Chapter 15 Lecture Notes PDF

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Chapter 15: Chemical Equilibrium Focus of this section:

Factors that affect reaction Solving equilibrium problems

Equilibrium • Reversible reaction: a reaction that proceeds simultaneously in both directions • most reactions are reversible

! !→ 2 NH (g ) N2 (g ) + 3 H 2 (g ) ← !! 3 H2O(l)

H2O(g)

(chem rxns)

(physical changes)

• At equilibrium: = forward and reverse reaction rates are equal

! !→ 2 NO (g ) N2O 4 (g ) ← !! 2 N2O4(g) ฀฀→ 2 NO2(g) 2 NO2(g) ฀฀→ N2O4(g)

equal rates for elementary steps!

• Chemical equilibria are dynamic, not static  the reactions do not stop EXAMPLE:

! !→ 2 NO (g ) N2O 4 (g ) ← !! 2

-- N2O4 is colorless gas NO2 is brown -- look at different experiments, each starting with different ratio of N2O4 and NO2

• Experiment 1: start with pure N2O4

!!! !→ 2 NO ( g ) N2O 4 (g ) ← 2

• Experiment 2: start with pure NO2

!!! !→ 2 NO ( g ) N2O 4 (g ) ← 2

• can write rate law for forward and reverse reactions: rateforward = kf [N2O4]

ratereverse = kr [NO2]2 Experiment 2

Experiment 1

!!! !→ 2 NO ( g ) N2O 4 (g ) ← 2 Experiment 1: initial: [N2O4] high, rateforward high [NO2] zero, ratereverse zero

as rxn proceeds: [N2O4] decreases, rateforward decreases [NO2] increases, ratereverse increases at equilibrium: [N2O4] constant, rateforward constant [NO2] constant, ratereverse constant rateforward = ratereverse !!! !→ 2 NO ( g ) N2O 4 (g ) ← 2 Experiment 2: initial: [NO2] high, ratereverse high [N2O4] zero, rateforward zero

as rxn proceeds: [NO2] decreases, ratereverse decreases [N2O4] increases, rateforward increases at equilibrium: [NO2] constant, ratereverse constant [N2O4]constant, rateforward constant rateforward = ratereverse

• equilibrium can be established with any mixture of reactants and products Equilibrium Constant:

! !→ 2 NO ( g ) N2O 4 (g ) ← !! 2

rate forward = ratereverse note: at equilibrium, [reactants] NOT equal to [products] and [reactants] and [products] not equal to zero

k f [N 2O 4 ] eq = k r[NO 2]2eq 2 kf [NO2 ]eq = = Kc k r [N 2 O 4 ]eq

Equilibrium expression

-- Kc = Equilibrium Constant “c” stands for concentration -- writing equilibrium expression: = products over reactants = exponents are coefficients from balanced chemical reaction

• for ANY BALANCED reaction, can write a reaction quotient (Qc) (also called mass action expression) = fraction with product concentrations in the numerator and reactant concentrations in the denominator -- each concentration raised to a power equal to the corresponding stoichiometric coefficient in the balanced chemical equation = looks identical to Kc!! (because SAME EXPRESSION) • For the general reaction:

-- when concentrations are non-equilibrium concentrations, it’s a Q -- when concentrations are equilibrium concentrations, it’s a K

-- value of Q changes as a reaction progresses:

• Important points about Kc -- Reaction coefficients become exponents. -- Equilibrium constants are temperature dependent. -- Equilibrium constants do not have units. -- magnitude of K can be used to predict where the equilibrium lies: = If Kc >> 1, products are favored (reaction goes nearly to completion). = If Kc K ⇒ reverse reaction favored Q = K ⇒ equilibrium present Q < K ⇒ forward reaction favored

--------------------------------------------At 448 °C, Kp = 51 for the reaction:

!!! !→ 2 HI( g) H 2 ( g ) + I2 ( g ) ← Predict the direction the reaction will proceed, if at 448 °C the pressures of HI, H2, and I2 are 1.3, 2.1 and 1.7 atm, respectively.

--------------------------------------------Equilibrium Calculations: • reactions occurring in solution  use Kc • reactions occurring in gas phase  use Kp or Kc • in general, problems come in 2 flavors: -- calculate value for Kc given equilibrium concentrations -- given Kc, calculate equilibrium concentrations • things to keep in mind when doing equilibrium problems: -- only equilibrium concentrations go into Kc expression -- use molarity for concentrations -- be sure your chemical reaction is BALANCED!

--------------------------------------------For the following reaction: H2(g) + I2(g)

HI(g)

Calculate the value of Kc given the following information: [H2]initial = 0.100 M [I2]initial = 0.100 M At equilibrium: [I2] = 0.0200 M

--------------------------------------------At 1130 °C, KP = 2.59 × 10–2 for:

! !→ 2 H (g ) + S (g ) 2 H2S(g ) ← !! 2 2 At equilibrium, PH2S = 0.557 atm and PH2 = 0.173 atm, calculate PS2 at 1130 °C.

---------------------------------------------

• Method for calculating equilibrium concentrations from Kc and initial concentrations: 1. Construct an equilibrium table (ICE table), and fill in the initial concentrations (including any that are zero) 2. Use initial concentrations to calculate the reaction quotient, Q, and compare Q to K to determine the direction in which the reaction will proceed. 3. Define x as the amount of a particular species consumed, and use the stoichiometry of the reaction to define the amount of other species consumed or produced in terms of x. 4. For each species in the equilibrium, add the change in concentration to the initial concentration to get the equilibrium concentration. 5. Use the equilibrium concentrations and the equilibrium expression to solve for x. 6. Using the calculated value of x, determine the concentrations of all species at equilibrium. 7. Check your work by plugging the calculated equilibrium concentrations into the equilibrium expression. The result should be very close to the Kc stated in the problem.

--------------------------------------------KP = 82.2 at 25 °C for the following reaction:

! !→ 2 ICl(g ) I2 (g ) + Cl2 (g ) ← !! Initially, PI2 = PCl2 = 2.00 atm and PICl = 0.00 atm. What are the equilibrium pressures of I2, Cl2, and ICl?

--------------------------------------------perfect square

--------------------------------------------KP = 82.2 at 25 °C for the following reaction:

! !→ 2 ICl(g ) I2 (g ) + Cl2 (g ) ← !! Initially, PI2 = PCl2 = 2.00 atm and PICl = 0.00 atm. What are the equilibrium pressures of I2, Cl2, and ICl?

quadratic equation:

x =

a x2 + b x + c = 0

−b ±

b 2 − 4ac 2a

--------------------------------------------quadratic equation

--------------------------------------------At 2200 °C, Kp = 0.050 for the reaction: N2(g) + O2(g)

2 NO(g)

What is the partial pressure of NO at equilibrium if we begin with initial pressures of N2 = 0.80 atm and O2 = 0.20 atm?

---------------------------------------------

--------------------------------------------The value of Kc = 7.3 x 10–18 at T = 1000 °C for the following reaction: 2 H2O(g) 2 H2(g) + O2(g) What is the [H2]equil if the initial concentration of H2O = 0.100 M?

--------------------------------------------• for approximation to be valid: value of x must be < 1000 times that of the value it’s being compared to

--------------------------------------------A pure NO2 sample reacts at 1000 K by the following rxn:

!!! !→ 2 NO(g ) + O (g ) 2 NO2 (g ) ← 2 KP is 158. If at 1000 K the equilibrium partial pressure of O2 is 0.25 atm, what are the equilibrium partial pressures of NO and NO2.

---------------------------------------------

Factors That Affect Chemical Equilibrium: • can make qualitative predictions about what happens to an equilibrium mixture when a change is made • Le Châtelier’s principle = When a stress is applied to a system at equilibrium, the system will respond by shifting in the direction that minimizes the effect of the stress -- Stress refers to a disturbance of the system at equilibrium • Types of disturbances: -- addition of a reactant or product -- removal of a reactant or product -- change in volume of the system, resulting in a change in concentration or partial pressure of the reactants and products -- change in temperature • Shifting refers to the favoring of either the forward or reverse reaction such that the stress is partially offset as the system re-establishes equilibrium -- shift to the right  more products are formed by the forward reaction -- shift to the left  more reactants are formed by the reverse reaction

1. change [reactant] or [product]: -- changes value of Q  no longer equal to K -- reaction will move forward if you add reactant = QK = by moving backward, consume added product Cu(H2O)42+(aq) + 4 Cl–(aq) (blue)

CuCl42–(aq) + 4 H2O(l) (yellow)

-- add Cl–  more CuCl42– forms -- add Ag+  form AgCl(s)  shift to left 2. Changing volume in gaseous reaction: -- changes molar concentration; also changes pressure 3 H2 (g) + N2 (g)

2 NH3 (g)

-- decrease volume  pressure increases = can oppose this by decreasing total #moles of gas (shift to right) H2 (g) + I2 (g)

2 HI (g)

-- won’t respond to changes in volume (pressure) • moderate pressure changes have no effect on reactions that involve only solids or liquids

3. changes to temperature: 3 H2(g) + N2(g)

2 NH3(g) + heat

-- raise temperature  shift to left (shifts reaction in direction that produces an endothermic change) -- when temp changes, [ ]’s change even though volume is constant = new equilibrium position obtained = new value of K Kc =

[NH3]2 [H2] 3 [N2]

= as temp increases, shift to left, so [NH3] decreases and [H2] and [N2] increases 4. effect of catalyst: -- no effect on equilibrium  affects BOTH the forward and reverse reactions equally -- lowers Ea of both reactions to same extent so reach equilibrium FASTER -- presence of a catalyst, therefore, does not alter the equilibrium constant, nor does it shift the position of an equilibrium system 5. addition of inert gas at constant volume: -- if doesn’t react, there will be no change in [reactants] and [products] -- no change in equilibrium position

--------------------------------------------Hydrogen (used in ammonia production) is produced by the endothermic reaction, Ni ! !→ + CH4 ( g ) H2O( g ) ←!! CO( g ) + 3H 2( g ) 750 °C

Assuming the reaction is initially at equilibrium, indicate the direction of the shift (left, right or none) if: • H2O(g) is removed. • The temperature is increased. • The quantity of Ni catalyst is increased. • An inert gas (e.g. He) is added. • H2(g) is removed. • The volume of the container is tripled. ---------------------------------------------...