Disturbing Equilibrium and Non-equilibrium conditions

Le Châtelier’s Principle, the van’t Hoff equation, the Reaction Quotient (Q)

Lesson ObjectivesKnow -Le Châtelier’s principle outlines how equilibrium is affected by

changes in a system-Equilibrium can be affected by changes in reactant/product amounts, changes in volume or pressure, or changes in temperature-temperature is the only factor that affects the equilibrium constant

Understand -when disturbed from equilibrium, a reaction will shift whichever way (to products or to reactants) to relieve “stress” in the system-the reaction quotient is used to determine which way a reaction needs to shift to reach equilibrium

Do -analyze changing conditions to determine which way a reaction needs to shift to regain equilibrium-solve the van’t Hoff equation for a missing variable, whether the variable is a K value, DH, or a temperature.-solve the Q (reaction quotient) equation for non-equilibrium conditions-use Q to determine which way a reaction will shift to regain equilibrium

LE CHÂTELIER’S PRINCIPLEDisturbing Equilibrium

Le Chatelier’s Principle If a system at equilibrium is disturbed, it will s

hift to partially counteract the change

Factors that can disturb equilibrium: Adding or removing reactant or product

(physically or chemically) Increasing or decreasing volume (which results in

a pressure change for gases) Increasing or decreasing temperature (adding or

removing heat)

Adding or removing reactant or product Add reactant or product: reaction shifts to use up

the added substance (stress) Remove reactant or product: reaction shifts to add

it back in Substances can be removed by reacting them out –

making precipitate, water, gas, etc. Only affects equilibrium if the added or removed

substance is: In the equation (inert or irrelevant substances have no

affect on equilibrium) Not a solid or liquid (solids and liquids don’t affect

equilibrium because their concentrations don’t change)

Iodine chloride decomposes at high temperatures to iodine and chlorine:

2 ICl(g) I2(g) + Cl2(g)

If some iodine condenses, which direction would this reaction shift from equilibrium?

Example 1: changing amounts

Example 2: changing amountsSulfur dioxide and oxygen can combine to make sulfur trioxide

2 SO2(g) + O2(g) 2SO3(g)

If more oxygen is introduced into this system while it is at equilibrium, which way will the reaction shift?

2H2(g) + CO(g) CH3OH(g) H = -90.2 kJ

If a mixture of H2, CO and CH3OH is at equilibrium, will the concentration of CO increase, decrease, or remain the same as equilibrium is reestablished after more H2 is added.

A. [CO] will increaseB. [CO] will decreaseC. [CO] will stay the same

2H2(g) + CO(g) CH3OH(g) H = -90.2 kJ

If a mixture of H2, CO and CH3OH is at equilibrium, will the concentration of CO increase, decrease, or remain the same as equilibrium is reestablished after CH3OH is added.

A. [CO] will increaseB. [CO] will decreaseC. [CO] will stay the same

Changing the volume

Increase volume (decrease pressure of gases) The system will shift to side with more moles of

gas to try to raise pressure Decrease volume (increase pressure of gases)

The system will shift to the side with fewer moles of gas to try to lower pressure

Example 3: changing volumeSulfur dioxide and oxygen can combine to make sulfur trioxide

2 SO2(g) + O2(g) 2SO3(g)

If the container holding this reaction is expanded (volume increases), which way will the reaction shift from equilibrium?

2H2(g) + CO(g) CH3OH(g) H = -90.2 kJ

If a mixture of H2, CO and CH3OH is at equilibrium, will the concentration of CO increase, decrease, or remain the same as equilibrium is reestablished after the volume of the system is decreased.

A. [CO] will increaseB. [CO] will decreaseC. [CO] will stay the same

Changing the temperature Increasing temp. is like adding (stressing) heat

System will shift in the endothermic direction Decreasing temp. is like removing heat

System will shift in the exothermic direction

If heat was shown in the equation, the system shifts to counteract the addition or removal of heat (just like reactants or products) Endothermic: ΔH = positive, written in reactants Exothermic: ΔH = negative, written in products

Example 4: changing temperaturesSulfur dioxide and oxygen can combine to make sulfur trioxide

2 SO2(g) + O2(g) 2SO3(g)

This reaction is highly exothermic, releasing nearly 200 kJ of energy per mole of product created. If the temperature of this reaction

was increased, which way would the reaction shift from equilibrium?

2H2(g) + CO(g) CH3OH(g) H = -90.2 kJ

If a mixture of H2, CO and CH3OH is at equilibrium, will the concentration of CO increase, decrease, or remain the same as equilibrium is reestablished after the temperature of the system is increased.

A. [CO] will increaseB. [CO] will decreaseC. [CO] will stay the same

2H2(g) + CO(g) CH3OH(g) H = -90.2 kJ

If a mixture of H2, CO and CH3OH is at equilibrium, will the concentration of CO increase, decrease, or remain the same as equilibrium is reestablished after more catalyst is added.

A. [CO] will increaseB. [CO] will decreaseC. [CO] will stay the same

Which of the following is NOT true according to LeChâtelier’s Principle?

A. A system at equilibrium that has been disturbed will shift away from the stressor

B. Removing a product from a system will cause the system to shift back to reactants

C. Changing amounts of solids or liquids in a reaction does not affect a system’s equilibrium

D. An increase in a gaseous system’s volume will shift the reaction toward the side with more moles of gas

E. A decrease in the temperature of a system at equilibrium will cause the reaction to shift in the exothermic direction.

Example 5: Calculating new values Iodine chloride decomposes at high temperatures:

2 ICl(g) I2(g) + Cl2(g) At equilibrium the partial pressures are (in atm)

ICl: 0.43, I2: 0.16 and Cl2: 0.27

a) Calculate Kb) If enough iodine condenses to drop its partial

pressure to 0.10 atm, what is the partial pressure of iodine gas when the equilibrium is reestablished?

CO(g) + Cl2(g) ⇄ COCl2 (g) + heat

THE REACTION QUOTIENT (Q)Non-equilibrium conditions

Q: the reaction quotient Q is just like K, except it is for ANY point in

time, not just at equilibrium. You will find Q (like finding K) and then compare Q to K

Q = [products][reactants]

COMPARING Q TO K Q<K

Q = [products]

[REACTANTS]reaction must shift to right to produce more products

Q>KQ = [PRODUCTS]

[reactants]

reaction must shift to left to produce more reactants

Q=Kreaction is AT EQUILIBRIUM

Example 6: Calculating and Using QFor the reaction:

2NO2 (g) 2NO (g) + O2 (g)K at a certain temperature is 0.50. Predict the

direction the system will move if one starts with:a) PO2

= PNO = PNO2 = 0.10 atm

b) PNO2 = 0.848 atm, PO2

= 0.0166atm

c) PNO2 = 0.20atm, PO2

= 0.010atm, PNO = 0.040atm

THE VAN’T HOFF EQUATIONEffect of Temperature on K

Effect of Temperature on K Of all types of ways to disturb equilibrium,

(adding/removing a substance, changing volume/pressure, changing temperature), temperature is the only factor to affect the equilibrium constant.

Use van’t Hoff equation to find new K , or if given new K, to find ΔH

ln K2 = ΔH 1 – 1

K1 R T1 T2

R = 8.31 J/mol·K

ln k2 = Ea 1 - 1

k1 R T1 T2

Look familiar?

Example 7: van’t Hoff equationFor the system

H2(g) + I2(g) 2HI(g) ΔH = -9.4 kJ and K= 62.5 at 800 K.

What is the equilibrium constant at 333°C?