Reaction EquilibriumReaction Equilibrium

Do any reactions truly go to Do any reactions truly go to completion??completion??

IntroductionIntroduction

Any reaction in a closed system never truly Any reaction in a closed system never truly stops reacting due to the fact that the stops reacting due to the fact that the molecules involved are still interacting, still molecules involved are still interacting, still colliding. Reactions in open systems colliding. Reactions in open systems cancan go to completion, however. go to completion, however.

Reactions in closed systems will eventually Reactions in closed systems will eventually appear to reach completion. At this point, appear to reach completion. At this point, the reaction has reached a point called the reaction has reached a point called equilibrium.equilibrium.

EquilibriumEquilibrium

A point in a reaction in which the rate of A point in a reaction in which the rate of the forward reaction is the same as the the forward reaction is the same as the rate of the reverse reaction. rate of the reverse reaction.

Another way to express this concept is Another way to express this concept is that equilibrium is the point at which that equilibrium is the point at which reactants becomes products reactants becomes products at the same at the same raterate as products become reactants. as products become reactants.

Refer to the graph below – NOTE: Refer to the graph below – NOTE: This shows reaction rate vs. timeThis shows reaction rate vs. time

Refer to the graph below – NOTE: Refer to the graph below – NOTE: This plots Concentration of substances vs. TimeThis plots Concentration of substances vs. Time

Refer to the graph below – NOTE: Refer to the graph below – NOTE: This plots Concentration of substances vs. TimeThis plots Concentration of substances vs. Time

Extra PracticeExtra Practice

Look at and answer question #10 on the Look at and answer question #10 on the practice test.practice test.

There are two main types of There are two main types of equilibriaequilibria

Homogeneous equilibriumHomogeneous equilibrium – has everything – has everything present in the same phase. The usual examples present in the same phase. The usual examples include reactions where everything is a gas, or include reactions where everything is a gas, or everything is in the same solution.everything is in the same solution.

Heterogeneous equilibriumHeterogeneous equilibrium – has substances – has substances present in more than one phase. The usual present in more than one phase. The usual examples include reactions involving solids & examples include reactions involving solids & liquids, or solids & gases.liquids, or solids & gases.

How do we represent a reaction at How do we represent a reaction at equilibrium?equilibrium?

Reactions at equilibrium are represented Reactions at equilibrium are represented using the double arrow.using the double arrow.

Examples:Examples: 2CO(g) 2CO(g) CO CO22(g) + C(s)(g) + C(s)

HH22(g) + I(g) + I22(g) (g) 2HI(g) 2HI(g)

NN22(g) + O(g) + O22(g) (g) 2NO(g) 2NO(g)

How do we write an equilibrium How do we write an equilibrium expression?expression?

If you recall from the Kinetics unit, the proposed If you recall from the Kinetics unit, the proposed rate of a reaction is dependent on the rate of a reaction is dependent on the concentration of the reactants raised to the concentration of the reactants raised to the power of the coefficients of each reactant in the power of the coefficients of each reactant in the equation. The equilibrium expression is treated equation. The equilibrium expression is treated the same way, except the rate of the reaction in the same way, except the rate of the reaction in both directions is considered.both directions is considered.

See the derivation of the expression in class, but See the derivation of the expression in class, but first…first…

Writing an equilibrium expression – for a Writing an equilibrium expression – for a homogeneous equilbrium expressionhomogeneous equilbrium expression

We are going to look at a general case with the equation:We are going to look at a general case with the equation:aA + bB aA + bB cC + dD cC + dD

No state symbols have been given, but they will be all (g), No state symbols have been given, but they will be all (g), or all (l), or all (aq) if the reaction was between substances or all (l), or all (aq) if the reaction was between substances in solution in water.in solution in water.

If you allow this reaction to reach equilibrium and then If you allow this reaction to reach equilibrium and then measure the equilibrium concentrations of everything, you measure the equilibrium concentrations of everything, you can combine these concentrations into an expression can combine these concentrations into an expression known as an known as an equilibrium constantequilibrium constant..

The equilibrium constant always has the same value The equilibrium constant always has the same value (provided you don't change the temperature), irrespective (provided you don't change the temperature), irrespective of the amounts of A, B, C and D you started with. It is also of the amounts of A, B, C and D you started with. It is also unaffected by a change in pressure or whether or not you unaffected by a change in pressure or whether or not you are using a catalyst.are using a catalyst.

Writing an equilibrium expression – for a Writing an equilibrium expression – for a heterogeneous equilibrium expressionheterogeneous equilibrium expression

The equilibrium expression is written the The equilibrium expression is written the same as that on the previous slide, except, same as that on the previous slide, except, pure solids are excluded from the pure solids are excluded from the expression.expression.

Example: CaCOExample: CaCO33(s) (s) CaO(s) + CO CaO(s) + CO22(g)(g)

Kc = [COKc = [CO22(g)](g)]

Homework!!Homework!!

On the practice test given in class, answer On the practice test given in class, answer questions 1 – 4.questions 1 – 4.

What does Kc or Kp tell us?What does Kc or Kp tell us?

First, you can use KFirst, you can use Kc c or Kor Kpp compared to Q, compared to Q, the reaction quotient, to determine the the reaction quotient, to determine the direction a reaction is going.direction a reaction is going.

Second, you can use KSecond, you can use Kcc or K or Kpp to determine to determine the extent to which a reaction will proceed.the extent to which a reaction will proceed.

Third, you can use Third, you can use Le’Chatlier’s PrincipleLe’Chatlier’s Principle to predict the direction of a reaction at to predict the direction of a reaction at equilibrium after changing one of the equilibrium after changing one of the reaction conditions.reaction conditions.

What is a reaction quotient?What is a reaction quotient?

A reaction quotient is found using the A reaction quotient is found using the same equation as is used to find the same equation as is used to find the equilibrium constant. The difference is equilibrium constant. The difference is that the reaction quotient uses actual that the reaction quotient uses actual concentrations and pressures, whereas concentrations and pressures, whereas equilibrium constants are calculated using equilibrium constants are calculated using concentrations and pressures of concentrations and pressures of substances at equilibrium.substances at equilibrium.

How to we compare equilibrium How to we compare equilibrium constants to reactions quotients?constants to reactions quotients?

If Q > Kc (or Kp), then the amount of products is If Q > Kc (or Kp), then the amount of products is too high and the reaction needs to go to the left too high and the reaction needs to go to the left to use up product and make more reactant.to use up product and make more reactant.

If Q = Kc (or Kp), the reaction is at equilibrium.If Q = Kc (or Kp), the reaction is at equilibrium.

If Q < Kc (or Kp), the amount of product is too If Q < Kc (or Kp), the amount of product is too small and the reaction is favoring the formation small and the reaction is favoring the formation of product, and therefore going to the right.of product, and therefore going to the right.

Let’s Practice comparing Q to KcLet’s Practice comparing Q to Kc Let’s look at the synthesis of Hydrogen Iodide from it’s elements:Let’s look at the synthesis of Hydrogen Iodide from it’s elements:

HH22(g) + I(g) + I22(g) (g) 2HI(g) 2HI(g)For each of the following sets of concentrations, determine whether the For each of the following sets of concentrations, determine whether the

reaction is at equilibrium. If it isn’t, decide what direction it must go reaction is at equilibrium. If it isn’t, decide what direction it must go to reach equilibrium.to reach equilibrium.

a) a) [H[H22] = [I] = [I22] = [HI] = 0.010 M] = [HI] = 0.010 M

b)b) [HI] = 0.30 M; [H[HI] = 0.30 M; [H22] = 0.01 M; [I] = 0.01 M; [I22] = 0.15 M] = 0.15 M

c)c) [H[H22] = [HI] = 0.10 M; [I] = [HI] = 0.10 M; [I22] = 0.0010 M] = 0.0010 M

Let’s try one more involving Q and KcLet’s try one more involving Q and Kc Nitrosyl Chloride is an orange gas that dissociates at Nitrosyl Chloride is an orange gas that dissociates at

high temperatures into chlorine and nitric oxide:high temperatures into chlorine and nitric oxide:

2NOCl(g) 2NOCl(g) 2NO(g) + Cl 2NO(g) + Cl22(g)(g)In a certain experiment, 2.0 moles No, 3.0 moles Cl2, and 1.4 moles In a certain experiment, 2.0 moles No, 3.0 moles Cl2, and 1.4 moles

NOCl were introduced into a 10 liter container.NOCl were introduced into a 10 liter container.

A)A) What is the value of the reaction quotient under these What is the value of the reaction quotient under these conditions?conditions?

B)B) If Kc for this reaction under the same temperature If Kc for this reaction under the same temperature conditions is 4.25, determine the direction of this conditions is 4.25, determine the direction of this reaction.reaction.

You Try!!!You Try!!!

Answer Question #5 on the practice test.Answer Question #5 on the practice test.

What about the extent a reaction is What about the extent a reaction is proceeding?proceeding?

If Kc or Kp is very large (>> 1), the reaction If Kc or Kp is very large (>> 1), the reaction favors the formation of product. What this favors the formation of product. What this means is that there is more product at means is that there is more product at equilibrium than reactant.equilibrium than reactant.

If Kc or Kp is small (<< 1), the reaction does not If Kc or Kp is small (<< 1), the reaction does not favor the formation of product. An example of a favor the formation of product. An example of a reaction of this type may be the following:reaction of this type may be the following:

NN22(g) + 2O(g) + 2O22(g) (g) 2NO 2NO22(g) Kc @ room (g) Kc @ room

temperature = 1 x 10temperature = 1 x 10-30-30

Do you understand??Do you understand??

Answer question 10 on the practice testAnswer question 10 on the practice test

What isLe’Chatlier’s Principle and how What isLe’Chatlier’s Principle and how is it used to determine shifts in is it used to determine shifts in

reactions at equilibrium?reactions at equilibrium?

There are three factors affecting the position of There are three factors affecting the position of equilibrium and hence Kc:  equilibrium and hence Kc:  TemperatureTemperature, , pressurepressure and and concentration.concentration.  If a reaction at   If a reaction at equilibrium is subjected to a change in any of equilibrium is subjected to a change in any of these factors the position of equilibrium will shift these factors the position of equilibrium will shift to counteract the change.  Think of it as a to counteract the change.  Think of it as a naughty child - ask a naughty child to sit down naughty child - ask a naughty child to sit down and it stands up!  This is known as and it stands up!  This is known as Le Le Chateliers PrincipleChateliers Principle

TemperatureTemperature

To predict what effect this will have on a reaction we will To predict what effect this will have on a reaction we will need to know the enthalpy change of the reaction.  In the need to know the enthalpy change of the reaction.  In the example below the enthalpy change for the reaction is a example below the enthalpy change for the reaction is a negative number, indicating that the forward reaction is negative number, indicating that the forward reaction is exothermic.  The backward reaction is therefore exothermic.  The backward reaction is therefore endothermic.endothermic.

Example:Example: NN22 (g)    +    3H (g)    +    3H22 (g)    (g)    2NH 2NH33 (g) + energy (g) + energy

An increase in temperature will result in the backward An increase in temperature will result in the backward reaction speeding up (to use up the excess heat energy) reaction speeding up (to use up the excess heat energy) so the equilibrium will shift to the left hand side, Kc will so the equilibrium will shift to the left hand side, Kc will decrease. decrease.

Pressure Pressure

As with temperature, it helps if you add some simple As with temperature, it helps if you add some simple labels to the equilibrium equation. labels to the equilibrium equation.

Example: NExample: N22 (g)    +    3H (g)    +    3H22 (g)   (g)     2NH   2NH33 (g) (g)

        HIGH PRESSURE              LOW PRESSUREHIGH PRESSURE              LOW PRESSURE This side has a total of  This side has a total of This side has a total of  This side has a total of  4 moles of reactants 2 moles of products 4 moles of reactants 2 moles of products

An increase in pressure would cause the equilibrium to An increase in pressure would cause the equilibrium to shift to the low pressure side, Kc would increase.  Note shift to the low pressure side, Kc would increase.  Note that pressure only affects a reaction involving gases. that pressure only affects a reaction involving gases.

Concentration Concentration If a substance becomes more concentrated in a reaction, If a substance becomes more concentrated in a reaction,

the equilibrium will shift to reduce its concentration. the equilibrium will shift to reduce its concentration. Example: NExample: N22 (g)    +    3H (g)    +    3H22 (g)  (g)     2NH    2NH33 (g) (g)

If the ammonia made in the above reaction is removed, If the ammonia made in the above reaction is removed, then the equilibrium position will shift to the right  hand then the equilibrium position will shift to the right  hand side to make additional ammonia to replace what was side to make additional ammonia to replace what was taken away.taken away.

If extra ammonia is added to the reaction, the reaction If extra ammonia is added to the reaction, the reaction shifts to the left to use up the excess ammonia gas.shifts to the left to use up the excess ammonia gas.

Catalyst Catalyst

A catalyst speeds up both the forward and A catalyst speeds up both the forward and backward reactions, it doesn't change the backward reactions, it doesn't change the position of equilibrium but does allow the position of equilibrium but does allow the reaction to reach equilibrium quicker. reaction to reach equilibrium quicker.

What is a practical application of What is a practical application of Le’Chatlier’s Principle??Le’Chatlier’s Principle??

What are the optimal conditions under which to What are the optimal conditions under which to make ammonia gas from it’s elements?make ammonia gas from it’s elements?

In the manufacture of ammonia in the Haber process, In the manufacture of ammonia in the Haber process, the conditions that would give the best yield of ammonia the conditions that would give the best yield of ammonia would be a low temperature and a high pressure.would be a low temperature and a high pressure.

In reality a temperature of 500 degrees centigrade and a In reality a temperature of 500 degrees centigrade and a pressure of 200 atmospheres is used.  The high pressure of 200 atmospheres is used.  The high temperature is used  because it speeds up the formation temperature is used  because it speeds up the formation of ammonia and the relatively low pressure is used of ammonia and the relatively low pressure is used because running a plant at a high pressures requires because running a plant at a high pressures requires specialized and expensive machinery.specialized and expensive machinery.

You Try!!!You Try!!!

On the practice test, complete question #6On the practice test, complete question #6

More sample questions involving More sample questions involving equilibrium constants!!equilibrium constants!!

Given the equilibrium constant and the Given the equilibrium constant and the initial concentrations of the reactants, can initial concentrations of the reactants, can you find the equilibrium concentrations of you find the equilibrium concentrations of the reactions?the reactions?

Can you look at solubility-product Can you look at solubility-product constants (Kconstants (Kspsp) and qualitatively determine ) and qualitatively determine

the extent of solubility?the extent of solubility?

How do I find Kc or the equilibrium How do I find Kc or the equilibrium concentrations?concentrations?

Calculating Kc from a known set of equilibrium Calculating Kc from a known set of equilibrium concentrations seems pretty clear. You just plug concentrations seems pretty clear. You just plug into the equilibrium expression and solve for Kc.into the equilibrium expression and solve for Kc.

Calculating equilibrium concentrations from a set Calculating equilibrium concentrations from a set

of initial conditions takes more calculation steps. of initial conditions takes more calculation steps. In this type of problem, the Kc value will be given In this type of problem, the Kc value will be given as well as the initial concentrations of reactants, as well as the initial concentrations of reactants, you simply need to find the EQUILIBRIUM you simply need to find the EQUILIBRIUM CONCENTRATIONS!! CONCENTRATIONS!!

The best way to explain how to find The best way to explain how to find equilibrium concentrations from initial values equilibrium concentrations from initial values

and Kc is by example.and Kc is by example.

Given this equation:Given this equation:

HH22(g) + I(g) + I22(g) (g) 2 HI(g) 2 HI(g)

Calculate all three equilibrium Calculate all three equilibrium concentrations when [Hconcentrations when [H22]]oo = [I = [I22]]oo = 0.200 M = 0.200 M

and Kc = 64.0.and Kc = 64.0.

Here’s the example:Here’s the example:

The solution technique involves the use of The solution technique involves the use of an ICE box. Here is an empty one:an ICE box. Here is an empty one:

        [H[H22(g)](g)] [I[I22(g)](g)] [HI][HI]

IInitialnitial

CChangehange

EEquilibriumquilibrium

Let’s keep working!!Let’s keep working!!

Fill in the first row with given data:Fill in the first row with given data: Note – the initial concentration of HI = 0.0 M as the reaction has Note – the initial concentration of HI = 0.0 M as the reaction has

not started yet!not started yet!

[H[H22(g)](g)] [I[I22(g)](g)] [HI][HI]

IInitialnitial 0.200 M0.200 M 0.200 M0.200 M 0.0 M0.0 M

CChangehange

EEquilibriumquilibrium

Keep it up!!!Keep it up!!!Now for the change row. This is the one that causes the most difficulty in understanding:

  [H2] [I2] [HI]

Initial 0.200 0.200 0

Change - x - x + 2x

Equilibrium      

The minus sign comes from the fact that the H2 and I2 amounts are going to go down as the

reaction proceeds.

X signifies that we know some H2 and I2 get used up, but we don't know how much. What we do

know is that an EQUAL amount of each will be used up. We know this from the coefficients of the equation. For every one H2 used up, one I2 is used up also.

The positive signifies that more HI is being made as the reaction proceeds on its way to equilibrium.The 2 is important. HI is being made twice as fast as either H2 or I2 are being used up.

In fact, always use the coefficients of the balanced equation as coefficients on the "x" terms.In problems such as this one, never use more than one unknown. Since we have only one equation (the equilibrium expression) we cannot have two unknowns.

Keep going!!!Keep going!!!

The equilibrium row should be easy. It is The equilibrium row should be easy. It is simply the initial conditions with the simply the initial conditions with the change applied to it:change applied to it:

[H[H22(g)](g)] [I[I22(g)](g)] [HI][HI]

IInitialnitial 0.200 M0.200 M 0.200 M0.200 M 0.0 M0.0 M

CChangehange - X- X -XX + 2X+ 2X

EEquilibriumquilibrium 0.200 – X 0.200 – X -.200 – X -.200 – X 2X2X

Keep going!!!Keep going!!!Now we are ready to put values into the equilibrium expression. For

convenience, here is the equation again:

HH22(g) + I(g) + I22(g) (g) 2 HI(g) 2 HI(g)

The equilibrium expression is:

Kc = [HI]2

[H2][I2]

Plugging values into the expression gives

64.0 = (2x)2 ((0.200 - x) (0.200 - x))

Keep going!!Keep going!!

Two points need to be made before going Two points need to be made before going on:on:

1) Where did the 64.0 value come from? 1) Where did the 64.0 value come from? It was given in the problem.It was given in the problem.

2) Make sure to write (2x)2) Make sure to write (2x)22 and not 2x2. As you and not 2x2. As you well know, they are different. This mistake well know, they are different. This mistake happens a LOT!! happens a LOT!!

Almost Done!!Almost Done!!

Both sides are perfect squares (done so Both sides are perfect squares (done so on purpose), so we square root both sides on purpose), so we square root both sides to get:to get:

8.00 = (2x) / (0.200 - x)8.00 = (2x) / (0.200 - x)

From there, the solution should be easy From there, the solution should be easy and results in x = 0.160 Mand results in x = 0.160 M

Not we are finished!!!Not we are finished!!!

This is not the end of the solution since the This is not the end of the solution since the question asked for the equilibrium question asked for the equilibrium concentrations, so:concentrations, so:

[H[H22] = 0.200 - 0.160 = 0.040 M] = 0.200 - 0.160 = 0.040 M

[I[I22] = 0.200 - 0.160 = 0.040 M] = 0.200 - 0.160 = 0.040 M

[HI] = 2 (0.160) = 0.320 M [HI] = 2 (0.160) = 0.320 M

What if both sides are NOT perfect What if both sides are NOT perfect squares??? squares???

This example will involve the use of the This example will involve the use of the quadratic formulaquadratic formula. . (Pause for whining and moaning. Go ahead, get (Pause for whining and moaning. Go ahead, get

it out of your system! Now, back to chemistry.)it out of your system! Now, back to chemistry.)

Given this equation:Given this equation:

PClPCl33 + Cl + Cl2 2 <===> PCl<===> PCl55

Calculate all three equilibrium concentrations Calculate all three equilibrium concentrations when Kc = 16.0 & [PClwhen Kc = 16.0 & [PCl55]]oo = 1.00 M. = 1.00 M.

Here is the completed Here is the completed ICEICE box: box:

[PCl[PCl33] ] [Cl [Cl22] [PCl] [PCl55] ]

IInitial nitial 0 M 0 M 0 M 0 M 1.00 M 1.00 M CChange + x + x - xhange + x + x - x EEquilibrium x x 1.00 - x quilibrium x x 1.00 - x

The equilibrium expression is:The equilibrium expression is:

Kc = Kc = [PCl[PCl55] ]

([PCl([PCl33] [Cl] [Cl22])])

Substituting gives:Substituting gives:

16.0 = 16.0 = 1.00 - x1.00 - x

(x)(x)(x)(x)

After suitable manipulation (which you can After suitable manipulation (which you can perform yourself), we arrive at this quadratic perform yourself), we arrive at this quadratic

equation in standard form:equation in standard form: 16x16x22 + x - 1 = 0 + x - 1 = 0

Using the quadratic formula, which is:Using the quadratic formula, which is: x = (- b ± square root[bx = (- b ± square root[b22 - 4ac]) / 2a, - 4ac]) / 2a,

we obtain:we obtain: x = (- 1 + square root[12 - (4) (16) (-1)]) / 32x = (- 1 + square root[12 - (4) (16) (-1)]) / 32

After suitable calculations, we find x = 0.221.After suitable calculations, we find x = 0.221.

Why was the one answer not Why was the one answer not used???used???

Please notice that the negative root was Please notice that the negative root was dropped, because negative b turned out to be dropped, because negative b turned out to be negative one. The answer obtained in this type negative one. The answer obtained in this type of problem CANNOT be negative.of problem CANNOT be negative.

Why?Why?

Because we are dealing with the amount of a Because we are dealing with the amount of a physical substance in mol / L. Amounts of physical substance in mol / L. Amounts of substances are always represented with positive substances are always represented with positive numbers. An amount of a substance with numbers. An amount of a substance with physical reality cannot be represented with physical reality cannot be represented with negative numbers.negative numbers.

What if we get two positive What if we get two positive answers??answers??

The question then becomes how to determine which root is the The question then becomes how to determine which root is the correct one to use?correct one to use?

Let’s look at one final example:Let’s look at one final example:

Given this equation:Given this equation:

COClCOCl22 <===> CO + Cl <===> CO + Cl22

Calculate all three equilibrium concentrations Calculate all three equilibrium concentrations when Kc = 0.680 with [CO]when Kc = 0.680 with [CO]oo = 0.500 and [Cl = 0.500 and [Cl22]]oo = =

1.00 M.1.00 M.

Here is the completed ICE box: Here is the completed ICE box:

    [COCl[COCl22] [CO] [Cl] [CO] [Cl22] ]

Initial 0 M 0.500M 1.00 M Initial 0 M 0.500M 1.00 M Change + x - x - xChange + x - x - x Equilibrium x 0.500 - x 1.00 - x Equilibrium x 0.500 - x 1.00 - x

Use your Kc expression:Use your Kc expression:

The equilibrium expression is:The equilibrium expression is:

Kc = Kc = ([CO] [Cl([CO] [Cl22])])

[COCl[COCl22]]

Substituting into the expression gives:Substituting into the expression gives:

0.680 = 0.680 = ((0.5 - x) (1 - x))((0.5 - x) (1 - x))

xx

After some manipulation, we arrive at this quadratic After some manipulation, we arrive at this quadratic equation, in standard form:equation, in standard form:

xx22 - 2.18x + 0.5 = 0 - 2.18x + 0.5 = 0

Using the quadratic formula, we have this to Using the quadratic formula, we have this to start:start:

x = (2.18 ± square rootx = (2.18 ± square root[[(2.18)(2.18)22 - (4) (1) (0.5) - (4) (1) (0.5)]] ) / 2 ) / 2

After some manipulation we arrive at:After some manipulation we arrive at:(2.18 ± 1.66)(2.18 ± 1.66)

22

Both roots yield positive values, so how do we Both roots yield positive values, so how do we pick the correct one?pick the correct one?

What is the final answer???What is the final answer???

The answer lies in the fact that x is not the The answer lies in the fact that x is not the final answer, whereas (0.5 - x) is. It is the final answer, whereas (0.5 - x) is. It is the term (0.5 - x) which must be positive.term (0.5 - x) which must be positive.

So the root of 1.92 is rejected in favor of So the root of 1.92 is rejected in favor of the 0.26 value and the day is saved!!!the 0.26 value and the day is saved!!!

You Try!!! You Try!!!

Answer questions 8 & 9 on the practice Answer questions 8 & 9 on the practice test!!!test!!!