Chemistry part 25, Julia Burdge,2e (2009) pps

2 Calculating Equilibrium Concentrations If we know the equilibrium constant for a reaction , we can calculate the concentrations in the equi-1ibrium mixture from the initial reactant

where D.n = b - a In general,

Equation 15 3 D.n = moles of gaseous products - moles of gaseous reactants

Because pressure s are usually expres se d in atmospheres , the gas constant R is 0 08206 L atmlmol K, and we can write the relationship between Kp and Kc as

Equation 15.4 Kp = K c [(0 08206 L atmlK mol) X T]~ n

Kp is equal to K c only in the special case where D.n = 0 , a s in the following equilibrium reaction:

Sample Problem s 15.5 and 15.6 let you practice writing Kp expre ss ions and illustrate the

conversion between K c and Kp

Write Kp expressions for (a) PCI 3 (g) + C I 2(g) ;;: • = ::!:' PCl s(g), (b) 0 2(g) + 2H2(g) :;: = ::!:'

2H20(I), and (c) Fig) + H 2(g) • ' 2HF(g)

Strategy Write equilibrium expressions for each equation, expressing the concentrations of the gases in partial pressures

Setup (a) All the species in this equation are gases, so they will all appear in the Kp expression

(b) Only the reactants are gases

(c) All species are gases

Solution

15 3.1 Select the correct equilibrium expression for the reaction 15.3.2 Select the correct equilibrium expression for the reaction

H + (aq) + OW (aq) • ' H 2 0( I) CaO(s) + CO 2 (g) • ' CaC03(s)

is 4.63 X 10- 3 at 25°C What is the value of Kp at this temperature?

Strategy Use Equation 15.4 to convert from K c to Kp Be sure to convert temperature in degrees

Practice Problem A Por the reaction

K c is 2.3 X lO- z at 375°C Calculate K p for the reaction at this temperature

Practice Problem B Kp = 2.79 X 10-5 for the reaction in Practice Problem A at 472°C What is

K c for this reaction at 472°C?

Given the following information,

HP(aq) • • H + (aq) + P - ( aq )

15 3 4 K c for the reaction

Think About It Note that we have essentially disregarded the units of Rand T so that the resulting equilibrium constant, K p ,

is unitless Equilibrium constants commonly are treated as unitless quantities

Br2(g) :;: =~ 2Br(g)

K c = 6.8 X 10- 4 (at 25°C)

H Z C Z 0 4 (aq) :;:.=~' 2H + (aq) + C10~ -(aq)

is 1.1 X 10-3 at 1280°C Calculate the value of K p for this

reaction at this temperature

605

\

\

Q c is calculat ed using the i nitial concentration s

of reactants and products Similarly, Q p can be calcul a ted u s ing the initial pa rti al pr essu r e s of reactants and products

Remem ber t hat calculati ng Qc is just like

ca l cul a t ing K c : pr oducts over r eact ants, each

r ai s ed to t h e appro priate power except t ha t the concentra ti ons we use to calcu late Qc are

t h e st arting concen trati ons T o calculate K c we must use e q uilibri u m concentratio n s

The co mpari s on of Q with K can ref er ei th e r to

Q c and K c or Q p and Kp

U sing Equilibrium Expressions to Solve Problems

We have alread y u s ed equilibrium expressions to determine the value of an equilibrium constant

u s ing equilibrium concentration s In this section, we will learn how to use equilibrium expre s sions

to predict the direction of a reaction and to calculate equilibrium concentrations

Predicting the Direction of a Reaction

If we s tart an exp e riment with only reactant s, we know that the reactant concentration s will decrea s e and the product concentrations will increa s e; that i s , the reaction must proceed in the

with only product s, we kn o w that the product concentration s will decrease and the reactant centration s will increa se In thi s case, the reaction mu s t proceed in the reverse direction to achieve

con-equilibrium, But often we mu s t predict the direction in which a reaction will proceed when we

s tart with a mixture of reactants and product s For thi s s ituation, we calculate the value of the

reac-

tion quotient , Q e, and c o mpare it to the value of the equilibrium constant, Ke

The equilibrium c o n s tant , Ke , for the ga s eou s formation of hydrogen iodide from molecular hydrogen and mol e cular iodine ,

i s 5 4 3 at 430 ° C If we wer e to conduct an experiment s tarting with a mixture of 0.243 mole of

H ? , 0.146 mole of 1 2, a nd 1.9 8 moles of III in a 1.00-L container at 430 ° C, would more HI form

or would III be con s umed and more H2 and 1 2 form ? U s in g the s tarting concentration s , we can

calC u la t e ' the reaction quotient a s follows:

( 1.98) 2

-:-::-:: ::-., :-:: :-: :::- = 111

( 0.243 )( 0.146)

where the s ub s cript " i " indicate s initial concentration B e cau s e the reaction quotient doe s not

e qual K e (Q e = Ill , K e = 54 3), the reaction i s not at equilibrium In order to e s tabli s h rium, the reaction will pro cee d to the left, con s uming III and producing H 2 and 1 2, decrea s ing the valu e of the numerator and incr e a s ing the value in the denominator until the value of the reaction quotient equal s that o f th e e quilibrium constant Thu s, the reaction proceed s in the reverse direc-

equilib-ti o n ( from right to le f t ) in o rder to reach equilibrium

There are three p oss ibilitie s when we compare Q with K:

Q < K The ratio of initial c oncentrations of product s to reactants is too small To reach

equilibrium, r e actant s mu s t be converted to product s The system proceed s in the forward dire c ti o n ( from left to right )

Q = K The initial concentration s are equilibrium c oncentrations The system i s already at

equilibrium , and there will be no net mov e m e nt in either direction

Q > K The ratio o f initial concentration s of product s to reactant s is too large To reach

equilibrium , produ c t s mu s t be converted t o reactant s , The system proceed s in the rever s e dir ec ti o n (f rom right to left)

Sample Problem 15.7 s how s how the value of Q i s used to determine the direction of a tion that is not at equilibrium

reac-Sample Problem 15.7

A t 37 5 ° C , th e equilibrium c o n s tant f o r the reaction

i s 1 2 At the s tart o f a re ac ti o n , the concentration s of N2, H z, a nd NH 3 are 0.071 M, 9.2 X 10 - 3 M,

and 1 8 3 X 10 - 4 M, re s pe c ti v el y D e termine whether thi s s y s tem i s at equilibrium, and if not, determine in w hich directi o n it mu s t proceed to e s tablish e quilibrium

Strategy Us e the initial c o nc e ntr a tion s to calculate Qc, and then compare Qc with K c

SECTION 15.4 Using Equilibrium Expressions to Solve Problems 607

Setup

(1.83 X 1O-4 ) z

- - -, -,- = 0.61 (0.071)(9.2 X 10 - 3) 3

Solution The calculated value of Q c is le s s than K c Therefore, the reaction i s not at equilibrium and

must proceed to the right to establish eqUilibrium

Practice Problem A The equilibrium constant, K c , for the formation of nitro s yl chloride from nitric

oxide and chlorine,

2NO(g) + Cl z (g ) +=, ~ 2NOCl(g)

is 6.5 X 104 at 35 ° C In which direction will the reaction proceed to reach equilibrium if the s tarting

concentrations of NO, Clz, and NOCI are 1.1 X 10 - 3 M, 3.5 X 10 - 4 M , and 1.9 M, re s pectively?

•

Practice Problem B Calculate Kp for the formation of nitrosyl chloride from nitric oxide and

chlorine at 35 ° C, and determine whether the reaction will proceed to the right or the left to achieve

equilibrium when the starting pre ss ure s are P NO = 1.01 atm, P CI = 0.42 atm , and P NOC I = 1.76 atm

2

Calculating Equilibrium Concentrations

If we know the equilibrium constant for a reaction , we can calculate the concentrations in the

equi-1ibrium mixture from the initial reactant concentrations Consider the following system involving

two organic compounds, cis- and trans-stilbene:

The equilibrium constant (K e ) for this s ystem is 24.0 at 200 ° C If we know that the starting

con-centration of cis-stilbene is 0.850 M, we can use the equilibrium expre ss ion to determine the

equilibrium concentrations of both specie s The s toichiometry of the reaction tell s us that for every

mole of cis-stilbene converted, 1 mole of trans- s tilbene i s produced We will let x be the

equi-librium concentration of trans-stilbene in mollL; therefore , the equilibrium concentration of c i s

-stilbene mu s t be (0.850 - x) mollL It is u s eful to summarize the s e change s in concentration s in

Think About It In proceeding

to the right, a reaction con s ume s

reactant s and produce s more produ c ts This increases the numerator in the reaction quotient and decreases the denominator

The result is an increase in Q c until

it is equal to K c , at which point equilibrium will be established

Think About It Always check

your answer by inserting the

calculated concentrations into the

equilibrium expression:

[HIl ~ q (0.378)2 [H2leq[I2leq (0.05 1)2

= 54.9 = K c

The small difference between the

calculated Kc and the one given

in the problem statement is due to

rounding

follows:

[cis-stilbene] = (0.850 - x) M = 0.034 M

[trans-stilbene] = x M = 0.816 M

A good way to check the answer to a problem such as this is to use the calculated equilibrium

con-centrations in the equilibrium expression and make sure that we get the correct Kc value

K = 0.816 = 24

c 0.034

Sample Problems 15.8 and 15.9 provide additional examples of this kind of problem

K c for the reaction of hydrogen and iodine to produce hydrogen iodide,

is 54.3 at 430°C What will the concentrations be at equilibrium if we start with 0.240 M concentrations of both H2 and 12')

Strategy Construct an equilibrium table to determine the equilibrium concentration of each species

in terms of an unknown ( x ) ; solve for x, and use it to calculate the equilibrium molar concentrations

Setup Insert the starting concentrations that we know into the equilibrium table:

H 2 (g) + I 2 (g) :;:: =::!:' 2HI(g)

Initial concentration (M): 0.240 0.240 0 Change in concentration (M):

Equilibrium concentration (M):

Solution We define the change in concentration of one of the reactants as x Because there is no product at the start of the reaction, the reactant concentration must decrease; that is, this reaction must proceed in the forward direction to reach equilibrium According to the stoichiometry of the chemical reaction, the reactant concentrations will both decrease by the same amount (x), and the product concentration will increase by twice that amount (2x) Combining the initial concentration and the change in concentration for each species, we get expressions (in terms of x) for the

SECTION 15.4 Using Equilibrium Expressions to Solve Problems 609

Practice Problem A Calculate the equilibrium concentrations of H2, 12, and HI at 430°C if the initial

concentrations are [H2l; = [I2li = 0 M, and [HIli = 0.525 M

Practice Problem B Calculate the equilibrium concentrations of H2o 120 and HI at 430°C if the initial

concentrations are [H2]; = [I2l; = 0.100 M, and [HIl; = 0.200 M

For the same reaction and temperature as in Sample Problem 15.8, calculate the equilibrium

concentrations of all three species if the starting concentrations are as follows: [H2l; = 0.00623 M,

[12]; = 0.00414 M, and [HIl; = 0.0424 M

Strategy Using the initial concentrations, calculate the reaction quotient, Qc, and compare it to the

value of Kc (given in the problem statement of Sample Problem 15.8) to determine which direction

the reaction will proceed in order to establish equilibrium Then, construct an equilibrium table to

determine the equilibrium concentrations ·

Solution Because we know the reaction must proceed from right to left, we know that the

concentration of HI will decrease and the concentrations of H2 and 12 will increase Therefore, the

table should be filled in as follows:

Next, we insert these expressions for the equilibrium concentrations into the equilibrium expression

and solve for x

5 4 3 = c::-::-::-::-' :: -, ,-:: -::-::-

-'-: c -c-(0.00623 + x)(0.00414 + x)

It isn't possible to solve this equation the way we did in Practice Problem 15.8 (by taking the square

root of both sides) because the concentrations of H2 and 12 are unequal Instead, we have to carry out

Think About It Checking this

result gives

[HI]2

Kc = [H2][12]

(0.0414i :-.,.-:- :: :c -: -:-:: :-:: = 54 3 (0.00676)(0.00467)

If Q < K the reaction will occur as wr i tten If

Q > K the reverse reaction will occur

For this reaction, 1M = 0; therefore, K c is equal

is 1.1 X 1O ~3 If the initial concentrations are [Br2] = 6.3 X 1O~2 M and [Br] = 1.2 X 1O~2 M,

calculate the concentrations of these two species at equilibrium

Here is a summary of the use of initial reactant concentrations to determine equilibrium concentrations:

1 Construct an equilibrium table, and fill in the initial concentrations (including any that are

zero)

2 Use initial concentrations to calculate the reaction quotient, Q, and compare Q to K to

deter- mine ' the ' cfirectloi11ll ' whiCh ' th ' e ' reacdoil will proceed

3 Define x as the amount of a particular species consumed, and use the stoichiometry of the

reaction to define (in terIllS of x) the amount of other species consumed or produced

4 For each species in the equilibrium, add the change in concentration to the initial

concentra-tion to get the equilibrium concentraconcentra-tion

5 Use the equilibrium concentrations and the equilibrium expression to solve for x

6 Using the calculated value of x, determine the concentrations of all species at equilibrium

7 Check your work by plugging the calculated equilibrium concentrations into the equilibrium

expression The result should be very close to the Kc stated in the problem

The same procedure applies to Kp Sample Problem 15.10 shows how to solve an equilibrium problem using partial pressures

Sample Problem 15.10

A mixture of 5.75 atm of H2 and 5.75 atm OfI2 is contained in a 1.0-L vessel at 430°C The

eqiiiili:iii'uiii con·s·t'ant' (Kp) for the reaction

at this temperature is 54.3 Determine the equilibrium partial pressures of Hb Ib and HI

Strategy Construct an equilibrium table to determine the equilibrium partial pressures

Setup The equilibrium table is

H2(g) + 12(g) ,

2HI(g)

•

Ini tial partial pressure (atm): 5.75 5.75 0

Change in partial pressure (atm): - x - x +2x

Equilibrium partial pressure (atm): 5.75 - x 5.75 - x 2x

SECTION 15.5 Factors That Affect Chemical Equilibrium 611

Solution Setting the equilibrium expression equal to K p ,

Practice Problem A Determine the equilibrium partial pressures of H2, 1 20 and HI if we begin the

experiment with 1.75 atm each of H 2 and 1 2 at 430°C ··· ···· ···· ····

Practice Problem B Determine the equilibrium partial pressures of H2, 12, and HI if we begin the

experiment with 2.75 atm HI at 430°C

Use the following information to answer questions 15.4.1 and 15.4.2: K c for the reaction

is 1.7 X 10-2 at 250°C

concentration of A, B, and C be at this

temperature if [Ali = [Bli = 0.750 M

Factors That Affect Chemical Equilibrium

One of the interesting and useful features of chemical equilibria i s that they can be manipulated in

s pecific ways to maximize production of a desired substance Consider , for example , the indu s trial

production of ammonia from its con s tituent elements by the Haber proce s s:

More than 100 million tons of ammonia i s produced annually by this reaction , with mo s t of the

resulting ammonia being used for fertilizers to enhance crop production Clearly it would be in the

best intere s t of industry to maximize the yield of NH 3 In thi s section, we willleam about the v

ari-0us ways in which an equilibrium can be manipulated in order to accompli s h thi s goal

Le Chiitelier's principle s tate s that when a stress i s applied to a sys tem at equilibrium , the system will respond by shifting in the direction that minimi z e s the effect of the s tre ss In thi s con-

text, "stress" refers to a disturbance of the s ystem at equilibrium by an y of th e following mean s :

calculated partial pressures into the equilibrium expression gives

(9.04)2 -:c = 54.0 (1.23)2

The small difference between this

result and the equilibrium constant

given in the problem statement is due to rounding

W h en a r eaction starts w ith r e actan ts a nd

p ro d u cts , be su r e to c al c ul at e Q a nd co m pa re i t

to K to d eterm i n e wh i ch d i r ecti on the re a cti on will proceed to r e ach e q uili b r i um

Re m e m ber that a t equ ili br i u m, the reactio n

q uot i ent, Q " i s eq u a l to t h e eq uilibrium

constan t , K ,

While r e e stabli sh ing e q uilib r i u m ca u se s a

decrease i n t h e N 2 conc en t ra tio n , t he fin a l

co nc entr at i on will still be h ig he r t h an t hat in

t h e o r ig inal e quilib rium mixt u r e Th e system

re sp ond s to t he stre s s o f the a dde d re actan t b y

cons umi ng part o f it

• The addition of a reactant or product

• The remo v al of a reactant or product

• A change in v olume of the s ystem, re s ulting in a change in concentration or partial pressure

of the reactant s and products

• A change in temperature

" Shifting " refer s to the occunence of either the forward or rever s e reaction such that the effect

of the stress i s partiall y offset as the s ystem reestabli s he s equiliblium An equilibrium that shifts

to the right i s one in which more product s are produced by the forward reaction An equi'librium that shift s to the left is one in which more reactants are produced by the reverse reaction Using

Le Chatelier ' s principle, we can predict the direction in which an equilibrium will shift, given the specific stress that i s applied

Addition or Removal of a Substance

Again u s ing the Haber proces s as an example,

con s ider a s ystem at 700 K, in which the equiliblium concentrations are as follows:

, .,

Using the s e concentration s in the reaction quotient expression, we can calculate the value of K c for the reaction at thi s temperature a s follows:

[NH 3 f [N2] [H2] 3

(1.52) 2

(2.05)(1.56) 3

If we were to apply stre ss to thi s system by adding more N 2 , increasing its concentration from

2.05 M to 3.51 M , the s y s tem would no longer be at equilibrium To see that this is true, use the new concentration of nitrogen in the reaction quotient expression The new calculated value of Qc

( 0.173 ) i s no longer equal to the v alue of K c (0.297)

For this system to reestabli s h equilibrium, the net reaction will have to shift in such a way that Qc

is again equal to K e , which is constant at a given temperature Recall from Section 15.4 that when

Q is les s than K, the reaction proceed s to the light in order to achieve equilibrium Likewise, an equilibrium that is s tre s sed in such a way that Q becomes less than K will shift to the right in order

to reestablish equiliblium Thi s means that the forward reaction, the consumption of N 2 and H2 to

produce NH 3, will occur The result will be a net decrease in the concentrations of N2 and H2 (thus making the denominator of the reaction quotient s maller), and a net increase in the concentration

of NH 3 ( thu s making the numerator larger) When the concentrations of all species are such that

Q c i s again equal to K c, the sy s tem will have established a new equilibrium position, meaning that

it will have shifted in one direction or the other , resulting in a new equilibrium concentration for each specie s Figure 15.5 shows how the concentrations of N2, H2, and NH 3 change when N2 is added to the original equiliblium mixture

Conversely , if we were to remove N 2 from the original equiliblium mixture , the lower centration in the denominator of the reaction quotient would result in Q c being greater than Ke

con-In thi s case the reaction will shift to the left That is, the reverse reaction will take place, thereby increa s ing the concentrations of N 2 and H 2 and decreasing the concentration of NH 3 until Q c is once again equal to Kc

The addition or removal of NH 3 will cause a shift in the equilibrium, too The addition of

NH 3 will cau s e a shift to the left; the removal of NH 3 will cause a shift to the right Figure 15.6(a)

s how s the additions and removal s that cause thi s equiliblium to shift to the right Figure 15.6(b) show s those that cause it to shift to the left

In es s ence, a system at equiliblium will respond to addition of a species by consuming some

of that s pecies , and it will respond to the removal of a species by producing more of that species

It is important to remember that addition or removal of a species from an equilibrium mixture does not change the value of the equiliblium constant, K Rather , it changes temporarily the value of the reaction quotient , Q Furthermore , in order to cause a shift in the equilibrium, the species added

or removed must be one that appear s in the reaction quotient expression In the case of a

heteroge-SECTION 1S.5 Factors That Affect Chemica l Equilibrium 613

2.0 1.5 1.0

0 5 0.0

-

-

Immedi atel y after a ddition of Nz

neous equilibrium, altering the amount of a solid or liquid species does not change the position of

the equilibrium because doing so does not change the value of Q

Sample Problem 15.11 shows the effects of stress on a system at equilibrium

Hydrogen sulfide ( H z S) is a contaminant commonly found in natural gas It is removed by reaction

with oxygen to produce elemental sulfur

2 HzS( g) + O zCg) :;:, ==' 2S(s ) + 2HzO(g)

For each of the following scenarios, determine whether the equilibrium will shift to the right, shift

to the left, or neither: (a) addition of O z (g), (b) removal of H2S( g), (c) removal of H zO(g), and (d)

addition of S(s)

Strategy Use Le Chiitelier's principle to predict the direction of shift for each case Remember that

the position of the equilibrium is only changed by the addition or removal of a species that appears in

the reaction quotient expression

Setup Begin by writing the reaction quotient expression:

Qc = ?

[J:lzSnO z ]

Because sulfur is a solid, it does not appear in the expression Changes in the concentration of any of

the other species will cause a change in the equilibrium position Addition of a reactant or removal of

a product that appears in the expression for Q c will shift the equilibrium to the right:

Figure 15.5 Adding more of a

causes the equilibrium position to shift

toward product The system responds to

the addition of Nz by consuming some

of the added Nz (and some of the other

reactant, H z) to produce more NH3

Figure 15.6 (a) Addition of a

reactant or removal of a product will cause an equilibrium to shift to the

right (b) Addition of a product or

equilibrium to shift to the left

Think About It In each case,

analyze the effect the change will

have on the value of Q c In part (a),

for example, O2 is added, so its

concentration increases Looking at

the reaction quotient expression, we

can see that a larger concentration

of oxygen corresponds to a larger

overall denominator-giving the

overall fraction a smaller value

Thus, Q will temporarily be smaller

than K and the reaction will have to

shift to the right, consuming some

of the added O2 (along with some

of the H2S in the mixture) in order

to reestablish equilibrium

decrease (pressure increase) on the

N 2 0 4 (g) • • 2N0 2 ( g) equilibrium

equilibrium is driven toward the side

with the smallest number of moles of

gas

Removal of a reactant or addition of a product that appears in the expression for Q c will shift the

equilibrium to the left:

Solution

(a) Shift to the right

(b) Shift to the left

(c) Shift to the right

Changes in Volume and Pressure

If we were to start with a gaseous sys tem at equilibrium in a cylinder with a movable piston, we

could change the volume of the system, thereby changing the concentrations of the reactants and products

Consider again the equilibrium between N 2 0 4 and N0 2:

At 25 ° C the equilibrium constant for thi s reaction i s 4.63 X 10 - 3 Suppose we have an equilibrium

mixture of 0.643 M N 2 0 4 and 0.0547 M N0 2 in a cylinder fitted with a movable piston If we push

down on the piston, the equilibrium will be disturbed and will shift in the direction that minimizes the effect of thi s di s turbance Consider what happen s to the concentrations of both species if we

decrease the volume of the cylinder by half Both concentrations are initially doubled: [N 2 0 4J =

1.286 M and [NO z J 0.1094 M If we plug the new concentrations into the reaction quotient expression, we get

-=-1.-=-:28=-=6- = 9.3 1 X 10

-which is not equal to K e , so the syste m i s no longer at equilibrium Because Q e is greater than Ke, the

equilibrium will have to shift to the left in order for equilibrium to be reestablished (Figure 15.7)

In general, a decrea se in volume of a reaction vessel will cause a shift in the equilibrium in the direction that minimi zes the total number of mole s of gas Conversely, an increase in volume will cause a shift in the direction that maximizes the total number of moles of gas

SECTION 1S.5 Factors That Affect Chemical Equil ibrium 615

Sample Problem 15 12 shows how to predict the equilibrium shift that will be caused by a volume change

Strategy Determine which direction minimized the number of moles of gas in the reaction Count

only moles of gas " "

Setup We have (a) 1 mole of gas on the reactant side and 2 moles of gas on the product side,

(b) 3 moles of gas on the reactant side and 2 moles of gas on the product side, and (c) 2 moles of gas

on each side

Solution

(a) Shift to the left

(b) Shift to the right

(c) No shift

Practice Problem A For each reaction, predict the direction of shift caused by increasing the volume

of the reaction vessel

(a) 2NOCI(g) += ===' 2NO(g)+Clz(g)

(b) CaC0 3 (s) ' CaO(s)+C02(g)

Practice Problem B For the following equilibrium, give an example of a stress that will cause a shift

to the right, a shift to the left, and one that will cause no shift:

It is possible to change the total pressure of a s ystem without changing it s volume by

add-ing an inert gas such as helium to the reaction vessel Because the total volume remains the same,

the concentrations of reactant and product ga ses do not change Therefore , the equilibrium is not

disturbed and no shift will occur

Changes in Temperature

A change in concentration or volume may alter the position of an equilibrium ( i.e , the relative

amounts of reactants and products), but it does not change the value of the equilibrium constant

Only a change in temperature can alter the value of the equilibrium constant To understand why,

consider the following reaction:

The forward reaction is endothermic (absorbs heat, (t: H > 0):

t: H O = 58.0 kJ/mol

If we treat heat as though it were a reactant, we can use Le Chil.teber's principle to predict what will

happen if we add or remove heat Increasing the temperature (adding heat) will shift the reaction in

the forward direction because heat appears on the reactant side Lowering the temperature (removing

heat) will shift the reaction in the reverse direction Consequently, the equilibrium con s tant, given by

?

[N02]eq

K = -:

: c [N20 4] eq increases when the system is heated and decrea ses when the system is cooled (F igure 15.8 ) A

It i s a common e rror to count all the species in

a readion to determine which side has fewer moles To determine what diredion shift a

volume change w ill cause, it is only the number

of moles of gas that matters

Think About It When there is no difference in the number of moles

of gas, changing the volume of the reaction vessel will change the concentrations of reactant(s) and product(s)-but the system will remain at equilibrium (Q will remain equal to K.)

Figure 15.8 (a) NZ0 4-NO z

equilibrium (b) Because the reaction i s

endothermic, at higher temperature , the

N Z 0 4 (g) • 2NO z (g) equilibrium

shifts toward product, making the

reaction mixture darker

Media P layer/MPEG Animation : Figures

15 10 and 15 11, Le Chiltelier's Principle,

to be a product Increasing the temperature of an exothermic reaction causes the equilibrium

con-stant to decrease , shifting the equilibrium toward reactants

Another example of this phenomenon is the equilibrium between the following ions:

CoCl ~ ~ + 6H20 • CO(H20)~ + + 4CI~ + Heat

various stresses on systems at equilibrium

Catalysis

A catalyst speeds up a reaction by lowering the reaction's activation energy [ ~~ Section 14.6]

However, a catalyst lowers the activation energy of the forward and reverse reactions to the same extent (see Figure 14.14) The presence of a catalyst, therefore, does not alter the equilibrium con-

stant, nor does it shift the position of an equilibrium system Adding a catalyst to a reaction

Figure 15.9 (a) An equilibrium mixture of CoCll~ ion s and Co(H20)~+ ion s appears violet (b) Heating favors the formation of CoCll~, making the solution look more blue (c) Cooling favors the formation of Co(HzO)~+ , making the solution look more pink