chapter 10 : (general equilibrium). mr belland presents - how my son and i play

CHAPTER 10 : (GENERAL EQUILIBRIUM)

MR BELLAND PRESENTS -HOW MY SON AND I PLAY

Chemical equilibrium is when two opposing reactions occur simultaneously and at the same rate.

For example:A + B → C + D

and C + D → A + B

This is usually written as a reversible equation using double arrows:

A + B ⇌ C + D

What equilibrium is not:•equal amounts of products and reactants•permanent•instantly establish•a stop in the chemical reaction•for reactions that go to completion (like dissociation of a strong acid)•the same for every condition of T

Example case 1:in a 1 L vesselR 2SO2 + O2 ⇌ 2SO3

I 0.400 mol 0.200 mol 0C -0.056 mol -0.028 mol + 0.056 molE 0.344 mol 0.172 mol 0.056 mol

in 1 L vessel the above mols are also M

Describe what occurred here.

Example case 2:in a 1 L vesselR 2SO2 + O2 ⇌ 2SO3

I 0 0 0.500 molC +0.424 mol +0.212 mol -0.424 molE 0.424 mol 0.212 mol 0.076 mol

in 1 L vessel the above mols are also M

Describe what occurred here.

Suppose we want a rate law:for case 1 (forward reaction)

ratef = kf [SO2]2[O2]for case 2 (reverse reaction)

rater = kr [SO3]2

(for only 1 step it is the slow step, so the coefficients = order)

At equilibrium ratef = rater (by definition)

If ratef = rater

then kf [SO2]2[O2] = kr [SO3]2

rearranging to group terms…

What is on top, and what is on bottom, in relationship to the forward reaction?

Kc is called the equilibrium (concentration) constant (capitol K) and the c is sometimes left off

kf= Kc =

[SO3]2

kr [SO2]2[O2]

What is the value of the equilibrium constant, Kc, for case 1?

Kc is unitless, so be sure the units on all your numbers match!

Kc =(0.056)2

= 0.15(0.344)2 (0.172)

What is the value of the equilibrium constant, Kc, for case 2?

What do you notice?

Kc =(0.076)2

= 0.15(0.424)2 (0.212)

Did you notice for both cases we used the forward reaction? If we used the reverse reaction, we would have gotten a different K, but it would still be related to the forward K, it is just the inverse. So Kforward = 1/Kreverse or Kreverse = 1/Kforward

Let’s try it…

Example case 2:in a 1 L vesselR 2SO3 ⇌ 2SO2 + O2 I 0.500 mol 0 0C -0.424 mol +0.424 mol +0.212 molE 0.076 mol 0.424 mol 0.212 mol

What is Kc and what is 1/Kc for this reaction?

One other thing, if perhaps many equilibrium reactions are involved in a mechanism, and you are given the K for each step, the total can be found as…

Ktotal = K1 K2 K3…

In general for aA + bB ⇌ cC + dD

How would this change if the number of products or reactants was different?

Remember, the assumption is this is the only step in the mechanism or is the slow step, otherwise the coefficients don’t necessarily match the exponents.

Kc =[C]c[D]d

[A]a[B]b

So Kc is a measure of how much the reaction has occurred (the activity of the reaction).

The bigger the Kc the more products and less reactants. Explain.

What would be Kc for no reaction?What would be Kc for a strong acid/base?

Practice:in a 5 L vesselR N2 + 3H2 ⇌ 2NH3

ICE 3.01 mol 2.10 mol 0.565 mol

Find Kc

Practice:in a 2 L vesselR 2N2O ⇌ 2N2 + O2

I 10.0 mol 0 0CE 2.20 mol

Find Kc

In the previous example the reactant and products were gases, but we used Molarity anyway. Sometimes when all the substances are gases it is easier to use their partial pressures instead of concentrations, so…

Why call it Kp? What would the units be?

Kp =(PC)c(PD)d

(PA)a(PB)b

Kp is not the same as Kc, because [] is only dependent on the moles and volume where P is also dependent on temperature and the gas constant, but if these are accounted for, then…

Kp = Kc(RT)Δn

where Δn = ngas products - ngas reactants

When would Kp = Kc ?

What if we don’t know if a reaction is at equilibrium yet?

We then determine the reaction quotient, Q

if Q < Kc , then it is not yet at equilibriumforward reaction will still be faster

if Q > Kc, then it is past equilibriumreverse reaction will now be faster

if Q = Kc, then it is at equilibrium

Q =[C]c[D]d

[A]a[B]b

Practice:For 2HI ⇌ H2 + I2 the Kc = 65.0if E 0.500M 0.280 M 3.40 Mis this at equilibrium?

Typically Kc is used to find the concentration of reactants and products at equilibrium. For example, what is the equilibrium concentrations if Kc = 49.0R A + B ⇌ C + DI 0.200 M 0.200 M 0 0

What is the equilibrium concentrations if Kc = 49.0R A + B ⇌ C + DI 0.300 M 0.100 M 0 0

Changes to equilibrium:•if [] or pressure/volume changes or a catalyst is added, then Q changes, but Kc

will not change, so the reaction rates will change to get Q = Kc again (LeChatelier’s can tell you how it will change, shifting right or left means speeding up forward or reverse reaction)•if T changes then Kc will change, so Q = Kc will happen, but it will be a new Kc

Let’s practice using LeChatelier’s Principle

If 2A(g) + 3B(g) ⇌ C(g) + 2D(g)

Then for each change, state how it will shift (forward or reverse) and why it will shift (Q less than K or Q greater than K)

If 2A(g) + 3B(g) ⇌ C(g) + 2D(g)

Pressure is increased:

What affect does that have?How will the equilibrium shift?Why will the equilibrium shift?

If 2A(g) + 3B(g) ⇌ C(g) + 2D(g)

Pressure is decreased:


If 2A(g) + 3B(g) ⇌ C(g) + 2D(g)

Volume is increased:


If 2A(g) + 3B(g) ⇌ C(g) + 2D(g)

Volume is decreased:


Explain why pressure and volume can only affect Kp and not Kc.

What change would affect Kc?

If 2A(g) + 3B(g) ⇌ C(g) + 2D(g)

[A] is increased:


If 2A(g) + 3B(g) ⇌ C(g) + 2D(g)

[A] is decreased:


If 2A(g) + 3B(g) ⇌ C(g) + 2D(g)

[D] is increased:


If 2A(g) + 3B(g) ⇌ C(g) + 2D(g)

[D] is decreased:


Why would catalysts not change the equilibrium?

Why would the addition of inert gases not change the equilibrium?

What affect does changing temperature have on equilibrium? What can we do with LeChatelier's principle?

If 2A(g) + 3B(g) ⇌ C(g) + 2D(g) + Heat

Temperature is increased:


If 2A(g) + 3B(g) ⇌ C(g) + 2D(g) + Heat

Temperature is decreased:


If 2A(g) + 3B(g) + Heat ⇌ C(g) + 2D(g)



If 2A(g) + 3B(g) ⇌ C(g) + 2D(g)∆H = + 76 kJ/mol



If 2A(g) + 3B(g) ⇌ C(g) + 2D(g)∆H = - 76 kJ/mol



Practice:in a 1 L vessel, H2 + I2 ⇌ 2HIat equilibrium [HI] = 0.490 M, [H2] = 0.080 M and [I2] = 0.060 M

then 0.300 more moles of HI is added, what will be the new equilibrium concentrations?

First, find the Kc

Next, find the new [] with the addition and find Q (we are sure we are not at equilibrium!)

Then, we set us a RICE Table with the new initial [] and the change that Q vs Kc told us would happen, finally solve for x

There are other things we can do with K:

1) Remember ΔG0rxn? ΔG0

rxn is the energy change as ALL reactants react to produce ONLY products. In an equilibrium we still have some reactants left over, so this can be accounted for…

ΔG0rxn = -RTlnK (can be Kc or Kp or Q)

Why would ΔG0rxn be less if there are still

some reactants left over?

Another form, on the equation pages:

∆G = ∆G° + RTlnQ

Why Q, not K?

Notice! For both of these equations R is used in Joules but G is typically measured in kiloJoules. Be careful that the units match!


2) The Nernst equation should actually look like this:

Ecell = E°cell - lnQ

Why Q, and not K?

R would be in volt coulombs mol-1 k-1!

RT

nF


3) As mentioned earlier, K varies with Temperature, but thanks to van’t Hoff’s equation we can calculate the K at a new temperature if ΔH0 is known.

ln (KT2

) =

ΔH0

(1

-1

)KT1 R T1 T2

chapter 10 : (general equilibrium). mr belland presents - how my son and i play

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