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Chemical Equilibria. Chapter 16. Chemical reactions. Can reverse (most of the time) Though might require a good deal of energy  or double arrows = reversible rxn It “swings both ways” Once fwd/rev rxns occur at equal rates = equilibrium (no net change seen). Equilibrium constant.

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chemical reactions
Chemical reactions
  • Can reverse (most of the time)
  • Though might require a good deal of energy
  •  or double arrows = reversible rxn
  • It “swings both ways”
  • Once fwd/rev rxns occur at equal rates
  • = equilibrium (no net change seen)
equilibrium constant
Equilibrium constant
  • aA + bB  cC + dD
  • Thus,
equilibrium constant in action
Give the equilibrium constant for:

H2(g) + I2(s) 2HI(g)

Initially, 0.0175 M of reactants

Decrease to 0.0037 M reactants

What is the concentration of HI formed?

Equilibrium constant in action
ice table
ICE Table
  • Initial Change Equilibrium Table
  • Great way to explain and show concentration changes
  • Let’s make an ICE table for the previous reaction on the board
problems
Problems
  • Express Keq for:

CH3OH(g) CO(g) + 2H2(g)

  • Express Keq for:

C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)

  • Solve Keq for the following:

A(g) 2B(g)

    • Given [A]i = 1.00M, [B]i = 0.00M, and

[A]eq =0.75M

caveats to k eq solids
Caveats to Keq: solids
  • Solids in reversible rxns are excluded from expression since concentration derived from constant densities:
  • S(s) + O2(g) SO2(g)
  • Keq = ?
caveats to k eq aqueous solns
Caveats to Keq: aqueous solns
  • Same rule for pure liquids
  • Ex: NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)
  • Keq = ?
caveats to k eq gases k p
Caveats to Keq: gases & Kp
  • For gases we can use partial pressures
    • Why?
      • Hint: Relationship between pressure and concentration
        • See ideal gas law
  • Given:

aA + bB  cC + dD

  • Kp not necessarily equal to Keq
problem
Problem
  • Given

2NO(g) + O2(g) 2NO2(g)

  • Kp = 2.2 x 1012 @ 25°C
  • Find Keq
determining equilibrium constant
Determining equilibrium constant
  • Given 2SO2(g) + O2(g) 2SO3(g)
  • And: [SO2]i = 1.00 M, [O2]i = 1.00 M & [SO3]f = 0.925 M
  • What is Keq?
more problem solving
More problem solving
  • Sulfuryl chloride (SO2Cl2) dissociates into sulfur dioxide and chlorine in the gas phase:

SO2Cl2(g) SO2(g) + Cl2(g)

  • In an experiment, 3.174 g of SO2Cl2 (MW = 134.96g/mol) is placed in a 1.000 L flask and is at 100.0°C. At equilibrium, the total pressure in the flask is 1.30 atm.

Calculate:

  • a) The partial pressures of each gas at equilibrium.
  • b) The Kp at 100.0°C for the reaction.
yet another
Yet another
  • Isopropanol can dissociate into acetone and hydrogen:

(CH3)2CHOH(g) (CH3)2CO(g) + H2(g)

  • At 179°C, the equilibrium constant is 0.444. Calculate the equilibrium partial pressures of all three gases if 10.00 g (MW = 60.10g/mol) of isopropanol are initially placed in a 10.00 L vessel.
another but with a twist
Another but with a twist
  • At 25°C, the equilibrium constant for the reaction below is 5.9 x 10-13.

2NO2(g) 2NO(g) + O2(g)

  • Suppose a container is filled with 0.89 atm of NO2
  • Calculate the equilibrium partial pressures of each gas
  • OK to approximate if “x  [A]0 < 5%“
    • “5% rule”
more on k eq
More on Keq
  • Is the rxn product or reactant favored?
    • I.e., Will it form more product or reactant?
  • If K  1, then prod concentration higher than reactant concentration  prod favored
    • Makes mathematical sense
      • See right
  • If K < 1, then reactant concentration higher than prod concentration  reactant favored
  • If K = 1, neither favored; both equal concentration
reaction quotient q
Reaction quotient, Q
  • When rxn not at equilibrium, use Q
  • Where

aA + bB  cC + dD

    • Then 
  • Remember, Q used for system when system NOT in equilibrium
q its benefits
Q – its benefits
  • Answers the question:

Is the system at equilibrium (i.e., does Q = K)?

  • If not, we can predict which way the reaction will continue to proceed
  • If Q < K, rxn still needs to go to prod side to achieve equilibrium (i.e., where Q = K)
    • In other words, insufficient product formed for equilibrium conditions
  • If Q > K, rxn has “overshot” K and needs to go to reactant side to achieve equilibrium (i.e., where Q = K)
    • In other words, an excess of product is formed for equilibrium conditions
work on this
Work on this…
  • N2(g) + 3H2(g)  2NH3(g)
  • The equilibrium constant at 400°C is Keq=0.5.
  • Suppose we make a mixture with the following concentrations:
  • [NH3] = 5.0M, [N2] = 3.5M, [H2] = 1.9M
  • In which direction will the reaction go?
  • a) products b) reactants
manipulating k eq
Manipulating Keq
  • C(s) + ½ O2(g) CO(g)
  • Kp’ = ?
  • Changing it: 2C(s) + O2(g) 2CO(g)
  • Kp” = ?
  • How are Kp’ and Kp” related mathematically?
more on k eq1
More on Keq
  • Given 2NO2(g) N2O4(g)
  • Keq = ?
  • What is Keq when rxn is reversed?
  • Therefore, what can we say about Keq fwd and Keq rev?
even more on k eq
Even more on Keq
  • Remember Hess’s Law?
  • 1) A+B  C
  • 2) B+C  D
  • What is the net rxn and it’s Knet using Hess’s Law?
  • Can one obtain the same values as above not using Hess’s Law?
  • What can we say about Knet using each rxn’s Keq?
disturbing chemical equilibria
Disturbing chemical equilibria
  • Le Châtelier’s Principle
  • Change one component of the rxn & the rxn will attempt to rectify it
  • Think of it this way:
  • If something is changed, how can it be undone or controlled so that equilibrium is achieved once again?
temperature variation on equilibrium
Temperature variation on equilibrium
  • 2NO2(g) N2O4(g) + heat
  • Kp = ?
  • H° = -57.1 kJ
  • @ 273 K, Kp = 1300, and @ 298 K, Kp = 170
  • Hence, if one raises the temp to 298 K, which way will it swing?
pressure volume change on equilibrium
Pressure & volume change on equilibrium
  • If volume decreased (pressure increased)  favors smaller # of molecules
  • If volume increased (pressure decreased)  favors larger # of molecules
  • If reversible rxn has = # of molecules on each side, a volume/pressure change will do nothing
question
Question
  • What change in equilibrium will be seen when one adds a solid and why?
how about this
How about this?
  • 2H2S(g) + O2(g) 2S(s) + 2H2O(g); ΔH = -221.19 kJ/mol
  • If O2(g) is added to the reaction vessel, what happens to the amount of S(s)?
    • a) It increases b) It decreases c) Nothing
  • If the volume of the vessel is cut in half, what happens to the ratio of PH2O/PH2S?
    • a) It increases b) It decreases c) Nothing
  • If the temperature is increased, what happens to the equilibrium constant K?
    • a) It increases b) It decreases c) Nothing
  • If S(s) is added to the reaction, what happens to PH2O?
    • a) It increases b) It decreases c) Nothing