Equilibrium Standard 7 Chapter 8. standard This topic illustrates that many chemical reactions are reversible and involve an equilibrium process. The.

Equilibrium Standard 7 Chapter 8

standard This topic illustrates that many chemical reactions are reversible and involve an equilibrium process. The consideration of the many factors that can affect an equilibrium is an important aspect of physical chemistry 7.1 chemical equilibria: reversible reactions; dynamic equilibrium 7.2 ionic equilibria

Learning outcome 7.1.a explain, in terms of rates of the forward and reverse reactions, what is meant by a reversible reaction and dynamic equilibrium 7.1.b state Le Chatelier’s principle and apply it to deduce qualitatively 7.1.c state whether changes in temperature, concentration or pressure of a catalyst affect the value of the equilibrium constant for a reaction 7.1.d deduce expressions for equilibrium constants in terms of concentrations, Kc, and partial pressure, Kp 7.1.e calculate the values of equilibrium constants in terms of concentrations or partial pressures from appropriate data 7.1.f calculate the quantities present at equilibrium, given appropriate data 7.1.g describe and explain the conditions used in the Haber process and the Contract process, as examples of the importance of an understanding of chemical equilibrium in the chemical industry

Learning outcomes 7.2.a show understanding of, and use, the Bronsted-Lowry theory of acids and bases, including the use of the acid-I base-I, acid-II base-II concept 7.2.b explain qualitatively the differences in behavior between strong and weak acids and bases and the pH values of their aqueous solutions in terms of the extend of dissociation

The Concept of Equilibrium Chemical equilibrium occurs when a reaction and its reverse reaction proceed at the same rate.

Irreversible reactions Most chemical into their reactions are considered irreversible – the products that are made cannot readily be changed back to the reactants. For example, when wood burns it is impossible to turn it back into unburnt wood again! Similarly, when magnesium reacts with hydrochloric acid to form magnesium chloride and hydrogen, it is not easy to reverse the reaction and obtain the magnesium.

What are reversible reactions? Reversible reactions occur when the backwards reaction (products  reactants) takes place relatively easily under certain conditions. The products turn back into the reactants. + ABCD + (reactants)(products) For example, during a reversible reaction reactants A and B react to make products C and D. However, products C and D can also undergo the reverse reaction, and react together to form reactants A and B.

Reversible and irreversible reactions What kind of reactions are reversible and irreversible?

Reversible biochemical reactions Many biochemical reactions (those that take place inside organisms) are reversible. For example, in the lungs, oxygen binds to haemoglobin (Hb) in red blood cells to create oxyhemoglobin When the red blood cells are transported to tissues, the oxyhemoglobin dissociates back to hemoglobin and oxygen There are also some very important industrial reactions, like the Haber process, that are reversible Hb + 4O 2 Hb.4O 2

Characteristics of equilibrium An equilibrium reaction has four particular features under constant conditions: 1.It is dynamic 2.The forward and reverse reactions occur at the same time 3.The concentrations of reactants and products remain constant at equilibrium 4.It requires a closed system

1 Dynamic equilibria Dynamic equilibrium means that the molecules or ions of reactants and products are continuously reacting. Reactants are continuously being changed to products and products are continuously being changed back to reactants. …going back and forth… …at the same time… …at the same rate…

2 The forward and backward reactions occur at the same time At equilibrium the rate of the forward reaction equals the rate of the backward reaction. Molecules or ions of reactants are becoming products, and those in the products are becoming reactants at the same time. Consider: aW + bX ↔ cY + dZ When the rate of the forward reaction is equal to the rate of the reverse reaction, the system has reached “dynamic equilibrium” Rate FWD = Rate REV

3. The concentrations of reactants and products remain constant at equilibrium At equilibrium, the quantities of everything present in the mixture remain constant, although the reactions are still continuing. This is because the rates of the forward and the back reactions are equal. Once equilibrium is achieved, the amount of each reactant and product remains constant.

In an ordinary reaction; all reactants end up as products; there is 100% conversion CONCENTRATION CHANGE IN A REACTION As the rate of reaction is dependant on the concentration of reactants... the forward reaction starts off fast but slows as the reactants get less concentrated FASTEST AT THE START SLOWS DOWN AS REACTANTS ARE USED UP TOTAL CONVERSION TO PRODUCTS THE STEEPER THE GRADIENT, THE FASTER THE REACTION

4. Equilibrium requires a closed system Closed system: one in which none of the reactants or products escapes from the reaction mixture Open system: matter is lost to the surroundings

DYNAMIC EQUILIBRIUM IMPORTANT REMINDERS a reversible chemical reaction is a dynamic process everything may appear stationary but the reactions are moving both ways the position of equilibrium can be varied by changing certain conditions Trying to get up a “down” escalator gives an idea of a non-chemical situation involving dynamic equilibrium.

DYNAMIC EQUILIBRIUM Summary When a chemical equilibrium is established... both the reactants and the products are present at all times the equilibrium can be approached from either side the reaction is dynamic - it is moving forwards and backwards the concentrations of reactants and products remain constant

What happens in a reversible reaction? Consider: aA + bB ↔ cC + dD Where : A reacts with B to produce C and D a, b, c, and d are the coefficients of the balanced equation 1)A is mixed with B and begins to react quickly A and B are at maximum concentrations C and D are not present at the beginning 2)The “forward” reaction (A + B) continues, but is slowing 3)When enough product is formed, the “reverse” reaction begins A and B concentrations are decreasing C and D concentrations are increasing

aA + bB ↔ cC + dD At the beginning of the reaction, the concentrations of A and B were at their maximum. That means that the rate of the reaction was at its fastest As A and B react, their concentrations fall. That means that they are less likely to collide and react, and so the rate of the forward reaction falls as time goes on

aA + bB ↔ cC + dD In the beginning, there isn't any C and D, so there can't be any reaction between them. As time goes on, though, their concentrations in the mixture increase and they are more likely to collide and react. With time, the rate of the reaction between C and D increases

aA + bB ↔ cC + dD Eventually, the rates of the two reactions will become equal. A and B will be converting into C and D at exactly the same rate as C and D convert back into A and B again At this point there won't be any further change in the amounts of A, B, C and D in the mixture. As fast as something is being removed, it is being replaced again by the reverse reaction. We have reached a position of dynamic equilibrium

Dynamic Equilibrium The reaction does not stop! Products are still being formed Products are still combining to reform the reactants Nothing “appears” to be happening Concentrations stop changing Color changes cease, etc… The only thing “equal” about equilibrium are the rates of the forward and reverse reactions Equilibrium might be reached when there is mostly products, mostly reactants, or maybe a 50/50 mix of both

Depicting Equilibrium In a system at equilibrium, both the forward and reverse reactions are being carried out; as a result, we write its equation with a double arrow N 2 O 4 (g) 2 NO 2 (g)

The Equilibrium Constant

Equilibrium Consider the reaction At equilibrium, the forward reaction: N 2 O 4 (g)  2 NO 2 (g), and the reverse reaction: 2 NO 2 (g)  N 2 O 4 (g) proceed at equal rates. Chemical equilibria are dynamic, not static – the reactions do not stop.

Equilibrium Let’s use 2 experiments to study the reaction each starting with a different reactant(s). Exp #2 pure NO 2 Exp #1 pure N 2 O 4

Equilibrium Experiment #1 Exp #1 pure N 2 O 4

Equilibrium Experiment #2 Exp #2 pure NO 2

Equilibrium Are the equilibrium pressures of NO 2 and N 2 O 4 related? Are they predictable?

The Equilibrium Constant Forward reaction: N 2 O 4 (g)  2 NO 2 (g) Rate law: Rate = k f [N 2 O 4 ] Reverse reaction: 2 NO 2 (g)  N 2 O 4 (g) Rate law: Rate = k r [NO 2 ] 2 N 2 O 4 (g) 2 NO 2 (g)

The Equilibrium Constant Therefore, at equilibrium Rate f = Rate r k f [N 2 O 4 ] = k r [NO 2 ] 2 Rewriting this, it becomes kfkrkfkr [NO 2 ] 2 [N 2 O 4 ] = N 2 O 4 (g) 2 NO 2 (g)

The Equilibrium Constant The ratio of the rate constants is a constant at that temperature, and the expression becomes K eq = kfkrkfkr [NO 2 ] 2 [N 2 O 4 ] =

The Equilibrium Constant To generalize this expression, consider the reaction The equilibrium expression for this reaction would be: where K c is the equilibrium constant K c = [C] c [D] d [A] a [B] b aA + bBcC + dD

THE EQUILIBRIUM CONSTANT K c for an equilibrium reaction of the form... aA + bB cC + dD then (at constant temperature) [C] c. [D] d = a constant, (K c ) [A] a. [B] b where [ ] denotes the equilibrium concentration in mol dm -3 K c is known as the Equilibrium Constant

THE EQUILIBRIUM CONSTANT K c for an equilibrium reaction of the form... aA + bB cC + dD then (at constant temperature) [C] c. [D] d = a constant, (K c ) [A] a. [B] b where [ ] denotes the equilibrium concentration in mol dm -3 K c is known as the Equilibrium Constant VALUE OF K c AFFECTED by: a change of temperature NOT AFFECTED by: a change in concentration of reactants or products a change of pressure adding a catalyst

Equilibrium Constant, K c For any equilibrium reaction: The rates of the forward and back reaction are equal when equilibrium is established. The reactions in an equilibrium process are elementary reactions. The rates can be described as: The equilibrium constant, K, is equal to the ratio of the rate constants of the forward and reverse reactions. It can be expressed in terms of either concentrations or pressures (subscripts are used to distinguish). KcKc

The value of K c is constant at a given temperature. Reactants and products that are pure solids or pure liquids DO NOT appear in the equilibrium constant expression. Values of K larger than 1 indicate that the forward reaction is faster and more favored. Values of K less than 1 indicate that the reverse reaction is faster and more favored. Equilibrium Constant, K c

The Equilibrium Constant This constant value is termed the equilibrium constant, K c, for this reaction at 25°C.

The Equilibrium Constant For the NO 2 / N 2 O 4 system: equilibrium constant expression equilibrium constant Note: at 100°C, K = 6.45

The Equilibrium Constant Because pressure is proportional to concentration for gases in a closed system, the equilibrium expression can also be written K p = (P C ) c (P D ) d (P A ) a (P B ) b aA + bBcC + dD

Relationship between K c and K p From the ideal gas law we know that Rearranging it, we get PV = nRT P = RT nVnV

Relationship between K c and K p Plugging this into the expression for K p for each substance, the relationship between K c and K p becomes Where K p = K c (RT)  n  n = (moles of gaseous product) − (moles of gaseous reactant) P = RT nVnV

Equilibrium Can Be Reached from Either Direction As you can see, the ratio of [NO 2 ] 2 to [N 2 O 4 ] remains constant at this temperature no matter what the initial concentrations of NO 2 and N 2 O 4 are.

Equilibrium Can Be Reached from Either Direction This is the data from the last two trials from the table on the previous slide.

Equilibrium Can Be Reached from Either Direction It does not matter whether we start with N 2 and H 2 or whether we start with NH 3. We will have the same proportions of all three substances at equilibrium.

Dynamic Equilibrium Note: the initial amounts of reactants and products do not matter Example: N 2 O 4(g) ↔ 2 NO 2(g) Start with only [N 2 O 4 ] ….. at eq., [N 2 O 4 ] = 0.83M and [NO 2 ] = 0.33M Start with only [NO 2 ]…. at eq., [N 2 O 4 ] = 0.83M and [NO 2 ] = 0.33M

Dynamic Equilibrium Note : 1)[products] on top and [reactants] on bottom 2)Coefficients become exponents

Dynamic Equilibrium K = the “equilibrium constant” K is unitless K is temperature dependent as long as the temperature is constant, so is K for a reversible reaction

Dynamic Equilibrium Note : aW (aq) + bX (s) ↔ cY (aq) + dZ (l) Pure substances (solids, liquids) do not have a changeable [molarity] and so drop out of the equilibrium expression

K – the equilibrium constant The size of K tells us something about the equilibrium “position” i.e. what are the concentrations of the reactants and products at equilibrium? Because the products are in the numerator, the larger the K value, the more products that are present at equilibrium and the fewer reactants that remain.

K – the equilibrium constant General rule: K<10 -4 the equilibrium mixture is mostly reactants The reaction does not “proceed” very far forward in order to reach equilibrium The smaller K is, the fewer products formed

K – the equilibrium constant General rule: 10 -4 <K<10 4 the equilibrium mixture has substantial amounts of reactants and products Not necessarily a “50/50” mix, but reasonably similar amounts of reactants and products

K – the equilibrium constant General rule: K>10 4 the equilibrium mixture is mostly products The larger K gets, the more the forward reaction “goes to completion”

Write equilibrium constant expressions of the following reactions White Board

Dynamic Equilibrium …going back and forth… …at the same time… …at the same rate…

LeChatelier’s Principle If a system at equilibrium is disturbed it will respond in the direction that counteracts the disturbance and re-establishes equilibrium Disturbed(?) 1.add/remove a chemical 2.change temperature 3.change pressure (gases) 4.add/remove catalyst

1. adding/removing a chemical If you add a chemical, the system tries to “remove” it This is done by reacting it away This uses up the chemicals on its “side” of the equation and making more of the chemicals on the other “side” Equilibrium is re-established (K), but the individual concentrations are different

1. adding/removing a chemical Consider: A + B ↔ C + D If you add more A… The system tries to remove it by reacting it away, which makes more products [C]  [D]  [B] ↓ It is said the equilibrium has “shifted to the right” or “shifted towards the products”

1. adding/removing a chemical Consider: A + B ↔ C + D If you add more C… The system tries to remove it by reacting it away, which makes more reactants [A]  [B]  [D] ↓ It is said the equilibrium has “shifted to the left” or “shifted towards the reactants”

1. adding/removing a chemical Consider: A + B ↔ C + D If you remove some B… The system tries to replace it by reacting to make more of it (and whatever else is on its side of the equation) [A]  [C] ↓ [D] ↓ It is said the equilibrium has “shifted to the left” or “shifted towards the reactants”

1. adding/removing a chemical Consider: A + B ↔ C + D If you remove some D… The system tries to replace it by reacting to make more of it (and whatever else is on its side of the equation) [C]  [A], [B] ↓ The reaction is driven forward in this case, or towards the products

Just how does one “remove” a chemical? Note: [D] is the concentration of D Solids do not have a molarity because they are not dissolved into anything If one product in an aqueous system is a solid, the solid is called a “precipitate” and is not in the equilibrium mixture This “drives” the reaction forward Double replacements

Just how does one “remove” a chemical? Same for gases in an open container They can bubble out of the mixture (leave) Ex: opening a soda bottle H 2 CO 3(aq) ↔ H 2 O (l) + CO 2(g) Ex: Mg (s) + 2 HCl (aq) ↔ H 2(g) + MgCl 2(aq) If the container is open, the reaction just keeps going forward

2. Changing the volume Remember Boyle’s Law Changing the volume of a container of gases changes their pressure as well Inverse relationship If V↓, P  If V , P↓

2. Changing the volume If V↓, P  The equilibrium will shift to try to make the P↓ How is this done? Shift to whichever side has less gas Fewer moles of a gas Coefficients in the balanced equation Less gas means lower pressure

2. Changing the volume Ex: N 2(g) + 3 H 2(g) ↔ 2 NH 3(g) 4 moles of gas in the reactants, 2 in products If V↓, P  …the system will try to make P↓ by shifting to the products (less gas) Every time the reaction proceeds forward, 4 moles of gas becomes 2…which means the P↓ Vice versa if V 

3. Changing the temperature Consider : A + B ↔ C + D + Heat For this system… The forward reaction is exothermic The reverse reaction is endothermic Treat heat as if it were a substance being added or removed Add heat, equilibrium shifts away from the side with heat [A],[B]  [C],[D]↓ Remove heat, equilibrium shifts toward the side with heat [A],[B]↓ [C],[D] 

4. Catalytic effect Adding or removing a catalyst has no effect on the value of K The activation energy is lowered for the forward and the reverse reaction, and they both speed up by the same amount, so Rate FWD still = Rate REV If not at equilibrium, it will be reached quicker is a catalyst is used.

CATALYSTS FACTORS AFFECTING THE POSITION OF EQUILIBRIUM EaEa MAXWELL-BOLTZMANN DISTRIBUTION OF MOLECULAR ENERGY EXTRA MOLECULES WITH SUFFICIENT ENERGY TO OVERCOME THE ENERGY BARRIER MOLECULAR ENERGY NUMBER OF MOLECUES WITH A PARTICULAR ENERGY Catalysts work by providing an alternative reaction pathway involving a lower activation energy.

LeChatelier’s Principle If a system at equilibrium is disturbed it will respond in the direction that counteracts the disturbance and re-establishes equilibrium Disturbed 1.add/remove a chemical 2.change temperature 3.change pressure (gases) 4.add/remove catalyst Discuss the following situations with others at your table

Now…lets try it out… Ex: PCl 3(g) + Cl 2(g) ↔ PCl 5(g) ΔH= -88kJ How will [Cl 2 ] at equilibrium be changed by… Adding some PCl 3 ? System will try to react the PCl 3 away More products are formed To do this, more Cl 2 is consumed [Cl 2 ] ↓

Now…lets try it out… Ex: PCl 3(g) + Cl 2(g) ↔ PCl 5(g) ΔH= -88kJ How will [Cl 2 ] at equilibrium be changed by… Adding some PCl 5 ? The system will try to react it away More reactants are formed [Cl 2 ] 

Now…lets try it out… Ex: PCl 3(g) + Cl 2(g) ↔ PCl 5(g) ΔH= -88kJ How will [Cl 2 ] at equilibrium be changed by… Increasing the temperature? Heat is added, the system shifts away from the side with heat ΔH = negative, so heat is a product System shifts towards the reactants [Cl 2 ] 

Now…lets try it out… Ex: PCl 3(g) + Cl 2(g) ↔ PCl 5(g) ΔH= -88kJ How will [Cl 2 ] at equilibrium be changed by… Decreasing the volume? If V↓, P  System shifts towards the side with less gas to make P↓ Product side has fewer moles of gas [Cl 2 ] ↓

Whenever a change is made to a reversible reaction in dynamic equilibrium, the equilibrium will shift to try and oppose the change. Increasing the temperature shifts the equilibrium in the direction that takes in heat. Increasing the concentration of a substance shifts the equilibrium in the direction that produces less of that substance. Increasing the pressure shifts the equilibrium in the direction that produces less gas. Temperature Concentration Pressure Condition Effect

I.C.E. Tables: Calculating Equilibrium Concentrations Suppose you are carrying out a reaction, and you wish to determine either the value of the equilibrium constant, or the equilibrium concentrations of your products and reactants (if K is given); To do this, you would plug the starting concentrations into an I.C.E (initial, change, equilibrium) table and tabulate the change in each concentration as the reaction proceeds

I.C.E. Table Calculations: Finding K A closed system initially contains 1.0 x 10 -3 M H 2 and 2 x 10 -3 M I 2 at 448 o C. The equilibrium concentration of HI is 1.87 x 10 -3 M. Calculate K c. Concentration (M)H2H2 I2I2 2 HI Initial.001.0020 Change Equilibrium To determine the equilibrium concentrations of H 2 and I 2, we must determine the change in the initial concentration. We use the stoichiometry to do this We see that the change in HI is twice the change in H 2 and I 2 (1:2 stiochiometry). -x +2x.001 - x.002 - x.00187

Concentration (M)H2H2 I2I2 2 HI Initial.001.0020 Change-(.000935) +2(.000935) Equilibrium.000065.001065.00187 I.C.E. Table Calculations x =.000935 M We can easily determine the equilibrium concentrations. Now, we can calculate K c The magnitude of K c (much greater than 1) suggests that there is substantially more product than reactant at equilibrium. The reaction is favored to the right

I.C.E Tables: Calculating Equilibrium Concentrations from K Suppose that 2.00 moles of N 2 O 4 are injected into a 1.00 L reaction vessel held at 100 o C. Calculate the equilibrium concentrations. Set up the I.C.E. table. We don’t know the equilibrium concentrations, so we leave them in terms of x. Concentration (M)N2O4N2O4 2NO 2 Initial2.000 Change Equilibrium -x+2x 2.00-x2x

We plug in our equilibrium concentrations and the given value of K c Expand and cross multiply to get all terms on one side of the equation. We are now left with a QUADRATIC EQUATION (ax 2 +bx+c = 0). Concentration (M)N2O4N2O4 2NO 2 Initial2.000 Change- x+ 2x Equilibrium2.00 - x2x

A Blast From the Past: Solving the Quadratic Equation To solve a quadratic equation, you must use the quadratic formula Since all quadratic formulas are of the form ax 2 + bx +c = 0, we identify these values and plug them into the formula. Quadratics yield two results. a = 4 b = 0.2 c = -0.4 Check your values of x to determine the correct one. Only one value will work. X

Concentration (M) N2O4N2O4 2NO 2 Initial2.000 Change- (.29)+ 2 (.29) Equilibrium1.71.58 Plug the calculated value of x back into the I.C.E. table to determine the equilibrium concentrations (x = 0.29 M). DO NOT FORGET THIS STEP! I.C.E Tables: Calculating Equilibrium Concentrations from K

Relating Concentrations to Partial Pressures Recall from the gas laws that concentration is related to pressure: Therefore, equilibrium constants can be expressed in terms of partial pressures. Since solids and liquids do not contribute to the overall pressure, they are not included in the expression (as we learned previously). When using pressures, we write the equilibrium expression as K p. The subscript p means that K is in terms of pressure.

Example A mixture of H 2 and N 2 is allowed to attain equilibrium at 472 o C. The equilibrium mixture was found to contain 7.38 atm H 2, 2.46 atm N 2, and 0.166 atm NH 3. Find K p ? The value of K p is very small (less than 1), which tells us that the reaction is favored to the left.

LeChatlier’s Principle LeChatlier’s Principle states that: If a chemical reaction at equilibrium is subjected to a change in conditions that displaces it from equilibrium, the reaction adjusts toward a new equilibrium state. The reaction will proceed in the direction that offsets the change.

LeChatlier’s Principle For example, let’s say we have the following reaction at 25 o C: Assume that our equilibrium concentration of CO is 6M, that of CO 2 is 4M. As we learned earlier, we can express K c as: Now, we add more CO 2 (g) to the system, until [CO 2 ] = 8 M. WE ARE NO LONGER IN EQUILIBRIUM !!!. Thus, the state of the system is no longer described by the equilibrium constant K c, but by the reaction quotient, Q. The subscript ‘o’ indicates a non-equilibrium concentration

Reaction Quotients, Q The direction of spontaneity is always toward equilibrium. The value of Q tells us the direction in which a system not at equilibrium will proceed to reach equilibrium.

Back to the Example At a given temperature, the equilibrium constant DOES NOT CHANGE. To reestablish equilibrium (K c = 9), the reaction shifts right to consume some of the added CO 2 This will increase [CO] and decrease [CO 2 ] until K c equals 9 again. When equilibrium is re-established, the new concentrations will be:

LeChatlier’s Principle Applied To Volume and Pressure When you have gaseous reactants and products, changes to the volume and pressure of the system induce shifts in the equilibrium. The general rule is, for any equilibrium involving gases, decreasing the volume drives the equilibrium toward the side with the smaller number of moles of gas. From Boyle’s Law, we know that pressure and volume are inversely proportional, so increasing pressure has the same effect as lowering volume. Equilibrium Decreasing V shifts reaction left Increasing V shifts reaction right kFkF kBkB

LeChatlier’s Principle Applied To Changing Temperature Remember, all reactions must be either endothermic or exothermic In an endothermic process, heat is absorbed into the system in order for the reaction to proceed. Therefore, since heat is used in the reaction, heat may be considered a reactant Let’s use our previous example. We can rewrite this reaction as: Increasing the temperature of an endothermic process drives the equilibrium to the right because you are adding a reactant.

LeChatlier’s Principle Applied To Changing Temperature The opposite must then be true for an exothermic process. Since heat leaves the system in an exothermic process, heat may be treated as a product For the following exothermic reaction: We can rewrite the reaction as: Increasing the temperature of an exothermic process drives the equilibrium to the left because you are adding a product.

Let’s take the reaction below. Assume that the reaction is at equilibrium. The equilibrium concentrations are given. We can calculate K c from the equilibrium values K c =.056 Now, additional CO 2 is added to the system, raising [CO 2 ] to 1.25 M. This will cause the equilibrium to shift right. But how far? If we can ‘freeze time’ right at the instant that the CO 2 is added (before the shift), we would have the following: Calculations Based On LeChatlier’s Principle 0.75 M0.60 M 0.10 M0.25 M Concentration(M)CO 2 H2H2 COH2OH2O Initial1.250.600.100.25 Change-x +x Equilibrium1.25-x0.60-x0.10+x0.25+x

Calculations Based On LeChatlier’s Principle Now we solve this as usual. Convert to quadratic form. Use quadratic formula. Solve for x New equilibrium concentrations: [CO 2 ]= 1.215 M, [H 2 ]= 0.5651 M [CO] =.1349 M, [H 2 O] =.2849 M x = 0.0349M Concentration(M)CO 2 H2H2 COH2OH2O Initial1.250.600.100.25 Change-x +x Equilibrium1.25-x0.60-x0.10+x0.25+x

Group Example The reaction above is enclosed in a vessel at 25 o C. At equilibrium, you have 0.171 atm of A and 4.89 atm of C. Then, the system is disturbed by a 35% decrease in the pressure of A. What are the new equilibrium pressures of A and C? When performing an equilibrium calculation involving LeChatlier’s principle, you must first answer the following 1.What is the equilibrium expression? 2.What is the value of K p ? 3.What are the initial pressures (at the instant before the reaction proceeds)? 4.In which direction does the disturbance shift the equilibrium?

PressureA2C Initial(.111)4.89 Change+x-2x Equilibrium.111+x4.89-2x No longer at equilibrium. Must shift LEFT to reestablish K p When we plug our new equilibrium pressures into K p, the value of K p MUST be 139.8 x =.0525 atm P A =.1635 atm P C = 4.785 atm

Haber: Chemical warfare Chemical warfare can be traced back to ancient Greece: Spartans crept up to Athens with noxious bundles of wood, pitch and sulfur; lit them; and fled to outside the city walls waiting for coughing Athenians to leave their homes unguarded. Thus the invention of a “stink bomb” Developing better chemical warfare continued up until 1899. The Hague Convention of 1899 banned chemical-based weapons in war. All scientifically advance countries except one signed the compact. Who didn’t sign? The US thought it was hypocritical for countries to sign this but take pleasure in mowing down 18 year olds with machine guns and sink ships with torpedoes in the dark and let the sailors drown in the dark sea.

Haber: Chemical warfare Since the Hague Pact only concerned itself with using lacrimators (a tear- producing substance as tear gas) in warfare it did not stop France from testing on its own citizens French government used ethyl bromoacetate to bring down a ring of Parisian bank robbers in 1912 1914 France lobbed bromine shells at advancing German troops, unfortunately for France it was a windy day and the German troops did not know that they were “attacked” Word of this spread and Germany used it to their advantage. Germans had a CO leak in their barracks and blamed it on France so they could justify their own chemical warfare program

Haber: Chemical warfare Fritz Haber was a famous scientist around 1900 when he discovered how to convert the nitrogen in the air into a fertilizer. Nitrogen is as important to plants as vitamin C is to humans. Nitrogen makes up about 80% of the air we breathe but it rarely reacts with anything and so rarely becomes “fixed” in the soil Haber reactions: Haber heated nitrogen to hundreds of degrees, injected hydrogen gas and then cranked up the pressure to 100 atm. Reaction required osmium as a catalyst and voila (always wanted to put this word into a ppt) you get ammonia NH3. Ammonia is the precursor to fertilizer By WWI, Haber had likely saved millions from Malthusian starvation and even today we can still thank him for feeding most of the worlds population

Haber: Chemical warfare Haber did not care about fertilizers, he was trying to build nitrogen explosives for the Germans similar to what Timothy McVeigh used to blow a hole in the Oklahoma City courthouse in 1995 German leaders recruited Haber for their gas warfare division, this division was called “the Haber office” and the military even promoted Haber, a 46 year old Jewish convert to Lutheran to captain. This position did not sit well with his wife Clara, who herself was brilliant. First woman to earn a PhD from Breslau (now Wroclaw) university. Marie Curie married an open-minded man like Pierre Curie who backed his wife’s work while Clara married Fritz Haber who put her down and did not allow her to have a career

Haber: Chemical warfare Clara spent her time translating her husbands work into English but she refused to work on the chemical warfare. Germany got around the pesky Hague Pact by interpreting a part of it to their advantage. By signing the pact, Germany had agreed to not use projectiles as the sole method of spreading the gas. They added shrapnel to the gas. They came up with a shell filled with xylyl bromide. They launched it against Russia but the Russian temperature was so cold that the xylyl bromide froze solid Lots of trial and error. Eventually the Germans gave up on bromine and set its sites on bromines cousin chlorine

Haber: Chemical warfare Chlorine turns its victims skin yellow, green and back. It glasses over the eyes with cataracts. People end up dying by drowning from the fluid buildup in their lungs. This chlorine gas had many names in history but known as mustard gas Haber also worked on a biological law which bares his name, Haber’s Rule which quantifies the relationship between gas concentration, exposure time and death rate. Clara demanded that Haber stop working on gas warfare. He did not listen to her. She ended up shooting herself with his army pistol. He did not bother to stay and make funeral arrangements but left the next morning to continue his work

Haber: Chemical warfare Germany did lose the war. Before the dust had settled on WWI, Haber had won the Nobel Prize in Chemistry A year later he was prosecuted for being an international war criminal. Germany and Haber had huge reparations to pay to the Allies. He invented an insecticide, Zyklon A Eventually a new regime with a short memory took over Germany and the Nazis soon exiled Haber for his Jewish roots. He died will seeking refuge. Work on Haber’s insecticide continued and within a few years the Nazis were gassing millions of Jews, including relatives of Haber, with the second-generation gas…. Zyklon B

Equilibrium Standard 7 Chapter 8. standard This topic illustrates that many chemical reactions are reversible and involve an equilibrium process. The.

Similar presentations

Presentation on theme: "Equilibrium Standard 7 Chapter 8. standard This topic illustrates that many chemical reactions are reversible and involve an equilibrium process. The."— Presentation transcript:

Similar presentations

About project

Feedback

Log in

Auth with social network:

Equilibrium Standard 7 Chapter 8. standard This topic illustrates that many chemical reactions are reversible and involve an equilibrium process. The.

Similar presentations

Presentation on theme: "Equilibrium Standard 7 Chapter 8. standard This topic illustrates that many chemical reactions are reversible and involve an equilibrium process. The."— Presentation transcript:

Similar presentations

About project

Feedback