mol H 2 O and x mol CO will be produced. Hence H 2(g) + CO 2(g)  H 2 O (g) + CO (g) At start 0.01 mol/L 0.01 mol/L Change - x - x + x + x at equilibrium x x X X

square root X = – X X= mol/l At equilibrium, therefore [H 2 ] = [CO 2 ] = 0.01 mol/L mol/L = mol/L [H 2 O] = [CO] = mol/L

Example 7: For the reaction C (s) + CO 2(g)  2 CO (g) K p is atm at 1000°C. What is the partial pressure of CO (g) in an equilibrium system in which the partial pressure ofCO 2(g) is 0.1atm? Solution: P CO2 = 16.8 P CO = 4.10 atm

Example 8: K p for the equilibrium: FeO (s) + CO (g)  Fe (s) + CO 2(g) at 1000°C is If CO (g) at a pressure of 1.0 atm, and excess FeO (s) are placed in a container at 1000°C, what are the pressures of CO (g) and CO 2(g) when equilibrium is attained?

FeO (s) + CO (g)  Fe (s) + CO 2(g) At start Change 1.0 atm - x x At equilibrium1.0 - x atmx Solution: Let x equal the partial pressure of CO 2 when equilibrium is attained X = P CO2 = atm 1.0 – X = P co =.713

If a change is made to an equilibrium, the equilibrium shifts in the direction that consumes the change – Case 1: Changing the amounts of reactants / products. – Case 2: Changing the volume by changing pressure. – Case 3: Changing the temperature.

If the concentration of substance is increased, the equilibrium will shift in a way that will decrease the concentration of the substance that was added. e.g: H 2(g) + I 2(9)  2 HI (g) Increase H 2 or I 2 → shift to to formation of HI Removal of H 2 or I 2 ← Reaction shift to decomposition of HI.

Increasing the pressure causes a shift in the direction that will decrease the number of moles of gas. 2 SO 2(g) + O 2(g)  2 SO 3(g) 3 moles2 moles When the pressure on an equilibrium mixture is increased (or the volume of the system decreased), the position of equilibrium shifts to the right., and vice versa. For reactions in which  n = 0, pressure changes have no effect on the position equilibrium. e.g: N 2(g) + O 2(g)  2 NO (g)

For the reaction N 2(g) + 3 H 2(g)  2 NH 3(g)  H = KJ Since  H is -ve, the reaction to the right evolves heat N 2(g) + 3 H 2(g)  2 NH 3(g) KJ The highest yields of NH 3 will be obtained at the lowest temperatures and high pressures.

Also consider the reaction CO 2(g) + H 2(g)  CO (g) + H 2 O (g)  H = KJ Since  H is + ve, we can write the equation 41.2 KJ + CO 2(g) + H 2(g) —— CO (g) + H 2 O (g) Increasing the temperature always favors the endothermic change, and decreasing the temperature always favors the exothermic change.