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## Chemical Equilibria

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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

- aA + bB cC + dD
- Thus,

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 actionICE 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

- 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 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 Keq: aqueous solns

- Same rule for pure liquids
- Ex: NH3(aq) + H2O(l) NH4+(aq) + OH-(aq)
- Keq = ?

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

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

- 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

- 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

- 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 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

- 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

- 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…

- 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 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 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 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

- 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

- 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

- 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

- What change in equilibrium will be seen when one adds a solid and why?

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

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