The Position of Equilibrium Section 7.2

The Equilibrium Constant The position of equilibrium for a particular reversible reaction can be defined by a constant This constant has a numerical value found by relating the equilibrium concentrations of the products to those of the reactants Denoted by Kc the equilibrium constant

Deducing the Equilibrium Constant Consider this standard reaction: aA + bB cC + dD Where a, b, c and d represent the stoichiometry of each substance, then:

Continued The concentrations must be those that occur at equilibrium, not the starting values Kc is constant for a given temperature Change the temperature, you change Kc

Equilibrium Law Equilibrium law: in any reversible reaction at a state of equilibrium, the rate of the forward reaction equals the rate of the reverse reaction

Problem #1 Consider the following reaction: N2(g) + 3H2(g) 2NH3(g) How is the Kc calculated?

Problem #1 Continued An experiment has this data at equilibrium at 500°C: [N2] = 0.922 mol dm-3 [H2] = 0.763 mol dm-3 [NH3] = 0.157 mol dm-3 Calculate the Kc for this experiment

Working Out 0.0602 A note about units: the concentrations calculated would give you varying units, Kc is unitless because you are using the “activity” which is close to concentration

Another Consideration H2(g) + I2(g) 2HI(g) Kc for the forward reaction: For the reverse reaction:

Continued The value of the equilibrium constant for the reverse reaction is the reciprocal of that for the forward reaction

How Far will a Reaction Go? Consider the following reactions and Kc H2(g) + I2(g) 2HI(g) Kc = 2 at 277°C H2(g) + Cl2(g) 2HCl(g) Kc = 1018 at 277°C The large difference illustrates how stable HCl is at this temperature compared to HI Virtually all of the reaction between H2 and Cl2 has produced HCl

Continued The magnitude of Kc is a good indicator of how far a reaction goes toward completion (under given conditions) If Kc>>1, then the reaction has gone virtually to completion If Kc<<1, then the reaction has hardly taken place at all

Continued For Kc = 1, there are equal amounts of reactants and products The Kc for a given reversible reaction at equilibrium only changes if the temperature changes

Other Considerations The concentrations of certain substances remain constant, so they are omitted from the equilibrium constant expression All solids have a fixed density, so these are omitted The concentration of pure liquid water is also constant (important to consider when solutions are aqueous, water omitted)

Continued If water is used as a liquid with other liquids ( not aqueous solutions), then it is included Water must be included if it is in the gas phase

Le Châtelier's Principle This principle is used to predict the qualitative effects of changes on the position of equilibrium When a system at equilibrium is disturbed, the equilibrium position will shift in the direction which tends to minimize, or counteract, the effect of the disturbance

More Possible changes in conditions that we need to consider are: Changes in the concentrations of either the reactants or the products Changes in pressure for gas phase reactions Changes in temperature The presence of a catalyst

Changes in Concentration Increasing the concentration of a reactant will move the position of equilibrium to the right This favors the forward reaction and will increase the equilibrium concentrations of the products

More The addition of more product to an equilibrium mixture would shift the position of equilibrium to the left This would favor the reverse reaction When new equilibrium concentrations are substituted into the equilibrium expression, the value of Kc remains unchanged

Example Fe(H2O)63+(aq) + SCN-(aq) [Fe(H2O)5SCN]2+ + H2O(l) yellow-brown colorless blood-red If the concentration of either the thiocyanate ion, or the iron (III) ion is increased, then the intensity of the color increases The shift in equilibrium to the right causes the concentration of the added reactant to fall again

Continued If the concentration of iron(III) ions is decreased by adding fluoride ions, which form a stable complex with the iron (III) ions, then the intensity of the coloration decreases The shift of the equilibrium to the left produces more aqueous iron (III) ions to counteract the reduction caused by the F-

Changes in Pressure Changing the pressure only affects reactions that involve gases Gas phase reactions will only be affected by a change of pressure if the reaction involves a change in the number of moles on the two sides of the equation

Increase in Pressure 2SO2(g) + O2(g) 2SO3(g) Increase the pressure and the equilibrium shifts to the side with the least moles of gas 2SO2(g) + O2(g) 2SO3(g) Increase P: equilibrium → , to side with fewer moles of gas C(s) + H2O(g) CO(g) + H2(g) Increase P: equilibrium ←, to side with fewer moles of gas (

Decrease in Pressure 2SO2(g) + O2(g) 2SO3(g) Decrease pressure and the equilibrium shifts towards the side with the most moles of gas 2SO2(g) + O2(g) 2SO3(g) Decrease P: equilibrium ←, to side with more moles of gas C(s) + H2O(g) CO(g) + H2(g) Decrease P: equilibrium →, to side with more moles of gas

More About Pressure H2(g) + I2(g) 2HI(g) Changing pressure has no effect as 2 moles of gas are converted to 2 moles of gas The changes in concentrations that result from the changes in pressure are such that the value of Kc remains unchanged

Temperature Change If the temperature of a system is increased, then the equilibrium shifts in the direction of the endothermic change, this will tend to lower the temperature N2(g) + O2(g) 2NO(g) ∆H = +180 kJ mol-1 Increased T: Kc increases, equilibrium → Decreased T: Kc decreases, equilibrium ←

Temperature Change If the temperature of a system is decreased the equilibrium shifts in the direction of the exothermic change, this is to generate heat and raise the temperature 2SO2(g) + O2(g) 2SO3(g) ∆H = -197 kJ mol-1 Decreased T, Kc increases, equilibrium → Increased T, Kc decreases, equilibrium ←

More About Temperature Changes in temperature affect the rate constants of the forward and reverse reactions to different extents This means that the actual value of Kc changes also

Presence of a Catalyst The presence of a catalyst reduces the activation energy of both the forward and reverse reactions by the same amount Both reactions are sped up by the same factor Equilibrium is established more rapidly, but the position of equilibrium and the value of Kc are not affected

SUMMARY

Industrial Processes Many industrial processes involve equilibria The goal is to produce the desired product efficiently and rapidly, but with the minimum amount of waste and the minimum input of energy Kinetics and equilibria have to be considered

Haber Process Produces ammonia from nitrogen and hydrogen gases A heated iron catalyst is used N2(g)+ 3H2(g) 2NH3(g) ∆H = -92 kJ mol-1 4 moles of gas are converted to 2 moles of gas A high pressure will favor the product

Continued The forward reaction is exothermic, so a low temperature would favor the products Issues with the process: Low temperature causes a low reaction rate High pressure is expensive to provide A compromise is chosen to maximize product

Compromise Typical conditions chosen for the Haber process are pressures in the range of 200-1000 atm and temperatures about 700K

Collecting Ammonia The reaction is not allowed to reach equilibrium because the reaction rate slows as equilibrium is approached Usually only about 20% of the reactants are converted to ammonia The ammonia is condensed and the reactants are recycled to form more ammonia

Uses for Ammonia The manufacture of fertilizers The manufacture of nitrogen-containing polymers such as nylon Can be oxidized to produce nitric acid Nitric acid is used to produce explosives such as TNT and dynamite Nitric acid is also used in making dyes

Contact Process The Contact process is the production of sulfuric acid by the oxidation of sulfur First, pure sulfur is burned in air S(s) + O2(g) SO2(g) Then sulfur dioxide is mixed with air and passed over a vanadium (V) oxide catalyst 2SO2(g) + O2(g) 2SO3(g) ∆H = -196 kJ mol-1

Continued High pressure would favor the formation of sulfur trioxide, but low pressure actually does a good job Low temperature favors the product, but not too low (slows the reaction) Compromise: 700-800K, the use of a catalyst (finely divided V2O5) and about 2 atm

Continued The result is over 90% conversion to sulfur trioxide The sulfur trioxide must be reacted with water to produce sulfuric acid SO3(g) + H2O(l) → H2SO4(l)

Uses of Sulfuric Acid Manufacture of: Fertilizers Polymers Detergents Paints and pigments Used in the petrochemical industry and in the processing of metals

Another use of Sulfuric Acid Used as an electrolyte in automobile batteries (battery acid)