Equilibrium All reactions are reversible In labs many
Equilibrium • All reactions are reversible • In labs, many products of chem. reactions can be directly reacted to form the reactants • Sometimes the conditions for the reaction to occur are similar for the forward and reverse reactions, other times they are different – Ex. Fe 3 O 4 + 4 H 2 3 Fe + 4 H 2 O(l) 3 Fe + 4 H 2 O(g) Fe 3 O 4 + 4 H 2 If steam is not allowed to escape
• Both reactions are taking place simultaneously at different rates • Eventually the two rates are equal • Chemical Equilibrium • Write one equation with a double arrow
Characteristics • State of balance between opposing changes can be physical • Vapor-liquid • Changes can be chemical • The original changes before an equilibrium exists are observable • Changes in • concentration • Temperature • State
• Definite relationship between concentrations of reactants and products when reversible reaction is in Equilibrium • Example: H 2 (g) + I 2(g) D 2 HI(g) at 4900 C [H 2]= 1 mol/dm 3 [I 2]= 1 mol/dm 3 When H and I come into contact, reaction starts Equil. is achieved when reaction rate = product rate
• The relationship between Reactants and Products is expressed as Keq and can be expressed as: • Kc(eq) = [Products] [Reactants] • Coefficients in the balanced equation are used as superscripts in the equation • The Equilibrium Constant (Kc) will not change in a closed system • It is affected by changes in temperature
• What does the Kc value determine? • Size of value determines the extent a reaction proceeds to the right hand side of a reversible reaction before an equilibrium is achieved • A constant value <1 indicates little product forming • Constant >1 shows product formation favored • The higher the value, the greater the amt. of product
• Reactions are sensitive to changes in conc. , temp. and press. • A change in one of these places stress on the system • Stress causes the equil. to Shift • A new equil. is eventually achieved with the new factors • Le. Chatlier’s Principle
• Le. Chatelier’s Principle – System in equilibrium subjected to stress( change in con. , temp. or press. ) the equilibrium will shift in the direction that tends to counteract the effect of the stress
Concentration changes • • • A+BD C+D Add A more AB collisions Forward reaction faster Equil. No longer exists B consumed and its conc. Decreases the reaction slows • C + D conc. Increases and its collisions increase and its reaction begins to speed up • A new equilibrium is established
Temperature changes • endo N 2(g) + 3 H 2(g) D 2 NH 3(g) + 92 k. J exo Haber Process • Every equilibrium has an endo and an exo component htemp. stress countered by favoring the endothermic reaction since it absorbs heat $temp. stress countered by favoring the exothermic reaction as heat is given off
Pressure changes • Applies mostly to reactions with gases endo N 2(g) + 3 H 2(g) D 2 NH 3(g) + 92 k. J exo Haber Process • Increase pressure- favors side with fewer moles ( # of particles) of substance • Decrease pressure favors side with more moles of substance
• Calculate an equilibrium constant K • Don’t always know conc. of all substances • Knowing conc. Of one, can stoichiometrically determine others • 1. Tabulate known initial and equil. conc. for all species in the equilibrium • 2. For species with known conc. of initial and equilibrium conc. , calculate change in conc. as reaction approaches equilibrium • 3. Use stoich to determine conc. of all others • 4. From init. and changes in conc. Calculate the equilibrium concs. Evaluate Kc
• Example: a mixture of 5. 000 x 10 -3 mol H 2 and 1. 000 x 10 -2 mol I 2 is placed in a 5. 000 L container at 448 o. C and allowed to come to equilibrium. Analysis of the equil. mixture shows the conc. of HI is 1. 87 x 10 -3 M. Calculate Kc at 448 o. C for the reaction. • First: Calculate concentrations of all reactants and products
• [H 2]init = 5. 000 x 10 -3 mol = 1. 000 x 10 -3 M 5. 000 L • [I 2]init = 1. 000 x 10 -2 mol = 2. 00 x 10 -3 M 5. 000 L • • Place above values into a ICE table I initial conc. C change in conc. E equilibrium conc.
• H 2(g) + Initial 1. 000 X 10 -3 M Change Equilibrium I 2(g) D 2. 00 x 10 -3 M 2 HI(g) 0 M 1. 87 x 10 -3 M
• Second: calculate change conc. of HI using initial and equil. values • Change is +1. 87 x 10 -3 M • Third: use stoichiometry to calculate changes in other substances • (1. 87 x 10 -3 mol HI)(1 mol H 2) = L 2 mol HI • 0. 935 x 10 -3 mol H 2/L • The value for I 2 is the same as H 2
• H 2(g) + I 2(g) D 2 HI(g) Initial 1. 000 x 10 -3 M 2. 000 x 10 -3 M 0 M Change -0. 935 x 10 -3 M +1. 87 x 10 -3 M Equilibrium 1. 87 x 10 -3 M
• Fourth: calculate equil. conc. using initial conc. and changes • Equilibrium conc. H 2 is init. minus consumed • [H 2] = 1. 000 x 10 -3 M – 0. 935 x 10 -3 M = 0. 065 x 10 -3 M • [I 2] = 2. 00 x 10 -3 M – 0. 935 x 10 -3 M = 1. 065 x 10 -3 M
• H 2(g) + I 2(g) D 2 HI(g) Initial 1. 000 x 10 -3 M 2. 000 x 10 -3 M 0 M Change -0. 935 x 10 -3 M +1. 87 x 10 -3 M Equilibrium 0. 065 x 10 -3 M 1. 87 x 10 -3 M • Finally: now know equilibrium concentrations of reactants and products • Use equilibrium expression to calculate Kc • Kc = [HI]2 = (1. 87 x 10 -3)2 [H 2][I 2] (0. 065 x 10 -3)(1. 065 x 10 -3) = 51
• For solutions that have solutes dissolved in water, the same principle can be applied • Ksp is the solubility constant • Ksp = [Products] • Example: • Mg. Cl 2 Mg+2(aq) + 2 Cl-1(aq) • Ksp = [Mg+2][Cl-1]2
K of acids and bases • • Ka Kb Ka = [H+][cation] = [H+]2 Kb = [anion][OH-] = [OH-]2 Strong acids and bases dissociate completely into H+ or OH- and use above equations • Weak acids and bases do not and are similar to equilibrium equations • Ka or Kb = [products] [reactants] use ICE chart
- Slides: 22