CHEM 163 Chapter 17 Spring 2009 Instructor: Alissa Agnello 1

What affects reaction rates?

Chemical Equilibrium Many reactions can go forward AND backwards 3 If the opposite reaction can occur, then this reaction is a reversible reaction

What is equilibrium? …in terms of reaction rates? …in terms of reactant and product concentrations? Has the forward reaction stopped? Has the reverse reaction stopped?

Chemical Equilibrium Reactions continue at equal (but opposite) rates No further changes in concentrations of reactants or products occurs 5

Chemical Equilibrium If fwd and rev reactions are both elementary steps, how would we write their rate laws? 6 rate fwd =k fwd [N 2 O 4 ] eq rate rev =k rev [NO 2 ] 2 eq k fwd [N 2 O 4 ] eq k rev [NO 2 ] 2 eq =

Equilibrium Constants (K) 7

Equilibrium Constants: Small K Small K value: – Greater concentration of reactants or products? – Reaction favors reactants 8

Equilibrium Constants: Large K Large K value: – Greater concentration of reactants or products? – Reaction favors products 9

Equilibrium Constants Will each of the following favor reactants or products? Reaction with K c = 2.9 x Reaction with K c = x

Reaction Quotient (Q) K derived from rates Q derived from concentrations At a given temperature, a system will always return to the same [product] : [reactant] ratio Q What if K = Q?

Reaction Quotients 12 reactantsproducts coefficients

2-minute practice Write a reaction quotient expression for the following: Compare your answer with your neighbors!

Q for Overall reaction Many reactions take place in multiple steps. Add together the steps to get overall reaction Multiply K for each step to get overall K Multiply Q for each step to get overall Q

Other situations… In a reversible reaction: When the coefficients are multiplied by a common factor (n):

If solids or liquids are present… heterogeneous equilibrium Q and K only related to concentration that change (gases)

Rearrange the ideal gas law, so that concentration is isolated on one side. What is the relationship between pressure and concentration? 2-minute exercise

Q p : using partial pressures concentration pressure

Shortcut for relating Q p to Q c What is the ∆n for the reaction? n reactants = 2 n products = 2 ∆ n = 0 Q p = Q c If ∆ n ≠ 0 Q p = Q c (RT) ∆n K p = K c K p = K c (RT) ∆n

Is a reaction at equilibrium? Compare Q and K! Q < K: – Too much reactant – Equilibrium “shifts” towards products Q > K: – Too much product – Equilibrium “shifts” towards reactants Q = K: – At equilibrium

Equilibrium Calculations If equilibrium concentrations and/or equilibrium constant is known: – Write Q expression, plug in concentrations, solve – At equilibrium Q = K Solving for K when concentrations not given: – Write Q expression – Set up table including: Initial concentrations (or pressures) Change Equilibrium concentrations (or pressures)

Mix together graphite and carbon dioxide (P = atm) to create carbon monoxide. Pressure (atm) CO 2 (g) + C (graphite) ↔ 2CO (g) Once equilibrium is reached, the pressure in the vessel (from CO 2 and CO) is atm. Initial Change Equilibrium x+ 2x x + 2x – x + 2x =0.757P total =

Solving for “x” To solve for x, you may need to use the quadratic formula. Set up your equation: a*x 2 + b*x + c = 0 You will end up with two values. Which value is right? Remember: K can’t be negative We can’t have a higher concentration of reactant than our initial concentration.

Steps for Solving 1.Write balanced equation 2.Set up table 3.Solve for x by… 1.Setting up Q expression 2.Setting total pressures equal to final pressure 4.Solve for equilibrium concentrations or pressures (using x) 5.To check: plug in solved concentration or pressures into Q expression and compare to known K value

3-minute Practice Consider the reaction: 0.45 mole H 2 S is placed in a 3.0 L container. Make a table for this situation. K c = 9.30 x at 700 °C H 2 S (g) ↔ H 2 (g) + S 2 (g) 22 Calculate the equilibrium concentration of H 2 (g) at 700 °C

Le Châtelier’s Principle At equilibrium, concentrations of substances do not change. If a stress is put on the reaction at equilibrium, the equilibrium will shift to relieve the stress. What changes count as stress? Concentration – Adding or removing reactant or product Volume (Pressure) Temperature 26

Effect of Concentration Changes 2NO 2 (g) NO (g) + O 2 (g) 27 What is the effect on the concentration of each substance? Add NO 2 ? Add NO? Add O 2 ? Remove NO 2 ? Remove NO? Remove O 2 ? 2

Calculations: adding/removing substances Make a table! Pressure (atm) CO 2 (g) + C (graphite) ↔ 2CO (g) Original Equil. Disturbance New Initial Change New Equil x - 2x x x Predict direction of shift

Effect of Volume Changes Each mole of gas exerts a certain pressure Decrease the volume… – Increase the pressure… – Shifts to side of reaction with FEWER moles of gas 29 2NO 2 (g) 2NO (g) + O 2 (g) What happens if we increase the volume? Shifts to side with MORE moles What happens if we increase the pressure? Shifts to side with FEWER moles

Effect of Volume Changes 30 With volume changes and concentration changes, Equilibrium shifts to relieve new stress …in order to return to equilibrium

Effect of Temperature Changes If T is increased, equilibrium shifts to remove the heat If T is decreased, equilibrium shifts to create heat K c changes! – Unlike for concentration and volume changes 31 A (g) B (g) + C (g) Endothermic: Heat + Increase the temperature? Shifts to the products (to use up heat) K c increases

Effect of T on K van’t Hoff Equation: Exothermic ( ∆ H° < 0): increasing T decreases K c Endothermic ( ∆ H° > 0): increasing T increases K c R = J/mol ∙ K

3-minute Practice Consider the exothermic reaction between nitrogen gas and hydrogen gas, creating ammonia gas (NH 3 ). Write a balanced equation for this reaction. Which direction will the equilibrium shift if: T is increased? Ammonia is removed? Volume of the container is decreased? A catalyst is used? Hydrogen is added? 33

Chapter 17 Homework Due: Tuesday, 4/14 #18, 23, 29, 34, 45, 55, 63, 76, 77, 91, 114