Chemical Equilibrium Be sure you download this powerpoint from the website

Chemical Equilibrium

Be sure you download this powerpoint from the website.

The Concept of Chemical Equilibrium

• Chemical equilibrium exists when the rates of the forward and reverse reactions are equal and the concentrations of the reactants and products remain constant. – In other words…

+ Breactant C + Dproduct

– Occurs in a closed system, and no observable changes are detected.

↔


• Chemical Equilibrium-involves different substances as reactants

• Physical equilibrium- involves equilibrium between the phase changes of the same substance. – Example: physical equilibrium of water in a closed

system:

Physical equilibrium

– Physical equilibrium of water in a closed system:

• How would this look in an open system?• At a high temperature (above freezing)?

• At a low temp (below freezing)?• At a REALLY low temp (really below freezing)?

The Equilibrium Constant (K)

• AP exam questions will either give you K to calculate a concentration of a substance, or give you concentrations to calculate K.– K is the mathematical expression of the law of

mass action– Law of mass action• For a reversible reaction at equilibrium and a constant

temperature, a certain ratio of reactant and product concentrations has a constant value, K.


• General Equation:K=

Where:aA + bB + dD

Mole coefficients are represented by a,b,c,d.Concentrations are expressed in Molarity (n/L)



• Explained in English: the equilibrium constant is equal to the concentration of products divided by the concentration of reactants.

• Concentrations are multiplied to the exponent of their mole coefficients (a, b, c, d).



• When K > 1, there are more products than reactants.

• When K<1, there are more reactants than products.

• When K=1, they are equal.


• For a homogenous equilibrium-all reacting species are in the same phase.

• Example: N2O4(g)2NO2(g)

Dinitrogen tetraoxide gas “dissociates” in to Nitrogen dioxide gas. A reversible reaction that

exists in equilibrium in a closed system.


• In a homogenous equilibrium:N2O4(g)2NO2(g)

= Subscript K indicates that concentrations are expressed

in molarity. When concentrations are expressed as partial pressures, concentration is represented by Kp.


• When concentrations are expressed as partial pressures, concentration is represented by Kp.

N2O4(g)2NO2(g)

Our equation then becomes:Kp =

Where Ppartial= Xi*PT


• Remember that is our mole fraction for a heterogeneous mixture of gases:

= • And Total pressure is the sum of the individual

partial pressures:Ptotal=∑P1∞

AND…

PV=nRT


• Therefore:N2O4(g)2NO2(g)

1. Kp =

2. Kp= =(RT)b-a

Where: b-a = ∆nA helpful hint is to read the equations to yourself

as if you were translating them from a foreign language to English.


• Since:

Kp= =(RT)b-a

• And because: b-a = ∆nThen:

Kp = (RT)∆n =Kc(RT) ∆n

because = And :

Kp = Kc (0.0821*T) ∆n

Homework…

• Read Read Read Read Read Read Read Read Read Read Read Read Read Read Read Read Read chapter on Gases and on Chemical Equilibrium!!!!!!

• You will have a quiz tomorrow on gas stoichiometry. I will assign you homework on equilibrium tomorrow. You will have a quiz on equilibrium Thursday.


Part II

Heterogeneous Equilibrium

• Heterogeneous Equilibrium-results when reactants and products are in different phases. – Example: calcium carbonate is heated in a closed

system, the following equilibrium is attained:CaCO3(s)CaO(s) + CO2(g)

The presence of two solids and one gas represents three separate (not different) phases.


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

• The concentration of a solid is dependent on its density, not how much of the substance we have. Therefore; We can use our conversion factors to calculate concentration. – For example, Copper (Cu) has density of 8.96 g/cm3

Therefore, [Cu] == .141mol/cm3

Or141 mol/L


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

The density of CaCO3(s) is 2.71 g/cm3

The density of CaO(s) is 3.35 g/cm3

And since BY DEFINITION the concentration of a solid is itself a constant AT CONSTANT TEMPERATURE and pressure;

K’c=, since they are constant:

K’c=Kc=[CO2]…wth…? CO2 is the only product that is a gas.


• Example:– (NH4)2Se(s) 2NH3(g) + H2Se(g)

• Remember that Kc=

• so: Kc= ?• And since the concentration of a solid is constant, we

disregard it, therefore:

Kc=

If the concentrations are given in molarity…if the concentrations are given in partial pressures, things

get real…


• If the concentrations are given in partial pressures(NH4)2Se(s) 2NH3(g) + H2Se(g)

Kp=P

From:Kc=

Since the solid has no partial pressure, we ignore it.

Just in case you were wondering why…

• If you were to calculate the pressure of a solid using the ideal gas equation PV=nRT where [A]=n/v– Lets use Copper again…

– Copper (Cu) has density of 8.96 g/cm3

– A concentration of 141 mol/L at STP

– Therefore; P/RT=n/v, assume STP, and one liter of substance.

P==3,160 atmThis means we would need 3160atm to convert a gas into a solid. In other words, you cant use the Ideal gas equation with solids…

Multiple Equilibria

• The reactions we have considered are relatively simple, a more complicated situation is one in which the product molecules in one equilibrium system are in coved in a second process.

aA + bB↔cC + dDcC + dD ↔ eE + fF

Multiple Equilibria

• When:aA + bB↔cC + dDcC + dD ↔ eE + fF

• Then:– Kc’ (for the first reaction)= – Kc’’ (for the second reaction)=

Multiple Equilibria

• When: aA + bB ↔ cC + dDcC + dD ↔ eE + fF

• (don’t write this down because it is confusing) The overall reaction is given by the sum of the two reactions:

aA + bB ↔ cC + dD K’ccC + dD ↔ eE + fF K’’caA + bB ↔ eE + fF Kc

Multiple Equilibria

• The overall reaction is given by the sum of the two reactions:

aA + bB ↔ cC + dD K’c cC + dD ↔ eE + fF K’’c aA + bB ↔ eE + fF Kc

• Then the equilibrium constant for the overall reaction =

Kc=

Multiple Equilibria

• Therefore: if a reaction can be expressed as the sum of two or more reactions, the equilibrium constant for the overall reaction is given by the product of the equilibrium constants of the individual reactions

Multiple Equilibria

• In other words:– Kc for multiple reactions:

Kc = (Kc’) x (Kc’’)

SO: Kc = Kc’ x Kc’’

The relationship between equilibrium and rate of reaction (kinetics)

• For equilibrium reactions, we have both forward and reverse reactions, therefore;

aA + bB ↔ cC + dD

• Where Kf is our forward equilibrium constant, and Kr is our reverse constant.

Ratef = Rater

Kf

Kr

The relationship between equilibrium and rate of reaction (kinetics)

• If, Ratef = Rater

• And Ratef = kf [A]a[B]b (reactants)

• And Rater = kr[C]c[D]d (products)– Then;

= kc =

Next Week

• We will cover – The reaction quotient (Q)– Le Chateliers Principle.

– Please read sections 13.5, 13.6, and 13.7 in your book.

Homeworkdue Thursday.

• Pgs 650-651, 17-35 odd #’s• Pgs 651-653, #’s 37-45 odd, 51, 55, 57.


Part III

The reaction quotient (Q)

• For reactions that have not reached equilibrium, we obtain the reaction quotient (Q) instead of the equilibrium constant (K).– Simply put, the reaction quotient uses the initial

concentrations of reactants and products in the same equation as K.

– You then compare Q to K to determine whether or not the reaction is at equilibrium. • Simple enough.

The reaction quotient (Q)

• Therefore, Q=

When we are given K and asked to find Q:Q<K, more reactants than productsQ=K, system is at equilibriumQ>K, more products than reactants.

The system must balance itself to get to equilibrium.

Example

• Consider the following reaction:N2(g) + 3H2(g) ↔2NH3(g)

• at the start of the reaction there are .249 mol of N2, 3.21 x 10-2 mol H2, and 6.42 x 10-4 mol NH3 in a 3.5L reaction vessel at 375°C, If the equilibrium constant (K) is 1.2, decide whether the system is in equilibrium. If it is not, predict which way the net reaction will occur.

Example

N2(g) + 3H2(g) ↔2NH3(g)

1. First calculate the concentrations (n/v)

2. Then calculate Q, Q=

3. compare Q to K(given)4. predict direction of reaction to reach

equilibrium

Example

• Q=.611• Q < K• Therefore, there are more reactants than

products, and the reaction must proceed from left to right to reach equilibrium.

LeChatlier’s Principle

• if an external stress is applied to a system at equilibrium, the system will adjust in such a way that the stress will is partially offset as the system reaches a new equilibrium position.

• Stresses include: change in Temp, volume, pressure, or a change in moles.

LeChatlier’s Principle

A classic example from Biology:• Your body is in a state of dynamic equilibrium with

your environment. When you dramatically change environments, you body feels the effects and works to return to equilibrium. This is a process called “acclimatization”.

A classic example from Biology

• The combination of oxygen with a hemoglobin molecule produces the polymer Oxyhemoglobin which transports oxygen to body tissues.

Hb(aq) + O2 ↔ HbO2(aq)


• At higher altitudes the concentration of oxygen decreases, thus leading to a state of disequilibrium. This is known as altitude sickness and can be fatal.

• Yuppies call this “the height hangover”.

So according to LeChatlier:


• A decrease in oxygen would shift the equilibrium from left to right. In other words, the left side of the arrow above would be bigger that the left; and Q >K. This would mean that there is more HbO2 than what the environment realistically allows as a result of a smaller O2 concentration.

• Your body (as the system) reacts by creating more Hb to capture more oxygen. This process takes weeks, even years.

Tomorrow

• We will wrap up this chapter – Calculating equilibrium concentrations– Addressing system stresses

Use the rest of the period to work on your homework.


Part IV

Calculating Equilibrium Concentrations

• If we know the equilibrium constant K for a particular reaction, we can calculate the equilibrium concentrations from the initial concentrations Using the ICE method– Initial, Change, Equilibrium

• We do this by solving algebraically.

Calculating Equilibrium Concentrations

• Consider the following reaction of cis-stilbene and tran-stilbene, two isomers that exist in equilibrium in a closed system.

• Kc for this system at 200° C is 24.0. if the initial concentration of cis-stilbene is .850 mol/L, and no trans-stilbene, what is the concentrations for both substances at equilibrium?– Solve algebraically using the ICE method

Cis-stilbene ↔ trans-stilbeneInitial (M): .850 0Change(M): -x +xEquilibrium (M): (.850 – x) x

So: Kc==

Solve for x

Example #2

• A mixture of .500 mol of H2 and .500 mol of I2 was placed in a 1L flask at 430°C. The equilibrium constant Kc for the reaction is 54.3 at this temperature. Calculate the concentrations of H2 and I2 at equilibrium.

H2(g) + I2(g) ↔ 2HI(g)

Example #2

H2(g) + I2(g) ↔ 2HI(g)

I (M): .500 .500 0C (M): -x -x +2xE (M):

So: Kc= 54.3, solve algebraically

Example #2

• X=.393 M• [H2]=.107M• [I2]=.107M• [HI]=.786M

Example #3same method, different execution

• For the same reaction as Example #2, the initial concentrations of .00623 mol of H2, .00414 mol of I2, and .0224 mol of HI. Calculate the concentrations at equilibrium.

• Your first instinct will be to calculate K, but you cant since you have the initial concentrations. In this case you have to calculate Q first and compare to K to determine the direction of the reaction.

Example #3

Qc==

• Qc===19.5

• Q>K, so more products than reactants when the system is at equilibrium. So the negative variables of change will be the reactants.

Example #3

H2(g) + I2(g) ↔ 2HI(g)

I (M): .00623 .00414 .0224C (M): -x -x +2xE (M):

so: Kc= 54.3, solve algebraically

Example #3

• By this time you have something like this:54.3(2.58 x 10-5 - .0104x + x2)=5.02 x 10-4 + .0896x + 4x2

Now collect terms, we get54.3x2 - .654x + 8.98 x 10-4 =0

You cant solve for x using the square method, so you have to use the quadratic formula:

X=

Example #3

X=X = .0114 M or .00156 M

.0114 is impossible because it exceeds our original concentrations.

Now calculate the equilibrium concentrations for each.

Addressing system stress…

• Here's what you need to know. The only thing that can act on a system and change its K, is temperature.

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Chemical Equilibrium Be sure you download this powerpoint from the website