A Pun It 14 Equilibrium

Unit 15 Equilibrium

I. The Concept of Equilibrium

Chemical equilibrium occurs when a reaction and its reverse reaction proceed at the same rate.

A. The Concept of Equilibrium

• When CO and H2O are mixed, they immediately begin to react to form H2 and CO2.This leads to a decrease in the concentrations of the reactants,

• Beyond a certain time, indicated by the dashed line, the concentrations of reactants and products no longer changes and equilibrium has been reached

• the concentrations of reactants never go to zero; the reactants will always be present in small but constant concentrations

B. A System at Equilibrium

Once equilibrium is achieved, the amount of each reactant and product remains constant.

C. 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

N2O4 (g) 2 NO2 (g)

II. Equilibrium constant• Expression that relates the concentration of

products to reactants in terms of a constant K

A. Variations of K • Kc for concentrations expressed in mol/L • Kp for partial pressures of gases• Ksp for solubility product, which has no

denominator because the reactants are solids• Ka is acid dissociation constant for weak acids• Kb dissociation constant for weak bases • Kw the ionization of water (Kw= 1 x 10−14)

B. Equilibrium expressionaA + bB → cC + dD

Keq= [C]c [D]d

[A]a [B]b

1) [A], [B], [C], and [D] are molar concentration or partial pressures at equilibrium2) Products are in the numerator, and reactants are in the denominator3) Coefficients in the balanced equation become exponents in the equilibrium expression4) Solids and pure liquids are ignored 5) Units are not given for Keq

What Are the Equilibrium Expressions for These Equilibria?

C. Calculating equilibrium constant If the concentrations at equilibrium are as follows: [O2]= 2.0 x 10−8M, [SO2]= 3.4 x10−9M and [SO3]= 0.971M then what is the equilibrium

constant?

Kc= (0.971)2

(2.0 x 10−8) x (3.4 x 10−9)2

= 4.2 x 1024

D. Keq and gases• Kc and Kp of gases can be related using following

formula: Kp= Kc (RT)∆n

Kp= partial pressure constant (atm)Kc= molar concentration constant (M)R= 0.0821 L atm/ mol KT= temperature (K)∆n= (moles of product gas −moles of reactant gas)

Question: For the following equilibrium Kp= 1.90

C (s) + CO2(g) 2 CO (g)Calculate Kc for this reaction at 25˚C.

Kc= Kp (RT)∆n

Kc= 1.90 . [(0.0821 L atm/ mol K)(298K)]2-1

=0.0777

III. What Does the Value of K Mean?

• If K >> 1, the reaction is product-favored; product predominates at equilibrium.

• If K << 1, the reaction is reactant-favored; reactant predominates at equilibrium.

A. K for reverse reactionThe equilibrium constant of a reaction in the reverse reaction is the reciprocal of the equilibrium constant of the forward reaction.

10.212

=

Kc = = 0.212 at 100C[NO2]2

[N2O4]N2O4 (g) 2 NO2 (g)

Kc = = 4.72 at 100C

[N2O4][NO2]2

N2O4 (g)2 NO2 (g)

B. Effect of increasing amount of chemicals on K

The equilibrium constant of a reaction that has been multiplied by a number is the equilibrium constant raised to a power that is equal to that number.

Kc = = 0.212 at 100C[NO2]2

[N2O4]N2O4 (g) 2 NO2 (g)

Kc = = (0.212)2 at 100C[NO2]4

[N2O4]22 N2O4 (g) 4 NO2 (g)

C. K for multi-step reactions• The equilibrium constant for a reaction made

up of two or more steps is the product of the equilibrium constants for the individual steps.

• Basically like Hess’s law with the way the reactions are added up but instead of adding the equilibrium constants, you multiply them to get equilibrium constant for final reaction

Example: HCrO4− + Ca2+ H+ + CaCrO4 

Rxn1: HCrO4− H+ + CrO4

2− Ka= 3 x 10−7

Rxn 2: CaCrO4 Ca2+ + CrO42− Ksp= 7.1 x 10−4

What is the K for the overall reaction if the K values for the multistep reaction is given.

Keq= (Ka) (1/Ksp)= Ka/Ksp since the second reaction needs to be flipped to get the final reaction.

Keq= 3 x 10−7 /7.1 x 10−4

= 4. 23 x 10−4

IV. Reaction Quotient• Determined exactly as equilibrium constant with initial

conditions instead of equilibrium conditions• Allows prediction of direction of reaction• For the reaction:

aA + bB cC + dDQ= [C]c [D]d

[A]a [B]b

• [A], [B], [C], and [D] are initial molar concentrations or partial pressures

• if Q is less than the calculated K for the reaction, rxn. proceeds fwd. generating products

• if Q is greater than K, the rxn. proceeds backward, generating reactants

• if Q=K, the reaction is already at equilibrium

If Q = K,

the system is at equilibrium.

If Q > K,there is too much product and the

equilibrium shifts to the left.

If Q < K,there is too much reactant, and the

equilibrium shifts to the right.

Question: Using the equilibrium constants and reactions below, determine whether or not each of the following systems is in equilibrium, predict whether it will proceed in the forward or reverse direction.1) H2(g) + I2 (g) 2HI (g) Kc= 49[H2] = 0.10M [I2]= 0.10M [HI]= 0.70MQ= 49 so reaction in equilibrium

2)HF(aq)+ H2O(l) F−(aq)+H3O+(aq) Kc=6.8x10−4

[HF]= 0.20M [F−]=2.0 x 10−4M [H3O+]=2.0 x 10−4M

Q= 2 x 10−7 Q < Kc so reaction must go in fwd. direction to reach in equilibrium

V. Equilibrium Calculations To solve equilibrium questions, we need to construct a equilibrium tableReactionInitial conditionChangeEquilibriumAnswer

•

• There are 4 types of questions you may be asked

Type 1 Solving for equilibrium constantNO2(g) + SO2(g) NO(g) + SO3(g)

Assume that a 4.00L flask is filled with 1 mole of each of the four compounds. The equilibrium concentration of NO2 is 0.261M. What is Kc equal?

The molarity of each compound is 1.00mol/4.00L= 0.250M .

Reaction NO2(g) SO2(g) NO(g) SO3(g)

Initial condition

0.250M 0.250M 0.250M 0.250M

Change +x +x −x −x

Equilibrium 0.250+x 0.250+x 0.250 −x 0.250 −x

Answer 0.261

Since 0.250+ x= 0.261 then x= 0.011MReaction NO2(g) SO2(g) NO(g) SO3(g)

Initial condition

0.250M 0.250M 0.250M 0.250M

Change +x +x −x −xEquilibrium 0.250+x 0.250+x 0.250 −x 0.250 −x

Answer 0.261 0.261 0.239 0.239

Kc= [NO] [SO3] = (0.239) (0.239) = 0.839 [NO2][ SO2] (0.261) (0.261)

Type 2 Finding equilibrium conc. of all speciesBr2 + Cl2 2BrCl

• The above reaction has an equilibrium constant of 6.90 and has 0.100 mol of BrCl introduced into a 500mL flask. What is the equilibrium concentration of all reactants and products?

• [BrCl]= 0.100mol/0.500L= 0.200M

Reaction Br2 Cl2 2BrCl

Initial condition

0 0 0.200M

Change +x +x −2xEquilibrium X X 0.200−2xAnswer

Kc= [BrCl]2

[Br2] [Cl2]

6.90= (0.200-2x)2

x2

• Take square root of both sides2.63= 0.200−2x→2.63x= 0.200 −2x→ 4.63x= 0.200 x

= 0.0432

Reaction Br2 Cl2 2BrCl

Initial condition

0 0 0.200M

Change +x +x −2xEquilibrium X X 0.200−2xAnswer 0.0432 0.0432 0.114

Type 3 Quadratic equation Br2 + Cl2 2BrCl

0.200M Br2 and 0.300M Cl2 is mixed with an

equilibrium constant of 6.90. What is the equilibrium concentrations of all species?

Reaction Br2 Cl2 2BrCl

Initial condition

0.200M 0.300M 0

Change -x -x +2x

Equilibrium 0.200−x 0.300−X 2x

Answer

Kc= [BrCl]2

[Br2] [Cl2]

6.90= (2x)2 . (0.200−x) (0.300−x)

Square root can’t be taken since denominator not anexact square. Multiplying denominator yields

6.90= (2x)2 . (x2 − 0.5x +0.06)

6.90 x2 − 3.45x + 0.414= 4x2

2.90x2 − 3.45x +0.414= 0

Using quadratic to solvex= −b ± √ b2 −4ac

2aSubstituting into equation getsx=−(–3.45) ±√(−3.45)2 − 4(2.90) (0.414)

2(2.90) x= 3.45±2.66 5.8

x= 1.05 or 0.136

Plugging in both values in for x we see that only 0.136 is reasonable since 1.05 gives a negative answer which is impossible

Reaction Br2 Cl2 2BrCl

Initial condition

0.200M 0.300M 0

Change -x -x +2x

Equilibrium 0.200−x 0.300−X 2x

Answer 0.064 0.164 0.272

Type 4 Assuming• If the equilibrium constant is less than 10−3 and

the initial concentrations of reactants are given we can avoid doing quadratics

2SO3 (g) 2SO2 (g) + O2(g)

• If 2.00 moles of SO3 is placed in a 1.00L flask and the equilibrium constant is 2.4 x 10−25 then what is the concentration of all species at equilibrium.

Reaction 2SO3 2SO2 O2

Initial condition

0.200M 0 0

Change -2x +2x +x

Equilibrium 0.200−2x 2x x

Answer

Kc= [SO2]2 [O2] [SO3]2

2.4 x 10−25= (2x)2 (x) (2.00M−2x)2

Since Kc is small we can assume that very little product is produced and that 2x subtracted from 2.00M will still be 2.00M

2.4 x 10−25= (2x)2 (x) (2.00M)2

9.6 x 10−25= 4x3

2.4 x 10−25= x3

6.2 x 10−9M= x

Checking to see assumption is correct:2.00M− 6.2 x 10−9M= 2.00M so assumption valid

Reaction 2SO3 2SO2 O2

Initial condition

0.200M 0 0

Change -2x +2x +x

Equilibrium 0.200−2x 2x x

Answer 2.00 1.24 x 10−8 6.2 x 10−9

Question: At 2000oC the equilibrium constant is 2400. 2NO(g) N2(g) + O2(g)

If [NO]init = .2M, determine the [NO]eq, [N2]eq and [O2]eq.

Reaction 2NO N2 O2

Initial condition

0.200M 0 0

Change -2x +x +x

Equilibrium 0.200−2x x x

Answer

Kc= [N2] [O2] [NO]2

2400= x2 .

(0.200−2x)2

48.99= x . 0.200−2x

48.99(0.200−2x)= x9.80 − 97.98x= x9.80= 98.98x0.099= x

Reaction 2NO N2 O2

Initial condition

0.200M 0 0

Change -2x +x +x

Equilibrium 0.200−2x x x

Answer 0.0022 0.099 0.099

VI. Le Châtelier’s Principle“If a system at equilibrium is disturbed by a change in temperature, pressure, or the concentration of one of the components, the system will shift its equilibrium position so as to counteract the effect of the disturbance.”•Physical stresses include changes in temperature, pressure and volume•Chemical stresses include the addition or removal of one or more of the products or reactants

A. Effect of Concentration• Even though concentrations may change, the

value of Kc or Kp would remain the same since the ratios of products to reactants is the same

1) Decreasing concentration• shift to produce more of whatever substance is

decreased causing concentration of other side of arrow to decrease

2) Increasing concentration• shift to consume substance that increased

which will increase concentration of substance on other side of arrow

N2 (g) + 3H2 (g) 2NH3 (g) 1) What would happen if you increase the concentration of hydrogen gas?

1) shift to right 2) shift to left 3) nothingshift to right to consume hydrogen2) What would happen to the concentration of ammonia?

1) increase 2) decrease 3) nothingincrease3) What would happen to the concentration of nitrogen?

1) increase 2) decrease 3) nothingdecrease4) What would happen if the concentration of nitrogen decreased?

1) shift to right 2) shift to left 3) nothingshift to left to increase nitrogen conc.5) What would happen to the concentration of hydrogen?

1) increase 2) decrease 3) nothing increase6) What would happen to the concentration of ammonia?

1) increase 2) decrease 3) nothingdecrease

B. Effect of Pressure • Only significant for gases• Pressure can be changed by changing the volume of

the container or by adding or removing either the reactants or products or adding an inert gas

• Nothing will happen if you have same number of moles of reactants as products

1) increasing pressure (decreasing volume)- reaction will go in direction that produces fewer moles of gas

2) decreasing pressure (increasing volume)-reaction will go in direction that produces more moles of gas

N2 (g) + 3H2 (g) 2NH3 (g) 1) Which direction will the reaction shift if the pressure increases?

1) right 2) leftif pressure increases than rxn. will shift to side with

fewer moles which is right side2) Which direction will the reaction shift if the pressure decreases?

1) right 2) leftRxn. will shift to left since fewer moles3) Which direction will the reaction shift if the volume decreases?

1) right 2) leftIf volume decreases than pressure increases and

reaction will shift to right

C. Temperature• Easiest way to handle is to treat temperature as

a chemical and put the word temperature on reactant side if the reaction is endothermic and on product side if the reaction is exothermic

Ex. N2 (g) + 3H2 (g) 2NH3 (g) + 300kJ

temperature + N2 (g) + 3H2 (g) 2NH3 (g)

• If the temperature is increased in above reaction then the reaction will shift to right

• if the temperature is decreased, the reaction shifts to left.

Question: The reaction below is exothermic in forward direction.

N2 (g) + 3H2 (g) 2NH3 (g)

1) Which direction will the reaction shift if the temperature is increased?

1) left 2) right2) Which direction will the reaction shift if the temperature is decreased?

1) left 2) right

Question: At elevated temperatures, SbCl5 gas

decomposes into SbCl3 gas and Cl2 gas as shown by

the following equation: SbCl5(g) SbCl3(g) + Cl2(g)

(a) An 89.7 gram sample of SbCl5 (molecular weight 299.0) is placed in an evacuated 15.0 litre container at 182ºC.

1.What is the concentration in moles per litre of SbCl5 in the container before any decomposition occurs?

89.7g SbCl5 x 1 mole = 0.300 mol

299 g

M= 0.300mol = 0.0200M 15L

2. What is the pressure in atmospheres of SbCl5 in the container before any decomposition occurs?

P = nRT = 0.300mol (0.0821) 455K V 15L

= 0.747 atm

(b)If the SbCl5 is 29.2 percent decomposed when equilibrium is established at 182ºC, calculate the value for either equilibrium constant Kp or Kc, for this decomposition reaction. Indicated whether you are calculating Kp or Kc.

SbCl5 ↔ SbCl3 + Cl2If 29.2% of the SbCl5 is decomposed than that means

that 29.2 % of it converts into the products and 70.8% of it stays as SbCl5[SbCl3] = [Cl2] = (0.0200 mol/L)(0.292)

= 5.84 x10-3M[SbCl5] = (0.0200 mol/L)(0.708) = 1.42 x10-2M

K c [ SbCl 3 ][ Cl 2 ]

[ SbCl 5 ]

( 5 . 84 10 3 ) 2

1 . 42 10 2 2 . 41 10 3

(c)In order to produce some SbCl5, a 1.00 mole sample of SbCl3 is first placed in an empty 2.00 litre container maintained at a temperature different from 182ºC. At this temperature, Kc, equals 0.117. How many moles of Cl2 must be added to this container to reduce the number of moles of SbCl3 to 0.700 mole at equilibrium?

Since you start with 1 mole of SbCl3 and end up with 0.700 at equilibrium then the amount of change in the reaction was 0.300 mol which you subtract from product side and add to reactant side

Reaction SbCl5 SbCl3 Cl2Initial condition

0 0.500M x

Change +0.150M -0.150M –0.150M

Equilibrium 0.150M 0.350M x –0.150M

Answer

[SbCl3] = 0.700mol/2l= 0.350M

[SbCl5]= 0.300/2L= 0.150M

Change = 0.300/2L= 0.150M

K= 0.117= (x –0.150M) (0.350M)0.150M

0.0176= 0.350x – 0.0525 0.0701=0.350x

x= 0.200M of Cl2

Since they are looking for moles we multiply by volume to get 0.200M(2L)= 0.400moles