Basic Music Guitar

7/30/2019 Basic Music Guitar

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17-1

Equilibrio Qumico


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Kinetics applies to the speedof a reaction, the concentration of

product that appears (or of reactant that disappears) per unit time.

Equilibrium applies to the extentof a reaction, the concentration

of product that has appeared after an unlimited time, or once no

further change occurs.

At equilibrium: rateforward= ratereverse

A system at equilibrium is dynamic on the molecular level;no further net change is observed because changes in one

direction are balanced by changes in the other.


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Reaching equilibrium on the macroscopic and molecular levels.

N2O4(g) 2NO2(g)


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Equilibrium Equal

the rates of the forward and reverse reactions areequal at equilibrium

but that does not mean the concentrations ofreactants and products are equal

some reactions reach equilibrium only after almostall the reactant molecules are consumed- we saythe position of equilibrium favors the products.

some reactions reach equilibrium only after smallpercentage of the reactant molecules areconsumed- we say the position of equilibriumfavors the reactants.


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If rateforward = ratereverse then

kforward[reactants]m = kreverse[products]

n

= = K the equilibrium constantkforward

kreverse

[products]n

[reactants]m

The values of m and n are those of the coefficients in the balanced

chemical equation. Note that this is equilibrium, not kinetics. The rates ofthe forward and reverse reactions are equal, NOT the concentrations of

reactants and products.

This is also known as the LAW OF MASS ACTION.


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The range of equilibrium constants

small K

large K

intermediate K

K = [Productos]m

[Reactivos]n


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Q - The Reaction Quotient

We use the molar concentrations of the substances in the

reaction. This is symbolised by using square brackets - [ ].

For a general reaction aA + bB cC + dD where a, b, c,

and d are the numerical coefficients in the balanced equation, Q

(and K) can be calculated as

Q = [C]c[D]d

[A]a[B]b

At any time, t, the system can be sampled to determine theamounts of reactants and products present.

Q is calculated in the same manner as K


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Initial and Equilibrium Concentration Ratios for the N2O4-NO2 System at 2000C(473 K)

Experiment

Initial

[N2O4] [NO2]

Equilibrium=K

[N2O4]eq [NO2]eq

1 0.1000 0.0000 3.57x10-3 0.193

2 0.10000.0000 9.24x10-4 9.82x10-2

3 0.05000.0500 2.04x10-3 0.146

4 0.02500.0750 2.75x10-3 0.170

[NO2]eq2

[N2O4]eq

10.4

10.4

10.4

10.5

Calculate K for each experiment K. Compare equilibriumCalculate K for each experiment K. Compare equilibrium

concentrations.concentrations.


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Sample Problem 17.1

PROBLEM:

SOLUTION:

Writing the Reaction Quotient from the Balanced

Equation

Write the reaction quotient, Qc, for each of the following reactions:

(a) The decomposition of dinitrogen pentoxide, N2O5(g) NO 2(g) + O2(g)

(b) The combustion of propane gas, C3H8(g) + O2(g) CO 2(g) + H2O(g)

PLAN: Be sure to balance the equations before writing the Qc expression.

42(a) N2O5(g) NO 2(g) + O2(g) Qc =

[NO2]4[O2]

[N2O5]2

3 45(b) C3H

8(g) + O

2(g) CO

2(g) + H

2O(g) Q

c

=

[CO2]3[H2O]

4

[C3H8][O2]5


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Calculating Variations on Q and K

aA + bB cC + dDQc =

[C]c

[D]d

[A]a[B]b

cC + dD aA + bBQ = 1/Qc

aA + bB cC + dDnQc = (Qc)

n

For a sequence of equilibria, Koverall = K1 x K2 x K3 x


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Sample Problem 17.2

PLAN:

Writing the Reaction Quotient for an Overall Reaction

PROBLEM: Understanding reactions involving N2 and O2, the most abundant

gases in air, is essential for solving problems dealing with

atmospheric pollution. Here is a reaction sequence between N2 and

O2 to form nitrogen dioxide, a toxic pollutant that contributes to

photochemical smog.

(a) Show that the Qc for the overall reaction sequence is the same as the

product of the Qcs of the individual reactions.

(1) N2(g) + O2(g) 2NO(g) Kc1 = 4.3 x 10-25

(2) 2NO(g) + O2(g) 2NO2(g) Kc2 = 6.4 x 109

(b) Calculate the Kc

for the overall reaction.

Write the sum of the overall reactions; write the Qc. Write the Qcs for

the individual reactions and then multiply the expressions.

We are given the Kcs for the individual reactions, so we multiply those

values.


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Sample Problem 17.2

SOLUTION:

Writing the Reaction Quotient for an Overall Reaction

N2(g) + 2O2(g) 2NO2(g)Qc =

[NO2]2

[N2][O2]2

Qc1

= [NO]2

[N2][O2]

Qc2 = [NO2]2

[NO]2[O2]

[NO]2

[N2][O2]

Qc1x Qc2 =[NO2]2

[NO]2[O2]

= [NO2]2

[N2][O2]

2

(2) 2NO(g) + O2(g) 2NO2(g)

(1) N2(g) + O2(g) 2NO(g)(a)

(b) Kc = Kc1 x Kc2 = (4.3 x 10-25) x (6.4 x 109) = 2.8 x 10-15


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Sample Problem 17.3

SOLUTION:

Determining the Equilibrium Constant for an Equation

Multiplied by a Common Factor

the equilibrium constant, Kc, is 2.4x10-3 at 1000K. If we change the coefficients

of the equation, which well call the reference (ref) equation, what are the

values of Kc for the following balanced equations?

PROBLEM: For the ammonia formation reaction N2(g) + 3H2(g) 2NH3(g)

(a) 1/3N2(g) + H2(g) 2/3NH3(g) (b) NH3(g)1/2N2(g) + 3/2H2(g)

PLAN: Compare each equation to the reference. Keep in mind that

changing the coefficients will be reflected in a power change in Kc

and a reversal of the equation will show up as an inversion of Kc.

(a) The reference equation is multiplied by 1/3, so Kc(ref) will be to the 1/3 power.Kc = [Kc(ref)]

1/3 = (2.4x10-3)1/3 = 0.13

(b) The reference equation is reversed and halved, so Kc(ref) is to the -1/2 power.

Kc = [Kc(ref)]-1/2 = (2.4x10-3)-1/2 = 20.


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Heterogeneous Equilibrium

When all reactants and products are in one phase, theequilibrium is homogeneous.

If one or more reactants or products are in a different

phase, the equilibrium is heterogeneous. Consider:

experimentally, the amount of CO2 does not seem to depend on

the amounts of CaO and CaCO3. Why?


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The reaction quotient for a heterogeneous system.

solids do not

change their

concentrations


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Heterogeneous Equilibrium

Neither density nor molar mass is a variable, theconcentrations of solids and pure liquids are constant.

(You cant find the concentration of something that

isnt a solution!)

We ignore the concentrations of pure liquids and pure

solids in equilibrium constant expressions.


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Expressing Equilibria with Pressure Terms

Kc and Kp

PV = nRT P =n

VRT = = M

P

RT

n

V

Qp =

P M so for 2NO(g) + O2(g) 2NO2(g)

p(NO2)2

p(NO)2 x p(O2)Qc =

[NO2]2

[NO]2 x [O2]

Kp = Kc (RT)n(gas)


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Sample Problem 17.4 Converting Between Kc and Kp

PROBLEM: A chemical engineer injects limestone (CaCO3) into the hot flue

gas of a coal-burning power plant for form lime (CaO), which

scrubs SO2 from the gas and forms gypsum. Find Kc for the

following reaction, if CO2 pressure is in atmospheres.

CaCO3(s) CaO(s) + CO2(g) Kp = 2.1x10-4 (at 1000K)

PLAN: We know Kp and can calculate Kc after finding ngas. R = 0.0821

L*atm/mol*K.

SOLUTION: ngas = 1 - 0 since there is only a gaseous product and no

gaseous reactants.

Kp = Kc(RT)n Kc = Kp/(RT)

n = (2.1x10-4)(0.0821 x 1000)-1 = 2.6x10-6


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Comparing Q with K tells us whether the

system: has come to equilibrium (Q = K) or

the reaction has to proceed further from reactants to

products (Q < K) or in the reverse direction from products to reactants(Q > K).


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17-21

Reaction direction and the relative sizes of Q and K.

Equilibrium:

no net changereactants products

Reaction

Progress

reactants products

Reaction

Progress


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Sample Problem 17.5

SOLUTION:

Comparing Q and K to Determine Reaction Direction

PROBLEM: For the reaction N2O4(g) 2NO2(g), Kc = 0.21 at 1000C. At a

point during the reaction, [N2O4] = 0.12M and [NO2] = 0.55M. Is

the reaction at equilibrium. If not, in which direction is it

progressing?

PLAN: Write an expression for Qc, substitute with the values given, and

compare the Qc with the given Kc.

Qc =[NO2]

2

[N2O4]=

(0.55)2

(0.12)= 2.5

Qc is > Kc, therefore the reaction is not at equilibrium and will

proceed from right to left, from products to reactants, until Qc = Kc.


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How to solve Equilibrium Problems

In order to study hydrogen halide decomposition, a researcher fills anevacuated 2.00 L flask with 0.200 mol of HI gas and allows the

reaction to proceed at 453 oC.

Concentration (M) 2 HI (g) H2

(g) + I2(g)

Initial 0.100 0 0Change () -2x + x + x

Equilibrium 0.100 -2x x x

At equilibrium, [HI] = 0.078 M, Calculate Kc

Kc = 0.020


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Using Equilibrium Constant to

Determine Quantities

Br2 (g) + Cl2(g) 2 BrCl (g)

Initial 0 0 0.200Change () + x + x -2 x

Equilibrium x x 0.200-2x

Si la constante de equilibrio es 6.90 y se aaden 0.100 moles deBrCl en un envase de 500 mL. Determina la conentracin enequilibrio para cada especie siguiendo la siguiente reaccin.

[Br2] = 0.0432 M, [Cl

2] = 0.0432 M, [BrCl] = 0.114 M

Concentration (M)


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Using Equilibrium Constant to

Determine Quantities

Br2 (g) + Cl2(g) 2 BrCl (g)

Initial 0.200 0.300 0Change () - x - x +2 x

Equilibrium 0.200-x 0.300- x 2x

Si la constante de equilibrio es 6.90 y se aaden 0.200 M de Br2

y 0.300 M de Cl2. Determina la conentracin en equilibrio para

cada especie siguiendo la siguiente reaccin.

[Br2] = 0.0432 M, [Cl

2] = 0.0432 M, [BrCl] = 0.114 M

Concentration (M)


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Kc=

[ BrCl]2

[ Br2 ] [Cl2 ]

6.90=2x

2

0.200x 0.300x

6.90=2x

2

x20.50 x0.06

6.90 x23.45 x0.414=4x2

2.90 x2

3.45x0.414=0

x=bb24ac

2a

x=1.05 or 0.136

Se puede eliminar

1.05 porque alutilizarlo nos da un

nmero negativo y no

hace sentido. Por lotanto x = 0.136 M

[Br2] =0.064 M, [Cl2] = 0.164 M, [BrCl] = 0.272 M


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2 SO3(g) 2SO

2(g) + O

2(g)

Initial 2.00 0 0

Change () -2x + 2x + x

Equilibrium 2.00 -2x 2x x

Si Kc = 2.4 x 10-25 y son colocados 2.00 moles de SO3

en un

envase de 1.00 L. Calcula la concentracin de todas las especies

en equilibrio.

Concentration (M)

Kc=

[SO2 ]2

[O2 ]

[SO3 ]2

2.41025=

2x2x

2.00x 2


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2.41025=2x2x

2.002

4x3=9.610

25

x3=2.41025

x=6.2109

Si asumimos que el cambio no cambia suficiente laconcentracin de SO

3se puede descartar x

[SO3] =2.00 M, [SO

2] = 1.24 x 10-9 M, [O

2] = 6.2 x 10-9 M

La asuncin es vlidasiempre y cuando el cambio

en la concentracin seamenor que 5%. Es decir:

% error = 6.2 x 10-9 =3.1 x 10-7 %2.00


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Sample Problem 17.9 Calculating Equilibrium Concentration with

Simplifying Assumptions

PROBLEM: Phosgene is a potent chemical warfare agent that is now outlawed

by international agreement. It decomposes by the reaction

COCl2(g) CO(g) + Cl2(g) Kc = 8.3x10-4 (at 3600C)

Calculate [CO], [Cl2], and [COCl2] when the following amounts of phosgene

decompose and reach equilibrium in a 10.0-L flask.

(a) 5.00 mol COCl2 (b) 0.100 mol COCl2

PLAN: After finding the concentration of starting material, write the expressions

for the equilibrium concentrations. When solving for the remaining

amount of reactant, see if you can make an assumption about the initial

and final concentrations which could simplifying the calculating by

ignoring the solution to a quadratic equation.

SOLUTION: (a) 5.00 mol/10.0 L = 0.500M (b) 0.100 mol/10.0 L = 0.0100M

Letx= [CO]eq = [Cl2]eq and 0.500-x and 0.0100-x= [COCl2]eq,

respectively, for(a) and (b).


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Sample Problem 17.9 Calculating Equilibrium Concentration with

Simplifying Assumptionscontinued

Kc =[CO][Cl2]

[COCl2]

(0.500 -x) = 4.8x10-2

assumexis


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Sample Problem 17.10

SOLUTION:

Predicting Reaction Direction and Calculating

Equilibrium Concentrations

PROBLEM: The research and development unit of a chemical company is studying

the reaction of CH4 and H2S, two components of natural gas.

CH4(g) + 2H2S(g) CS 2(g) + 4H2(g)

In one experiment, 1.00mol of CH4, 1.00mol of CS2, 2.00mol of H2S, and 2.00mol of

H2 are mixed in a 250-mL vessel at 9600C. At this temperature, Kc = 0.036

(a) In which direction will the reaction proceed to reach equilibrium?

(b) If [CH4] = 5.56M at equilibrium, what are the equilibrium concentrations of the

other substances?

PLAN: Find the initial molar concentrations of all components and use these to

calculate a Qc. Compare Qc to Kc, determine in which direction the

reaction will progress, and draw up expressions for equilibriumconcentrations.

[CH4]initial=1.00mol/0.25 L = 4.0M

[H2S]initial=2.00mol/0.25 L = 8.0M [H2]initial=2.00mol/0.25 L = 8.0M

[CS2]initial=1.00mol/0.25 L = 4.0M


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Sample Problem 17.10 Predicting Reaction Direction and Calculating

Equilibrium Concentrationscontinued

Qc =[CS2][H2]

4

[CH4][H2S]2

=[4.0][8.0]4

[4.0][8.0]2= 64

A Qc of 64 is >> than Kc = 0.036

The reaction will progress to the

left.

CH4(g) + 2H2S(g) CS2(g) + 4H2(g)concentrations

initial

change

equilibrium

4.0 8.0 4.0 8.0

+x + 2x - 4x

4.0 +x 8.0 + 2x

-x

4.0 -x 8.0 - 4x

At equilibrium [CH4] = 5.56M, so 5.56 = 4.0 +xandx= 1.56M

Therefore - [H2S] = 8.0 + 2x= 11.12M

[CS2] = 4.0 -x= 2.44M

[H2] = 8.0 - 4x= 1.76M


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Steps in solving equilibrium problems

PRELIMINARY SETTING UP WORKING ON A REACTION TABLE

SOLVING FORXAND EQUILIBRIUM QUANTITIES

1. Write the balanced equation.

2. Write the reaction quotient, Q.

3. Convert all of the amounts into

the correct units (M or atm).

1. When reaction direction is not

known compare Q with K.

2. Construct a reaction table.

Check the sign ofx, the change

in the quantity.

1. Substitute the quantities into Q.

To simplify the math, assume thatxis

negligible. [A]ini -x= [A]eq = [A]ini Solve forx.

5. Find the equilibrium quantities

Check to see that calculated

values give the known K.

Check that assumption is

justified (


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LeChateliers PrincipleLeChateliers Principle

When a chemical system at equilibrium

is subjected to a stress,

the system will return to equilibrium

by shifting to reduce the stress.

If the concentration increases, the system reacts to consume some of it.

If the concentration decreases, the system reacts to produce some of it.


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PCl3(g) + Cl2(g) PCl5(g)

Aumentar [] Disminuir []


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The Effect of Added Cl2 on the PCl3-Cl2-PCl5 System

Concentration (M) PCl3(g) + Cl2(g) PCl5(g)

Disturbance

New initial

Change

New equilibrium

*Experimentally determined value.

Original equilibrium 0.200 0.125 0.600

+0.075

0.200 0.200 0.600

-x -x +x

0.200 - x 0.200 - x 0.200 + x

(0.637)*


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SOLUTION:

Predicting the Effect of a Change in Concentration

on the Equilibrium Position

PROBLEM: To improve air quality and obtain a useful product, chemists oftenremove sulfur from coal and natural gas by treating the fuel

contaminant hydrogen sulfide with O2;

2H2S(g) + O2(g) 2S(s) + 2H2O(g)

What happens to

(a) [H2O] if O2 is added?(b) [H2S] if O2 is added?

(c) [O2] if H2S is removed?(d) [H2S] if sulfur is added?

PLAN: Write an expression for Q and compare it to K when the system is

disturbed to see in which direction the reaction will progress.

Q = [H2O]2

[H2S]2[O2]

(a) When O2 is added, Q decreases and the reaction progresses

to the right to come back to K. So [H2O] increases.


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Sample Problem 17.11 Predicting the Effect of a Change in Concentration

on the Equilibrium Positioncontinued

(b) When O2 is added, Q decreases and the reaction progresses

to the right to come back to K. So [H2S] decreases.

Q = [H2O]2

[H2S]2[O2]

(c) When H2S is removed, Q increases and the reactionprogresses to the left to come back to K. So [O2] increases.

(d) Sulfur is not part of the Q (K) expression because it is a solid.

Therefore, as long as some sulfur is present the reaction is

unaffected. [H2S] is unchanged.


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Effect of a Change in Pressure

(Volume) Pressure changes can occur in three ways:

Changing the concentration of a gaseous component

Adding and innert gas (ine that foes not take part in

the reaction)

Changing the volume of the reaction vessel


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Adding an Inert Gas

Adding an inert gas has no effect on theequilibrium position.

Adding an inert gas does not change thevolume, so all reactant and productconcentrations remain the same.


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Changing Pressure by Changing

Volume

A change in volume results in a change inconcentration

A change in pressure due to a change in volumedoes not alter K

c.


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If the volume become smaller (pressure is

higher), the reaction shifts so that the totalnumber of gas molecules decreases.

If the volume become larger (pressure is lower),the reactions shifts so that the total number of

gas molecules increases.

Ifngas

= 0, there is no effect on the equilibrium

position.


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The effect of pressure (volume) on an equilibrium system.

+

lower P

(higher V)

more moles

of gas higher P

(lower V)

fewer moles

of gas


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SOLUTION:

Predicting the Effect of a Change in Volume

(Pressure) on the Equilibrium Position

PROBLEM: How would you change the volume of each of the following

reactions to increase the yield of the products.

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

(b) S(s) + 3F2(g) SF6(g)

(c) Cl2(g) + I2(g) 2ICl(g)

PLAN: When gases are present a change in volume will affect the

concentration of the gas. If the volume decreases (pressure

increases), the reaction will shift to fewer moles of gas and vice versa.

(a) CO2 is the only gas present. To increase its yield, we

should increase the volume (decrease the pressure).(b) There are more moles of gaseous reactants than products, so we should

decrease the volume (increase the pressure) to shift the reaction to the right.

(c) There are an equal number of moles of gases on both sides of the

reaction, therefore a change in volume will have no effect.

The Effect of a Change in Temperature on an Equilibrium


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The Effect of a Change in Temperature on an Equilibrium

Only temperature changes can alter K.

Consider heat as a product or a reactant.Consider heat as a product or a reactant.

A temperature rise will increase Kc for a system with a

positive H0rxn.

A temperature rise will decrease Kc for a system with a

negative H0rxn.

N2H

6CO (s) 2 NH

3(g) + CO

2(g) H

rxn= +33 kJ/mol

N2H

6CO (s) + calor 2 NH

3(g) + CO

2(g)

CH4

(g) + 4 Cl2

(g) CCl4(l) + 4 HCl (g) H

rxn= -397 kJ/mol

CH4

(g) + 4 Cl2

(g) CCl4(l) + 4 HCl (g) + calor


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SOLUTION:

Predicting the Effect of a Change in Temperature on

the Equilibrium Position

PROBLEM: How does an increase in temperature affect the concentration of

the underlined substance and Kc for the following reactions?(a) CaO(s) + H2O(l) Ca(OH)2(aq) H

0 = -82kJ

(b) CaCO3(s) CaO(s) + CO2(g) H0 = 178kJ

(c) SO2(g) S(s) + O2(g) H0 = 297kJ

PLAN: Express the heat of reaction as a reactant or a product. Then considerthe increase in temperature and its effect on Kc.

(a) CaO(s) + H2O(l) Ca(OH)2(aq) heatAn increase in temperature will shift the reaction to the left, decrease

[Ca(OH)2], and decrease Kc.

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

The reaction will shift right resulting in an increase in [CO2] and increase in Kc.

(c) SO2(g) + heat S(s) + O2(g)

The reaction will shift right resulting in an decrease in [SO2] and increase in Kc.


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The vant Hoff Equation: The Effect of T on K

ln K2

K1

= - H0

rxn

R

1

T2

1

T2

-

R = universal gas constant

= 8.314 J/mol*K

K1 is the equilibrium constant at T1


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Effect of Various Disturbances on a System at Equilibrium

Disturbance Net direction of Reaction Effect on Value of K

Concentration

Increase [reactant] None

Decrease [reactant] None

Increase [product] None

Decrease [product] None

Pressure

Increase Pressure (decrease V) Toward formation of fewer mole gas None

Decrease Pressure (increase V) Toward formation of more moles gas None

None, concentrations unchanged None

Temperature

Increase T Toward absorption of heat Increase if endotermic

Decrease if exotermic

Decrease T Toward release of heat Increase if exotermic

Decrease if endotermic

Catalyst added

None

Increase Pressure (adding inert gas,

no change V)

one; orwar an reverse

equilibrium attained sooner; rates

increase equally


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Sample Problem 17.14 Determining Equilibrium Parameters from

Molecular Species

PROBLEM: For the reaction X(g) + Y2(g) XY(g) + Y(g) H>0

the following molecular scenes depict different reaction mixtures.(X = green, Y = purple)

(a) If Kc = 2 at the temperature of the reaction, which scene represents the

mixture at equilibrium?

(b) Will the reaction mixtures in the other two scenes proceed toward

reactant or toward products to reach equilibrium?

For the mixture at equilibrium, how will a rise in temperature affect [Y2]?

SOLUTION: The equilibrium constant, K, is[XY][Y]

[X][Y2].

scene 1: Qc = (5)(3)/(1)(1) = 15

scene 2: Qc = (4)(2)/(2)(2) = 2.0 scene 3: Qc = (3)(1)/(3)(3) = 0.33


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Sample Problem 17.14 Determining Equilibrium Parameters from

Molecular Species

continued

Qc

= 15 Qc

= 2.0 Qc

= 0.33

(b) In scene 1 Qc > Kc, so the system will proceed to reactants to

reach equilibrium while in scene 3 Qc < Kc, so the system will proceed

to products.

(a) In scene 2 Qc = Kc, so it represents the system at equilibrium.

(c) IfH > 0, heat is a reactant (endothermic). A rise in temperature

will favor products and [Y2] will decrease as the system shifts to

products.

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