Equilibrium 2009-20101 Chemical Equilibrium. Equilibrium 2009-20102 Chemical Equilibrium Few chemical reactions proceed in only one direction. Most are

Equilibrium 2009-2010 1

Chemical Equilibrium


Chemical Equilibrium

• Few chemical reactions proceed in only one direction. Most are reversible, at least to some extent. At the start of a reversible process, the reaction proceeds toward the formation of products. As soon as some product molecules are formed, the reverse process begins to take place and reactant molecules are formed from product molecules. Chemical equilibrium is achieved when the rates of the forward and reverse reactions are equal and the concentrations of the reactants and products remain constant.


Equilibrium is a dynamic process.

• Physical equilibrium

H2O(s) H2O(l)

At equilibrium, liquid water is reforming ice as fast as ice is melting.


Chemical equilibrium

N2O4(g) 2NO2(g)

colorless brown

Equilibrium is a state in which there are no observable changes as time goes by. When a chemical reaction has reached the equilibrium state, the concentrations of reactants and products remain constant over time, and there are no visible changes in the system.


The Law of Mass ActionFor a reversible reaction at equilibrium

and at a constant temperature, the ratio of reactant and product concentrations has a constant value, k (the equilibrium constant).

• Note that although the concentrations may vary, as long as a given reaction is at equilibrium and the temperature does not change, according to the law of mass action, the value of k remains constant.

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We can generalize this phenomenon with the following reversible reaction:

aA + bB cC + dD

where a, b, c and d are the stoichiometric coefficients for the reacting species A, B, C, and D.

k = [C]c[D]d

[A]a[B]b

•kc indicates that the concentrations of the reacting species are expressed in molarity or moles per liter.

•kp is used when equilibrium concentrations are expressed in terms of pressure.

7

k = [NO2]2

[N2O4]

For the reaction,

N2O4(g) 2NO2(g)

Note that the exponent 2 for [NO2] in this expression is the same as the stoichiometric coefficient for NO2 in the balanced chemical equation.

•k is defined as having no units.

at 25 oC


Changes in the concentrations of NO2 and N2O4 with time, in three situations.

(a)Initially only NO2 is present

(b)Initially only N2O4 is present

(c) Initially a mixture of NO2 and N2O4 is present

Note that even though equilibrium is reached in all cases, the equilibrium concentrations of NO2 and N2O4 are not the same as in the previous trials.


Solving for k, the equilibrium constant

The following table shows some experimental data for the NO2-N2O4 system at 25 oC. The gas concentrations are expressed in molarity, which can be calculated from the number of moles of the gases present initially and at equilibrium and the volume of the flask in liters.

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The NO2-N2O4 System at 25 oC

Initial

Concentrations

(M)

Equilibrium

Concentrations

(M)

Ratio of Concentrations at

Equilibrium

[NO2] [N2O4] [NO2] [N2O4][NO2]

[N2O4]

[NO2]2

[N2O4]

0.000 0.670 0.0547 0.643

0.0500 0.446 0.0457 0.448

0.0300 0.500 0.0475 0.491

0.0400 0.600 0.0523 0.594

0.200 0.000 0.0204 0.0898

0.0851

0.102

0.0967

0.0880

0.227

Analysis of the equilibrium data shows that the ratio [NO2]/[N2O4] gives scattered values.

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The NO2-N2O4 System at 25 oC

Initial

Concentrations

(M)

Equilibrium

Concentrations

(M)

Ratio of Concentrations at

Equilibrium

[NO2] [N2O4] [NO2] [N2O4][NO2]

[N2O4]

[NO2]2

[N2O4]

0.000 0.670 0.0547 0.643

0.0500 0.446 0.0457 0.448

0.0300 0.500 0.0475 0.491

0.0400 0.600 0.0523 0.594

0.200 0.000 0.0204 0.0898

0.0851

0.102

0.0967

0.0880

0.227

Analysis of the equilibrium data shows that the ratio [NO2]2/[N2O4] gives a nearly constant value that averages 4.63 x 10-3

4.65 x 10-3

4.66 x 10-3

4.60 x 10-3

4.60 x 10-3

4.63 x 10-3


The magnitude of the equilibrium constant tells us whether an equilibrium reaction favors the products or reactants.

• If k is much greater than 1 (in this context, any number greater than 10 is considered to be much greater than 1) the equilibrium will lie to the right and favors the products.

• Conversely, if the equilibrium constant is much smaller than 1 (any number less than 0.1 is much less than 1), the equilibrium will lie to the left and favor the reactants.

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The concentrations of reactants and products in gaseous reactions can also be expressed in terms of their partial pressures.

For the equilibrium process

N2O4(g) 2NO2(g)

we can write

kp = (PNO2)2

(PN2O4)

Where PNO2 and PN2O4 are the equilibrium partial pressures (in atm) of NO2 and N2O4, respectively.

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Example – Write the equilibrium constant expression kc and kp if applicable, for the following system:

(NH4)2Se(s) 2NH3(g) + H2Se(g)

• The “concentration” of a solid or a liquid is a constant, (an intensive property that does not depend on how much of the substance is present) and can be omitted when writing equilibrium expressions.

kc = [NH3]2[H2Se]

k = [NH3]2[H2Se]

[(NH4)2Se]

kp = (PNH3)2(PH2Se)


In general, kc is not equal to kp, since the partial pressures of reactants and products are not equal to their concentrations expressed in moles per liter.

• A simple relationship exists between kp and kc kp = kc(RT)n

where

n = moles of gaseous product – moles of gaseous reactants

R = the gas constant 0.0821 L.atm/mol. K

and T = temperature in Kelvins


Determine kp for the following reaction at 220 oC. The kc for this reaction at 220 oC is 10.5

kp = kc(RT)n

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

kp = (10.5)((.0821)(493))

kp = 6.41 x 10-3


Multiple Equilibria

The reactions we have considered so far are all relatively simple. A more complicated situation is one in which the product molecules in one equilibrium system are involved in a second equilibrium process:

aA + bB cC + dD k’c

cC + dD eE + fF k”c

aA + bB eE + fF kc

kc = k’c . k”c


Among the many known examples of multiple equilibra is the ionization of diprotic acids in aqueous solutions. The following equilibrium constants have been determined for carbonic acid at 25 oC.

H2CO3(aq) H+(aq) + HCO3 -(aq) k’c= 4.2

x 10-7

HCO3-(aq) H+(aq) + CO3

-2(aq) k’’c= 4.8 x 10-11

H2CO3(aq) 2H+(aq) + CO3 -2(aq) kc= ?

kc = k’c . k”c

kc = (4.2 x 10-7) . (4.8 x 10-11) = 2.0 x 10-

17


• If a reversible reaction is written in the opposite direction, the equilibrium constant becomes the reciprocal of the original equilibrium constant.

• The value of k also depends on how the equilibrium equation is balanced. (If you double the chemical equation throughout, the corresponding equilibrium constant will be the square of the original value.)

The balanced chemical equation must be written when quoting the numerical value of an equilibrium constant.


The equilibrium constant, kc, for the reaction

2HCl(g) H2(g) + Cl2(g)

Is 4.17 x 10-34 at 25 oC. What is the equilibrium constant for the reaction

½H2(g) + ½Cl2(g) HCl(g)

at the same temperature?

4.17 x 10-34

1= 4.90 x 1016


Q, the Reaction QuotientThe equilibrium constant kc for the formation of hydrogen iodide from molecular hydrogen and molecular iodine in the gas phase

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

is 54.3 at 430 oC. Suppose that in a certain experiment we place 0.243 mole of H2, 0.146 mole of I2, and 1.98 moles of HI all in a 1.00 L container at 430 oC. Will there be a net reaction to form more H2 and I2 or more HI?

[1.98]2

[0.243][0.146]

= = 111

[HI]2 [H2][I2]

Q=


Because the quotient [HI]2/[H2][I2] is greater than kc, this system is not at equilibrium. Consequently, some of the HI will react to form more H2 and I2 (decreasing the value of the quotient). Thus the net reaction proceeds from right to left to reach equilibrium.

• For reactions that have not reached equilibrium, such as the formation of HI considered above, we obtain the reaction quotient (Qc), instead of the equilibrium constant by substituting the initial concentrations into the equilibrium constant expression.


To determine the direction in which the net reaction will proceed to achieve equilibrium, we compare the values of Qc and kc.

• If Qc > kc, the reaction proceeds from right to left (consuming products, forming reactants) to reach equilibrium

• If Qc < kc, the reaction proceeds from left to right (consuming reactants, forming products) to reach equilibrium.

• If Qc = kc, the initial concentrations are equilibrium concentrations. The system is at equilibrium.


Calculating concentrations in an equilibrium system If we know the equilibrium constant for a particular reaction, we can calculate the concentrations in the equilibrium mixture from the initial concentrations.

Let’s consider the following system involving two organic compounds, cis-stilbene and trans-stilbene

cis-stilbene trans-stilbene

The equilibrium constant for this system is 24.0 at 200 oC.


Suppose that initially only cis-stilbene is present at a concentration of 0.850 mol/L.

How do we calculate the concentrations of cis and trans-stilbene at equilibrium?

Initial 0.850 0.000

Change

Equilbrium

cis- trans- stilbene stilbene


When the system proceeds towards equilibrium, cis-stilbene will decrease by x, and trans-stilbene will increase by the same value, x, (because of the 1:1 ratio in the BCE).

The equilibrium concentration of cis-stilbene is initial concentration minus x; trans-stilbene will be initial concentration plus x.

Initial 0.850 0.000

Change -x +x

Equilbrium (0.850 – x) x

cis- trans- stilbene stilbene

27

Next, we set up the equilibrium constant expression

kc = [trans-stilbene]

[cis-stilbene]

24.0

=

x

0.850 - x

x = 0.816 M c

[cis-stilbene] = (0.850- 0.816) M = 0.034 M

[trans-stilbene] = 0.816 M

20.4 – 24.0 x = x

20.4 = 25.0 x


Factors that Affect Equilibrium

Chemical equilibrium represents a balance between forward and reverse reactions. In most cases, this balance is quite delicate. Changes in experimental conditions (concentration, pressure, volume and temperature) may disturb the balance and shift the equilibrium position so that more or less of the desired product is formed. When we say that an equilibrium position shifts to the right, for example, we mean that the net reaction proceeds from left to right.


Le Chatelier’s PrincipleThere is a general rule that helps us to

predict the direction in which an equilibrium reaction will move when a change in concentration, pressure, volume, or temperature occurs. The rule, known as Le Chatelier’s principle, states that if an external stress is applied to a system at equilibrium, the system adjusts in such a way that the stress is partially offset.

The word “stress” here means a change in concentration, pressure, volume, or temperature that removes a system from the equilibrium state.


Changes in concentration

FeSCN+2(aq) Fe+3(aq) + SCN-(aq) red pale yellow colorless

An aqueous solution of Fe(SCN)3


In this case, the stress applied to the equilibrium system is an increase in the concentration of SCN- (from the dissociation of NaSCN). To offset this stress, some Fe+3 ions react with the added SCN- ions, and the equilibrium shifts from right to left, and the red color of the solution deepens.

What happens if we add some sodium thiocyanate, NaSCN, to the solution?



Similarly, if we added iron (III) nitrate to the original solution, the red color would also deepen because the additional Fe+3 ions (from the Fe(NO3)3 solution) would shift the equilibrium from right to left.



Now suppose we add some oxalic acid, H2C2O4, to the original solution. Oxalic acid ionizes in water to form the oxalate ion, C2O4

-2, which binds strongly to the Fe+3 ions. This formation of Fe(C2O4)3

-3 removes free Fe+3 ions in solution. Consequently, more FeSCN+2 units dissociate and the equilibrium shifts from left to right, and the solution will turn yellow.



Changes in pressure do not affect the concentrations of solids, liquids or aqueous solutions because they are virtually incompressible. On the other hand, concentrations of gases are greatly affected by changes in pressure.

Changes in Volume and Pressure


Suppose that the equilibrium system

N2O4(g) 2NO2(g)

is in a cylinder fitted with a movable piston.

What happens if we increase the pressure on the gases by pushing down on the piston at constant temperature?

Brown NO2(g) and colorless N2O4(g) in equilibrium.


Since the volume decreases, the concentration of both NO2 and N2O4 increases. Because the concentration of NO2 is squared in the equilibrium constant expression, the increase in pressure increases the numerator more than the denominator. The system is no longer at equilibrium, so we write

[NO2]2

[N2O4]Qc =

37

Qc > Kc, and the net reaction will shift to the left until Qc = kc.

(a)Brown NO2(g) and colorless N2O4(g) in equilibrium.

(b)When the volume in the syringe is suddenly decreased, the concentration of NO2 and N2O4 are both increased, (indicated by the darker brown color).

(c)A few seconds after the sudden volume decrease, the color is much lighter brown as the equilibrium shifts from the brown NO2(g) to colorless N2O4(g).

(a) (b) (c)


In general, an increase in pressure (decrease in volume) favors the net reaction that decreases the total number of moles of gases (the reverse reaction, in this case), and a decrease in pressure (increase in volume) favors the reaction that increases the total number of moles of gases.

For the reactions in which there is no change in the number of moles of gases, a pressure (or volume) change has no effect on the position of equilibrium.


We can also increase the pressure of a system without changing its volume by adding an inert gas (helium, for example) to the equilibrium system.

Adding helium to the equilibrium mixture at constant volume increases the total gas pressure and decreases the mole fractions of both NO2 and N2O4; but the partial pressure of each gas, given by the product of its mole fraction and total pressure, does not change. Thus the presence of an inert gas does not affect the equilibrium.


Changes in Temperature

A change in concentration, pressure, or volume may alter the equilibrium position, but it DOES NOT change the value of the equilibrium constant (k).

Only a change in temperature can alter the equilibrium constant.


N2O4(g) 2NO2(g) H = 58.0 kJ

The formation of NO2 from N2O4 is an endothermic process. Another way to write this reaction would be

The formation of N2O4 from NO2 is an exothermic process. This process could be written

2NO2(g) N2O4(g) H = -58.0 kJ

N2O4(g) + heat 2NO2(g)

2NO2(g) N2O4(g) + heat

or


What will happen if the following system

N2O4(g) 2NO2(g)

is heated at constant volume?


Since endothermic processes absorb heat from the surroundings, heating favors the dissociation of colorless N2O4 into brown NO2 molecules, (an endothermic process).

N2O4(g) 2NO2(g)


Another way of looking at this is that the stress is too much heat, so the reaction will shift to the right, away from the heat.



Cooling favors the exothermic process (the formation of colorless N2O4).

N2O4(g) 2NO2(g)


In other words, cooling the system results in too little heat, so the shift would be towards the heat.



The Effect of a Catalyst

A catalyst enhances the rate of a reaction by lowering the reaction’s activation energy of the forward and reverse reaction to the same extent. We can therefore conclude that the presence of a catalyst does not alter the equilibrium constant, nor does it shift the position of an equilibrium system.


Adding a catalyst to a reaction mixture that is not at equilibrium will simply cause the mixture to reach equilibrium sooner. The same equilibrium mixture could be obtained without the catalyst, but we might have to wait much longer for it to happen.


ksp, Solubility Equilibria

Precipitation reactions are important in industry, medicine, and everyday life.

Although useful, the solubility rules do not allow us to make quantitative predictions about how much of a given ionic compound will dissolve in water.


Consider a saturated solution of silver chloride that is in contact with solid silver chloride. The solubility equilibrium can be represented as

AgCl(s) Ag+(aq) + Cl-(aq)

Because salts are considered strong electrolytes, all the AgCl that dissolves in water is assumed to dissociate completely into Ag+ and Cl- ions.


Because the concentration of a solid is constant, the equilibrium constant for the dissolution of AgCl can be written as

ksp = [Ag+][Cl-]

where ksp is called the solubility product. In general, the solubility product of a compound is the product of the molar concentrations of the constituent ions, each raised to the power of its stoichiometric coefficient in the equilibrium expression.


Write the equilibrium expressions for the dissolution of

•magnesium fluoride

MgF2(s) Mg+2(aq) + 2F-(aq) ksp = [Mg+2][F-]2

•silver carbonate

•calcium phosphate

Ag2CO3(s) 2Ag+(aq) + CO3-2(aq) ksp =

[Ag+]2[CO3-2]

Ca3(PO4)2(s) 3Ca+2(aq) + 2PO4-3(aq) ksp = [Ca+2]3[PO4

-3]2


The value of ksp indicates the solubility of an ionic compound - the smaller the value, the less soluble the compound in water. However, in using ksp values to compare solubilities, you should choose compounds that have similar formulas, such as AgCl and ZnS, or CaF2 and Fe(OH)2.


For equilibrium reactions involving an ionic solid in aqueous solution, any one of the following conditions may exist:

• the solution is unsaturated, (a solution that contains less solute than it has the capacity to dissolve)

• the solution is saturated, (at a given temperature, the solution that results when the maximum amount of a substance has dissolved in a solvent), or

• the solution is supersaturated, (a solution that contains more of the solute than is present in a saturated solution).


Q, the ion productFor concentrations that do not correspond to

equilibrium conditions, we use the reaction quotient which in this case is called the ion product (Q), to predict whether a precipitate will form.


• For example, if we mix a solution containing Ag+ ions with one containing Cl- ions, then the ion product is given by

Q = [Ag+][Cl-]

• The Q reminds us that these concentrations do not necessarily correspond to those at equilibrium.


ksp for AgCl is 1.6 x 10-10. The possible relationships between Q and ksp for silver chloride are

Q< ksp [Ag+] [Cl-] < 1.6 x 10-10

Unsaturated solution

Q = ksp [Ag+] [Cl-] = 1.6 x 10-10

Saturated solution

Q > ksp [Ag+] [Cl-] > 1.6 x 10-10

Supersaturated solution; AgCl will precipitate out until the product of the ionic concentrations is equal to 1.6 x 10-10

58

Given the ksp of CaF2 is 3.2 x 10-11, will a ppt form when 0.100 L of 0.30 M Ca(NO3)2 is mixed with 0.200 L of 0.060 M NaF?

(In other words, is Q > ksp?)

0.100 L Ca+2 x 0.30 mol = .030 mol/.300 L = .10 M Ca+2

1 L

0.200 L F- x 0.060mol = .012 mol/.300 L = .040 M F-

1 L [Ca+2][F-]2 = Qsp

Qsp = (.10)(.040)2 = 1.6 x 10-4

Q > ksp, so yes, a ppt will form


There are two other ways to express a substance’s solubility:

• molar solubility, which is the number of moles of solute in one liter of a saturated solution (mol/L), and

• solubility, which is the number of grams in one liter of a saturated solution (g/L).

• Both solubility and molar solubility can be used to determine ksp.


The solubility of calcium sulfate is found experimentally to be 0.67 g/L. Calculate the value of ksp for calcium sulfate.

First we calculate the number of moles of CaSO4 dissolved in 1 L of solution:

0.67 g CaSO4 1 mol CaSO4

1 L soln 136.24g CaSO4

= 4.9 x 10-3 mol/Lx

CaSO4 Ca+2(aq) + SO4-2(aq),

thus at equilibrium,

[Ca+2] = 4.9 x 10-3 M and [SO4-2] = 4.9 x 10-3 M


Now we can calculate ksp:

ksp = [Ca+2][SO4-2]

= (4.9 x 10-3)(4.9 x 10-3)= 2.4 x 10-5


Sometimes we are given the value of ksp for a compound and are asked to calculate the compound’s molar solubility.

For example, the ksp of silver bromide is 7.7 x 10-13. Calculate the molar solubility of silver bromide.


First, we must identify the species present at equilibrium. Here we have Ag+ and Br- ions.

Let s be the molar solubility (in mol/L) of AgBr. Since one unit of AgBr yields one Ag+ and one Br- ion, at equilibrium both [Ag+] and [Br-] are equal to s.


We summarize the changes in concentrations as follows:

AgBr Ag+(aq) + Br-(aq)

Initial 0.00 0.00

Change +s +s

Equilibrium s s

65

ksp = [Ag+][Br-]

7.7 x 10-13 = (s)(s) = s2

s = 7.7 x 10-13 = 8.8 x 10-7 M

Therefore, at equilibrium

[Ag+] = 8.8 x 10-7 M

[Br-] = 8.8 x 10-7 M

Because 1 AgBr dissociates into 1

Ag+ and 1 Br-, the molar solubility of AgBr is also 8.8 x 10-7 M.


Common Ion EffectThe solubility product, ksp, is an

equilibrium constant; precipitation of an ionic compound from solution occurs whenever the ion product exceeds ksp for that substance.

In a saturated solution of AgCl, for example, the ion product [Ag+][Cl-] is, of course, equal to ksp. Furthermore, simple stoichiometry tells us that [Ag+]=[Cl-]. But this equality does not hold true in all situations.


Suppose we study a solution containing two dissolved substances that share a common ion, say, AgCl and AgNO3. In addition to the dissociation of AgCl, the following process also contributes to the total concentration of the common silver ions in solution:

AgNO3(s) Ag+(aq) + NO3-(aq)


If AgNO3 is added to a saturated AgCl solution, the increase in [Ag+] will make the ion product greater than the solubility product:

Q = [Ag+] [Cl-] > ksp

To reestablish equilibrium, some AgCl will precipitate out of the solution, as LeChatelier’s principle would predict, until the ion product is one again equal to ksp.

The effect of adding a common ion, then, is a decrease in the solubility of the salt (AgCl) in solution. Note that in this case [Ag+] is no longer equal to [Cl-] at equilibrium; rather, [Ag+]>[Cl-].


pH and Solubility

The solubilities of many substances also depend on the pH of the solution.

Consider the solubility equilibrium of magnesium hydroxide:

Mg(OH)2(s) Mg+2(aq) + 2OH-(aq)


Adding OH- ions (increasing the pH) shifts the equilibrium from right to left, thereby decreasing the solubility of Mg(OH)2.

On the other hand, adding H+ ions (decreasing the pH) shifts the equilibrium from left to right, (because H+ and OH- will form water, thus removing the OH- from solution) increasing the solubility of Mg(OH)2. Thus, insoluble bases tend to dissolve in acidic solutions.


FYI - Lewis acids and base reactions in which a metal cation (electron pair acceptor) combines with a Lewis base (electron pair donor) result in the formation of complex ions:

Ag+(aq) + 2NH3(aq) Ag(NH3)2+Lewis acid Lewis base

Transition metals have a particular tendency to form complex ions. For example, a solution of cobalt (II) chloride is pink because of the presence of the Co(H2O)6

+2 ions. When HCl is added, the solution turns blue as a result of the formation of the complex ion CoCl4-2

Co+2(aq) + 4Cl-(aq) CoCl4-2


Separation of Ions by Fractional Precipitation

In chemical analysis, it is sometimes desirable to remove one type of ion from solution by precipitation while leaving other ions in solution. For instance, the addition of sulfate ions to a solution containing both potassium and barium ions causes BaSO4 to precipitate out, thereby removing most of the Ba+2 ions from the solution. The other “product” K2SO4, is soluble and will remain in solution.

The BaSO4 precipitate can be separated from the solution by filtration or decanting.


Application of the Solubility Product Principle to Qualitative Analysis

Qualitative analysis is the determination of the types of ions present in a solution. (We will focus on the cations.)

• There are some twenty cations that can be analyzed readily in aqueous solution.

• These cations can be divided into groups according to the solubility products of their insoluble salts.


The separation of cations into groups is made as selective as possible; that is, the anions that are added as reagents must be such that they will precipitate the fewest types of cations.

The separation of cations at each step must be carried out as completely as possible.

Since an unknown solution may contain from one to all twenty ions, any analysis must be carried out systematically from Group I through group III.


A flow chart for the separation of cations in qualitative analysis

Solution containing ions of all cation groups

Solution containing ions of remaining groups

Solution containing ions of remaining groups

Solution contains

Na+, K+ and NH4+ ions

Group I precipitates

AgCl, Hg2Cl2, and PbCl2

Group II precipitates

CoS, FeS, MnS, NiS, ZnS, Al(OH)3 and Cr(OH)3

Group III precipitates

BaCO3, CaCO3 MgCO3 and SrCO3

HCl

NH4OH and (NH4)2S

decanting

decanting

decanting(NH4)2CO3


The End

Documents

Equilibrium 2009-20101 Chemical Equilibrium. Equilibrium 2009-20102 Chemical Equilibrium Few chemical reactions proceed in only one direction. Most are