Equilibrium Powerpoint Part 1

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Reversible Reactions & Equilibrium

In a closed system, reactions are reversible!

The conversion of reactants to products

(forward reaction) & the conversion of products to reactants (reverse reaction) occur simultaneously.

Reversible Reactions:

A + B AB

Equilibrium:

When RATE of forward reaction is the same as the RATE of reverse reaction EQUILIBRIUM

Only REVERSIBLE reactions can achieve equilibrium

Must be closed system & constant temp

A + B AB

Analogy:Hockey players on a team & equilibrium:

Players on bench rotate with players on ice

Player leave bench at same rate as players come back on bench

Same players on the team (never any new or different players on the team)

Ex.Heat + 2HI H2 + I2

Initially: only HI At Equi: decomp of HI equals rate of

synthesis of HI

H2 + I2

Ex.2SO2 + O2 2SO3 ∆H = -197kJ

Initially: only SO2 & O2

At Equi: syn of SO3 equals rate of syn of SO2 & O2

Equilibrium Constant – Keq

Keq: expresses the concentrations of reactants &

products at equilibrium

constant for every reversible reactions at equilibrium at a given temp & pressure

aA + bB cC + dD

Keq = [C]c[D]d

[A]a[B]b

Given:

2NO2 N2O4

Calculate Keq, if the conc of N2O4 at equi is 0.00140 M & the conc of NO2 at equi is 0.0172M

Keq = [N2O4][NO2]2

Keq = [0.0014][0.0172]2

Keq = 4.73

Given:

2NO2 N2O4

Calculate Keq, if the conc of N2O4 at equi is 0.00452 M & the conc of NO2 at equi is 0.031M

Keq = [N2O4][NO2]2

Keq = [0.00452][0.031]2

Keq = 4.70 Same Keq!!!

The value of Keq tells you exactly what is happening with a reaction

If: Keq = 1

At equi, conc of products & reactants are the same

Keq > 1 At equi, greater conc of products, than reactants

Keq < 1 At equi, greater conc of reactants, than products

Keq is the same for a given reaction at equilibrium at the same temp no matter what the initial conc were.

Keq does change with temp

Do not include pure solids and pure liquids in the Keq expression because their concentrations vary little.

Everything else must be included

Calculating

Consider:H2 (g) + I 2 (g) 2HI (g)

Find the Keq if the conc of H2 is 0.46, I2 is 0.39 M & HI is 3.0 M at equilibrium

Keq = [HI]2__ [H2][I2]

Keq = __[3]2__ [0.46][0.39]

Keq = 50

Conc of products greater at equi

Consider:PCl5 (g) PCl3 (g) + Cl2 (g)

Find the Keq if the initial conc of PCl5 is 0.70M & the final conc of Cl2 is 0.15M

Keq expression for equi conc only!!!

Use an ICE table

PCl5 PCl3 Cl2

+ 0.15+ 0.15- 0.15

0.15 0.15

Keq = [PCl3][Cl2]__ [PCl5]

Keq = [0.15][0.15]__ [0.55]

Keq = 0.041

Consider:2NH3 (g) 3H2 (g) + N2 (g)

Initially a 5.0L flask contains 0.2M NH3 & 0.08M N2. After equi [NH3] is 0.156M.

NH3 H2 N2

0.2 0.080

- 0.044 + 0.066 + 0.022

0.066 0.102

Keq = [H2]3[N2]__ [NH3]2

Keq = [0.066]3[0.102]__ [0.156]2

Keq = 0.0012

At equi, conc of reactants greater

Consider:H2 (g) + I2 (g) 2HI (g)

Find the [HI] at equi, if [H2] at equi is 0.50M and the [I2] at equi is 0.50M and Keq = 50.

H2 I2 HI

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

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

Keq = [HI]2

[H2] [I2]

50 = [x]2__ [0.5][0.5]

[HI] = 3.54 M at equi

What are the equi conc of each substance if a flask initially contains only 0.5M of H2 and 0.5M of I2? Keq = 50.

H2 I2 HI

0.5 00.5

0.5 - x

- x - x + 2x

0.5 - x 2x

Keq = [HI]2

[H2] [I2]

50 = [2x]2__ [0.5-x][0.5-x]

At Equi:

[H2] = 0.11M[I2] = 0.11M[HI] = 0.78M

7.07 = [2x]_ [0.5-x]

7.07(0.5 – x) = 2x

3.54 - 7.07x = 2x

x = 0.39

What are the equi conc of each substance if a 0.5L flask initially contains 2 moles H2 and 2 moles of I2?

H2 I2 HI

- x - x + 2x

4 - x 2x

Keq = [HI]2

[H2] [I2]

50 = [2x]2__ [4-x][4-x]

At Equi:

[H2] = 0.88M[I2] = 0.88M[HI] = 6.24M

7.07 = [2x]_ [4-x]

7.07(4 – x) = 2x

28. 28 - 7.07x = 2x

x = 3.12

If not same, multiply out then

use quadratic formula

Le Chatelier’s Principle

Le Chatelier’s Principle:

• If a closed system at equilibrium is subject to a change (stress), processes will occur to counteract the stress.

Factors that will affect Equi:

1. Concentration

2. Temperature

3. Pressure & Volume

Given:N2O4 (g) + 59.0kJ 2NO2 (g)

1. Concentration:• Adding more reactant pushes reaction in

direction of product• Removing reactant pushes reaction in direction

of reactants• Adding more product pushes reaction in

direction of reactants• Removing product pushes reaction in direction

of products

More products, so shift equi to counteract, so more N2O4 produced

What would happen if added more NO2?

Add NO2

1. Temperature:• Increasing temp causes the equi to favor the

endothermic side• Why? Increasing temp will increase the rate of

the endo reaction so excess heat is used up.• Decreasing temperature causes the equi to

favor the exothermic side (so that heat is produced/given off)

Increase temp favors endo side, so equi shifts to products, producing more NO2

What would happen if added heat?

Increase Temp

1. Pressure & Volume:• Affects an equi with an unequal number of

moles of gaseous reactants and products

• If increase press (decrease vol), equi shifts to side with less moles

• If decrease press (increase volume), equi shifts to side with more moles

Increase press favors side with less moles, so equi shifts to reactants, producing more N2O4

What would happen if increased the press / lowered vol?

Increase Pressure

• CATALYSTS:– Increase RATE– But have NO affect on equilibrium

• Adding an inert gas to a system will not affect the equi.

• Solids do not affect an equi, because their concentration doesn’t change

Factors Affecting Keq

• Only TEMPERATURE will affect the value of Keq

• Why?– When you increase temp, rate increases– But endo rate will increase more than the exo

rate. – Changing the ratio of products to reactants &

changing Keq!

2C6H6 + 15O2 12CO2 + 6H2O + heat

What would happen to Keq if we increased the temp?

Increase temp, favors endo reaction So, more reactants produced Keq = [products] / [reactants]

Reactants will ↑ & products will ↓ Keq will decrease

Equilibrium Powerpoint Part 1

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