Chemical Equilibrium Basics - York University · York University CHEM 1001 3.0 Chemical Equilibrium - 2 Chemical Equilibrium All chemical reactions are reversible; that is, if a reaction

York University CHEM 1001 3.0 Chemical Equilibrium - 1

Chemical Equilibrium Basics

Reading: Chapter 16 of Petrucci, Harwood and Herring (8thedition)

Problem Set: Chapter 16 questions 25, 27, 31, 33, 35, 43, 71


Chemical Equilibrium

All chemical reactions are reversible; that is, if a reactioncan occur in one direction:

CO(g) + 2H2(g) 6 CH3OH(g)

then the reverse reaction can also occur:

CH3OH(g) 6 CO(g) + 2H2(g)

This applies to both elementary and overall reactions.

As a reaction proceeds, the rate of the forward reactiondecreases and the reverse reaction increases until they areequal. This balance is called equilibrium.


Approach to Equilibrium - Example

Elementary reactions:

(1) N2O4(g) 6 2NO2(g) (forward)

(2) 2NO2(g) 6 N2O4(g) (reverse)

rate(1) = k1[N2O4] rate(2) = k2[NO2]2

At t=0 let [N2O4] > 0 and [NO2] = 0, then rate(1) > rate(2).

C As time passes, [N2O4] decreases and [NO2] increases.

C Rate(1) decreases and rate(2) increases until they are equal.

C When rate(1) = rate(2), we have equilibrium.


Approach to Equilibrium - continued

When rate(1) = rate(2), we have equilibrium. Therefore:

k1[N2O4] = k2[NO2]2

= equilibrium constant

At equilibrium:

C Concentrations remain constant.

C Both reactions continue to occur.

C This is a dynamic equilibrium.


Constant Conditions

In a static system, nothing happens. Example: Neon in theEarth's atmosphere is neither produced nor consumed.

At steady-state, the rate of production is balanced by therate of consumption. Example: Oxygen in the atmosphere isproduced by photosynthesis and consumed by respiration.

Dynamic equilibrium is a steady-state in which theconsumption process is the exact opposite of the productionprocess. Example: N2O4(g) W 2NO2(g).

People sometimes use equilibrium for all three of these.This is unacceptable in chemistry.


Examples of Equilibrium

In a closed container, H2O(l) willcome into equilibrium with H2O(g).

(But H2O(g) and H2O(l) are not inequilibrium in the atmosphere.)

(a) I2 in H2O(l) (yellow).

(b) I2 equilibrium between H2O(l)(top) and CCl4(l) (bottom, purple).Most I2 is in the CCl4.


Evidence of Dynamic Equilibrium

AgI(s) W Ag+(aq) + I-(aq)

Make up a saturatedsolution with excess solidAgI.

Add a small amount ofradioactive 131I-(aq) to thesolution.

The radioactivity willgradually appear in thesolid AgI.


The Equilibrium Constant

Elementary reaction: aA + bB + ... W gG + hH + ...

At equilibrium: kf[A]a[B]b... = kb[G]g[H]h...

so = equilibrium constant

This applies to every step in a reaction mechanism.For the overall reaction, KC has the same form:

C products in the numerator, reactants in the denominator

C multiply concentrations raised to powers

C powers are the stoichiometric coefficients


Equilibrium Constant vs. Rate Laws

These are similar, but different.

C Both involve products of concentrations.

C The rate law may or may not include reactants,products, and other species.

C The equilibrium constant includes all reactants andproducts, but only reactants and products.

C In the equilibrium constant, the powers are thestoichiometric coefficients. Not so for rate laws(except elementary reactions).


Equilibrium Constant - Examples

CO(g) + 2H2(g) W CH3OH(g)

½CO(g) + H2(g) W ½CH3OH(g)

CH3OH(g) W CO(g) + 2H2(g)

The correct expression for the equilibrium constant dependson how the reaction is written.


Approach to Equilibrium


C Three differentinitial conditions.

C Equilibriumreached in eachcase.

C Different finalconditions.

C All satisfy theequilibriumcondition.


Final State - Equilibrium

Y 14.5

Y 14.4

Y 14.5

8

KC


Beyond Concentrations

C In reality, rate laws do not always give the exactdependence of reaction rate on concentration.

C Also, equilibrium constants using concentrations, KC,are not constant (sometimes not nearly so).

C The expressions we use here are approximations.

C The thermodynamic equilibrium constant, Keq, isdefined using activities instead of concentrations.

C We use concentrations as approximations to activities.


Activities

C Thermodynamic equilibrium constants use activities.

C Activities are usually defined to be dimensionless.

C Thermodynamic equilibrium constants are dimensionless.

C Activities of pure solids and pure liquids are unity (1.0...).

C For dilute solutions:

solute activities approximately equal concentrations,

the solvent activity is approximately unity.

C For ideal gases, activities equal partial pressures.

C We will treat solutions as dilute and gases as ideal.


Equilibrium Constant - More Examples

2H2(g) + O2(g) W 2H2O(l)

CaCO3(s) W CaO(s) + CO2(g)

2H2O(l) W H3O+(aq) + OH-(aq) Keq . [H3O

+][OH-]

HCl(aq) + H2O(l) W H3O+(aq) + Cl-(aq)


Equilibrium Constants Involving Gases


For this reaction at 483 K, we found that:

The thermodynamic equilibrium constant is:

These are not the same.


Converting between KC and KP

From the ideal gas law: PV = nRT

So, for CO: PCO = (nCO/V)RT = [CO]RT

So:

and

KC = 14.5 at T = 483 K. R = 0.08206 L atm mol-1 K-1

So RT = 39.6 L atm mol-1 and KP = 9.23×10-3


Magnitudes of Equilibrium Constants

C Equilibrium constants have a very wide range of values.

C Equilibrium constants can change enormously withchanges in temperature.


Significance of Magnitude of K2H2(g) + O2(g) W 2H2O(l)

= 1.4×1083 at 298 K

If pO2 = 0.21 bar, what is the equilibrium pH2?

Y

Y PH2 = 5.8×10-42 bar Y < 1 molecule in the entireatmosphere


Significance of K - continued

C All reactions are reversible. So there are always somereactants left at equilibrium.

C A small K means that a reaction can not proceed in thedirection written (to the right). It is unfavorable.

C A reasonably large K means that the reaction may proceed.It is favorable. But it might be very slow.

C If K is so large that the equilibrium concentration of areactant is negligible, we say that the reaction "goes tocompletion".

C If K is very small, the reverse reaction goes to completion.


Significance of K - example

C N2(g) + O2(g) W 2NO(g) KP = 4.7×10-31 at 298 K

The reaction can not proceed in the direction written.At 298 K, no significant NO can form from air.

C N2(g) + O2(g) W 2NO(g) KP = 1.3×10-4 at 1800 K

At 1800 K, significant NO may form. It turns out thatthe rate is fast at high temperatures.

C 2NO(g) W N2(g) + O2(g) KP = 2.1×1030 at 298 K

At 298 K, the NO may convert completely back to N2

and O2. But the rate is very slow.


The Equilibrium Constant - Summary

C All reactions are reversible. When the rates of the forwardand reverse reactions are equal, we have equilibrium.

C Equilibrium is a dynamic state, both forward and reversereactions continue to occur.

C The equilibrium constant, K, gives the condition thatmust obtain at equilibrium.

C K can be used to determine the direction in which a reactionis allowed to proceed. But the reaction might be very slow.

C K is expressed in terms of activities. Using concentrationsor partial pressures is an approximation.


Predicting the Direction of Change


C If we start with only CO and H2, the reaction will initiallyproceed to the right.

C If we start with only CH3OH, the reaction will initiallyproceed to the left.

C What if we start with a mixture of reactants and products?In what direction will the reaction proceed?

The answer depends on where the reaction stops - so itdepends on the equilibrium constant. But how?


The Reaction Quotient

C The reaction quotient, Q, is defined to be the same asthe equilibrium constant, but using initial conditionsinstead of equilibrium conditions.

Q is used to decide the direction of change from aspecific set of initial conditions.

C If Q < K, the reaction will proceed to the right.

If Q > K, the reaction will proceed to the left.

If Q = K, the reaction is at equilibrium.

C QC is defined using concentrations; QP is defined usingpartial pressures. Must be consistent with KC or KP.


Using the Reaction Quotient

CO(g) + 2H2(g) W CH3OH(g) KC = 14.5 at 483 K

Initial conditions: [CO] = 1.0 M,

[H2] = 0.30 M,

[CH3OH] = 2.5 M

? In which direction will the reaction proceed?

QC > KC Y The reaction proceeds to the left.


Using the Reaction Quotient - continued

What happens as the reaction proceeds?


We have QC > KC, so the reaction proceeds to the left.

In other words, the rate of the reverse reaction is greaterthan the rate of the forward reaction.

Thus: [CH3OH] decreases, [CO] increases, [H2] increases

Y QC decreases until QC = KC

When QC = KC, equilibrium isreached. The forward and reverserates are equal.


More on Using the Reaction Quotient

What if conditions change?


At equilibrium, QC = KC. What if we then add more CO?

C Adding CO makes QC smaller.

C Then QC < KC.

C The reaction proceeds to the right.

As the reaction proceeds [CH3OH] 8, [CO] 9, [H2] 9, QC 8.

Eventually QC = KC and we have a new equilibrium.All concentrations are different from our first equilibrium.


Le Châtelier’s Principle

When an equilibrium system is subjected to a change, itresponds by attaining a new equilibrium that partiallyoffsets the impact of the change.

Input (Change) Response

increase inside pressure increase volume

increase outside pressure reduce volume

increase in temperature absorb heat

increase in concentration shift away from that species(as directed by Q)


Le Châtelier’s Principle - Example

Closed container with liquid water andwater vapor in equilibrium:

H2O(l) W H2O(g)

C Increase external pressure:Volume decreases, some H2O(g)changes to H2O(l) (shift to left).

C Add He(g) to increase internal pressure: Volume increases,some H2O(l) changes to H2O(g) (shift to right).

C Increase temperature: Some H2O(l) changes to H2O(g)(shift to right absorbs heat).


Adding/Removing Reactants or Products

The equilibrium will shift to partially offset the change.


C Change: adding CH3OH

Response: equilibrium shifts to left

C Change: adding CO

Response: equilibrium shifts to right

C Change: removing CH3OH



Effect of Changing the Volume



C A shift to the right reduces the pressure (replaces 3 molesof gas with 1 mole of gas).

C A shift to the left increases the pressure.

C Change: decreased volume (increased pressure)


C Change: increased volume (decreased pressure)

Response: equilibrium shifts to left


Effect of Inert Gas


(1) Add He(g) while keeping the volume fixed.

C No effect on the reactants or products.

C No effect on the equilibrium.

(2) Add He(g) while keeping the total pressure fixed.

C Volume increases.

C Partial pressures of reactants and products decrease.

C Equilibrium shifts to left.


Exothermic and Endothermic Reactions

C Breaking chemical bonds requires energy. Formingbonds releases energy.

C Chemical reactions may absorb or release energy. Thisenergy is often transferred as heat.

C Exothermic reactions release energy as heat:

CH4 + 2O2 6 CO2 + 2H2O

H2SO4(l) + 2H2O(l) 6 2H3O+(aq) + SO4

2-(aq)

C Endothermic reactions absorb energy as heat:

H2O(l) 6 H2O(g)

NaOH(s) 6 Na+(aq) + OH-(aq)


Effect of Change in Temperature


C For endothermic reactions, the equilibrium constantincreases when the temperature goes up.

C For exothermic reactions, the equilibrium constantdecreases when the temperature goes up.

Examples: T (K) KP

C CaCO3(s) W CaO(s) + CO2(g) 298 1.0×10-23

This reaction is endothermic. 1200 1.0

C CH4 + 2O2 6 CO2 + 2H2O

The equilibrium constant will be smaller at higher T.


Effect of a Catalyst

A catalyst is a substance that increases the rate of achemical reaction without being consumed by the reaction.

A catalyst:

C Does not change the position of equilibrium.

C Can decrease the time it takes to reach equilibrium.

C Will not make an unfavorable reaction occur.

C Can help a favorable reaction to proceed faster.


The Direction of Change - Summary

C Equilibrium is not achieved for just one set ofconcentrations. It is achieved for any concentrations thatsatisfy the equilibrium constant.

C The reaction quotient, Q, gives the direction of changefrom a specific set of initial conditions.

C Le Châtelier’s Principle: An equilibrium system respondsto a change in conditions by partially offsetting the change.

C The effect of changes in amount, pressure, or volume canalso be determined by using Q.

C The effect of temperature on K depends on whether thereaction is endothermic or exothermic.


Equilibrium Calculations

Many types of reactions rapidly reach equilibrium:

C acid/base reactions

C salts entering solution

C formation of complex ions

We need to learn how to calculate:

C equilibrium constants from composition data

C composition from equilibrium constants

Some of these calculations require the use ofstoichiometry and mass balance.


Determining KC from Equilibrium Data

Example: N2O4(g) W 2NO2(g)

At equilibrium at 25 C, a 3.00 L vessel contains 7.64 gof N2O4 and 1.56 g of NO2. Find KC.

Solution:

Calculate the concentrations of reactants and products:

[N2O4] = 0.0277 M [NO2] = 0.0113 M

Evaluate the equilibrium constant:

KC = [NO2]2/ [N2O4] = 4.61×10-3


KP from Initial and Final Amounts of a Substance

Example: 2SO3(g) W 2SO2(g) + O2(g)

Initially, 0.0200 mol of SO3 are put into an evacuated 1.52 Lvessel. At equilibrium at 900 K, there are 0.0142 mol SO3.Find KP.

Solution:

(1) Calculate equilibrium moles of all reactants andproducts.

(2) Calculate partial pressures of all species.

(3) Evaluate the equilibrium constant: KP = P2SO2PO2/P

2SO3


KP from Initial and Final Amounts - continued

(1) The ICE Table

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

Initial 0.0200 mol 0.0000 mol 0.0000 mol

Change -0.0058 mol 0.0058 mol 0.0029 mol

Equilibrium 0.0142 mol 0.0058 mol 0.0029 mol

(2) Partial pressures: 0.690 atm 0.282 atm 0.141 atm

This used the ideal gas law: P = nRT/V = (48.6 atm mol-1)×n

(3) Evaluate KP = P2SO2PO2/P

2SO3

KP = 0.0235 = 0.024 (2 significant figures)


Variations in the ICE Table

C The ICE table is based on moles.

C Other quantities (concentrations, partial pressures) maybe used if they are strictly proportional to moles.

Examples:

Concentration = may be used if V is fixed.

Partial pressure may be used if T and V are fixed:


Equilibrium Partial Pressures from KP

Example: NH4HS(s) W NH3(g) + H2S(g)

At 25 °C, KP = 0.108

Excess solid NH4HS is placed in an evacuated flask. Findthe partial pressures of NH3 and H2S at equilibrium.

Solution: KP = PNH3 PH2S

Reaction NH4HS(s) W NH3(g) + H2S(g)

Initial - 0.0000 atm 0.0000 atm

Change - x atm x atm

Equilibrium - x atm x atm

x2 = KP = 0.108 Y PNH3 = 0.329 atm and PH2S = 0.329 atm


Equilibrium from Initial Conditions and KC

Example: N2O4(g) W 2NO2(g)

At 25 °C, KC = 4.61×10-3

0.0240 mole of N2O4 is placed in a 0.372 L flask.Find the moles of N2O4 at equilibrium.

Solution: KC = [NO2]2/ [N2O4]

Reaction N2O4(g) W 2NO2(g)

Initial 0.0240 mol 0.0000 mol

Change -x mol +2x mol

Equilibrium (0.0240-x) mol +2x mol

KC = (2x/V)2 / ((0.0240-x)/V)


Solving for the Unknown Concentration

From the previous slide: VKC (0.0240-x) = 4x2

Method 1: If only a little N2O4 dissociates, then

xn0.0240 Y 0.0240-x . 0.0240 Y 4x2 . 0.0240VKC

So x.0.00321 and nN2O4 = 0.0240-x = 0.0208 mol|

Check: x/0.0240 = 0.13, so assumption may be OK.

Method 2: Use the result from Method 1 to improve theapproximation: 0.0240-x . 0.0240-0.0032 = 0.0208

So 4x2 . 0.0208VKC Y x.0.00299 and nN2O4 = 0.0210 mol|

Can repeat until a constant value is reached.


Solving for Unknown Concentration - continued

Method 3: Use quadratic equation.

If 0 = ax2 + bx + c the

0 = 4x2 + VKCx - 0.0240VKC

a = 4 b = VKC c = -0.0240VKC

Y x = 0.00300 and nN2O4 = 0.0210 mol|

Check result: KC = [NO2]2/ [N2O4]

[N2O4] = nN2O4/V = 0.0564 M, [NO2] = 2x/V = 0.016 M

KC = 4.61×10-3 which is what it should be


Initial Concentrations Given for Multiple Species

Example: V3+(aq) + Cr2+(aq) W V2+(aq) + Cr3+(aq)

[V3+] = [Cr2+] = 0.0100 M [V2+] = [Cr3+] = 0.150 M

KC = 720 Find the equilibrium concentrations.

Solution:

Decide in which direction the reaction will proceed:

Since QC < KC, the reaction will proceed to the right.


Initial Concen. for Multiple Species - continued

Reaction V3+(aq) + Cr2+(aq) W V2+(aq) + Cr3+(aq)

Initial 0.0100 M 0.0100 M 0.150 M 0.150 M

Change -x M -x M x M x M

Equil. (0.0100-x) (0.0100-x) (0.150+x) (0.150+x)

Y

Y Y x = 4.25×10--3

Y [V3+] = [Cr2+] = 0.0057 M [V2+] = [Cr3+] = 0.154 M


Equilibrium Calculations - Summary

To calculate equilibrium constants from composition data:

C Calculate equilibrium amounts all species.

C Evaluate the equilibrium constant.

To calculate equilibrium composition from equilibriumconstants and initial composition:

C Construct the ICE table to get the all the equilibriumamounts in terms of a single unknown, x.

C Substitute into the expression for K, then solve for x.

C Evaluate all the equilibrium amounts.

Amounts can be moles, concentrations, or partial pressures.

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Chemical Equilibrium Basics - York University · York University CHEM 1001 3.0 Chemical Equilibrium - 2 Chemical Equilibrium All chemical reactions are reversible; that is, if a reaction