Chemical Equilibrium

Collision Theory of ReactionsA chemical reaction occurs whenCollisions between molecules have sufficient energy to break the bonds in the reactants. Molecules collide with the proper orientation.Bonds between atoms of the reactants (N2 and O2) are broken, and new bonds (NO) form. Copyright 2007 by Pearson Education, Inc.Publishing as Benjamin Cummings

Collision Theory of Reactions A chemical reaction does not take place if the Collisions between molecules do not have sufficient energy to break the bonds in the reactants. Molecules are not properly aligned.

N2 O2 N2 O2Copyright 2007 by Pearson Education, Inc.Publishing as Benjamin Cummings

Activation EnergyThe activation energyIs the minimum energy needed for a reaction to take place upon proper collision of reactants. Copyright 2007 by Pearson Education, Inc.Publishing as Benjamin Cummings

Reaction Rate and Temperature Reaction rate Is the speed at which reactant is used up.Is the speed at which product forms.Increases when temperature rises because reacting molecules move faster, providing more colliding molecules with energy of activation.

Reaction Rate and ConcentrationIncreasing theconcentration of reactantsIncreases the number of collisions.Increases the reaction rate.

Copyright 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings

Reaction Rate and CatalystsA catalyst Speeds up the rate of a reaction.Lowers the energy of activation. Is not used up during the reaction.Copyright 2007 by Pearson Education, Inc.Publishing as Benjamin Cummings

Reversible ReactionsIn a reversible reaction, there is both a forward and areverse reaction.Suppose SO2 and O2 are present initially. As they collide, the forward reaction begins. 2SO2(g) + O2(g) 2SO3 (g)As SO3 molecules form, they also collide in the reverse reaction that forms reactants. The reversible reaction is written with a double arrow. forward2SO2(g) + O2 (g) 2SO3(g) reverse

Chemical EquilibriumAt equilibrium The rate of the forward reaction becomes equal to the rate of the reverse reaction.The forward and reverse reactions continue at equal rates in both directions.Copyright 2007 by Pearson Education, Inc.Publishing as Benjamin Cummings

EquilibriumAt equilibriumThe forward reaction of N2 and O2 forms NO.The reverse reaction of 2NO forms N2 and O2. The amounts of N2, O2, and NO remain constant. N2(g) + O2(g) 2NO(g)


Chapter 9 Chemical Equilibrium

9.3Equilibrium Constants

Copyright 2007 by Pearson Education, Inc.Publishing as Benjamin Cummings

Equilibrium ConstantsFor the reaction aA bBThe equilibrium constant expression, Kc, gives the concentrations of the reactants and products at equilibrium.Kc = [B]b = [Products][A]a [Reactants]The square brackets indicate the moles/liter of each substance. The coefficients b and a are written as superscripts that raise the moles/liter to a specific power.

Writing a Kc ExpressionIn the Kc expression for the following reaction at equilibrium 2CO(g) + O2(g) 2CO2(g) STEP 1The products are shown in the numerator and the reactants are shown in the denominator.Kc= [CO2] [products] STEP 2 [CO] [O2] [reactants]

The coefficients are written as superscripts. STEP 3Kc= [CO2]2 [CO]2 [O2]

Heterogeneous EquilibriumIn heterogeneous equilibriumSolid and/or liquid states may be part of a reaction. 2NaHCO3(s) Na2CO3(s) + CO2(g) + H2O(g)The concentration of solids and/or liquids is constant.The Kc expression is written with only the gases.

Kc = [CO2][H2O]

Calculating Equilibrium ConstantsWhat is the Kc for the following reaction? H2(g) + I2(g) 2HI(g) Equilibrium concentrations: H2 1.2 mole/L, I2 1.2 mole/L, and HI 0.35 mole/L. STEP 1 Write the Kc expression Kc = [HI]2 [H2][I2]STEP 2 Enter equilibrium concentrations.Kc= (0.35)2 = 8.5 x 10-2 (1.2)(1.2)

Chapter 9 Chemical Equilibrium

9.4 Using Equilibrium ConstantsCopyright 2007 by Pearson Education, Inc.Publishing as Benjamin Cummings

Reaching Chemical EquilibriumA container filled with SO2 and O2 or only SO3Contains mostly SO3 and small amounts of O2 and SO3 at equilibriumReaches equilibrium in both situations. Copyright 2007 by Pearson Education, Inc.Publishing as Benjamin Cummings

Equilibrium can Favor ProductIf equilibrium is reached after most of the forward reaction has occurred,The system favors the product.Copyright 2007 by Pearson Education, Inc.Publishing as Benjamin Cummings

Equilibrium with a Large KcAt equilibrium, A reaction with a large Kc produces a large amount of product. Very little of the reactants remain. Kc=[NCl3]2 = 3.2 x 1011 [N2][Cl2]3A large Kc favors the products. N2(g) + 3Cl2(g) 2NCl3(g)

When this reaction reaches equilibrium, it will essentially consist of the product NCl3.

Equilibrium can Favor Reactant If equilibrium is reached when very little of the forward reaction has occurred,The reaction favors the reactants.


Equilibrium with a Small KcAt equilibrium,A reaction that produces only a small amount of product has a small Kc. Kc = [NO]2 =2.3 x 10-9 [N2] [O2]A small Kc favors the reactants.N2(g) + O2(g) 2NO(g)

Summary of Kc Values A reactionThat favors products has a large Kc.With about equal concentrations of products and reactants has a Kc close to 1.That favors reactants has a small Kc.


Calculating Equilibrium ConcentrationsWhen the Kc is known and the concentrations of reactants or products except for one are given The concentration of the remaining substance can be calculated.STEP 1 Write the Kc expression.STEP 2 Solve the Kc expression for the unknown.STEP 3 Substitute the known concentrations. STEP 4 Check by using concentrations to solve for Kc.

Using Kc to Solve for Equilibrium ConcentrationAt equilibrium, the reaction PCl5(g) PCl3(g) + Cl2(g)has a Kc of 4.2 x 10-2 and contains [PCl3] = [Cl2] = 0.10 M.

What is the equilibrium concentration of PCl5?

Using Kc to Solve for Equilibrium ConcentrationSTEP 1 Write the Kc expression. Kc = [PCl3][Cl2] [PCl5]STEP 2 Solve for the unknown. [PCl5] = [PCl3][Cl2] Kc STEP 3 Substitute known concentrations and solve. [PCl5] = (0.10)(0.10) = 0.24M 4.2 x 10-2 STEP 4 Check by placing concentrations in Kc. Kc = (0.10)(0.10) = 4.2 x 10-2 (0.24)

LeChteliers PrincipleLeChteliers principle states that Any change in equilibrium conditions upsets the equilibrium of the system.A system at equilibrium under stress will shift to relieve the stress.There will be a change in the rate of the forward or reverse reaction to return the system to equilibrium.

Effect of Adding ReactantConsider the following reaction at equilibrium H2(g) + F2(g) 2HF(g)If more reactant (H2 or F2) is added, there is an increase in the number of collisions.The rate of the forward reaction increases and forms more HF product until new equilibrium concentrations equal Kc again.The effect of adding a reactant shifts the equilibrium towards the products.

Adding ReactantFor the reaction A + B C at equilibrium Adding more A upsets the equilibrium. The rate of forward reaction increases to re-establish Kc A + B C


Effect of Adding ProductConsider the following reaction at equilibrium. H2(g) + F2(g) 2HF(g)

When more HF is added, there is an increase in collisions of HF molecules.The rate of the reverse reaction increases and forms more H2 and F2 reactants.The effect of adding a product shifts the equilibrium towards the reactants.

Adding Reactant or ProductThe equilibrium shifts towardsProducts when H2 or F2 is added.Reactants when HF is added.

H2(g) + F2(g) 2HF(g)

Add H2 or F2Add HF

Effect of Removing ReactantRemoving reactant, H2 or F2, from the followingreaction at equilibrium. H2(g) + F2(g) 2HF(g) Decreases the collisions between reactants.Decreases the rate of the forward reaction. Shifts the equilibrium towards the reactants.

H2(g) + F2(g) 2HF(g)

Remove H2 or F2

Effect of Removing ProductWhen HF is removed from the following reaction at equilibrium H2(g) + F2(g) 2HF(g) The rate of the reverse reaction decreases.The rate of the forward reaction is greater.Equilibrium shifts towards the products. H2(g) + F2(g) 2HF(g)Remove HF

Concentration Changes and Equilibrium Copyright 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings

TABLE 9.4

Effect of Decreasing the VolumeWhen a reaction at equilibrium contains different numbersof moles of reactants than products, A decrease in volumeIncreases the concentration (mole/L) upsetting the equilibrium.Shifts the equilibrium towards the fewer number of moles. N2(g) + 3H2(g) 2NH3(g)

Decrease volumeMore molesFewer moles

Volume Decrease and EquilibriumA volume decreaseShifts the equilibrium towards the side with the smaller number of moles.Shifts the equilibrium towards reactant A in this example.Copyright 2007 by Pearson Education, Inc.Publishing as Benjamin Cummings

Effect of Increasing the VolumeWhen a reaction at equilibrium contains different numbers of moles of reactants than products, An increase in volumeDecreases the concentration (mol/L) upsetting the equilibrium.Shifts the equilibrium towards the greater number of moles. N2(g) + 3H2(g) 2NH3(g)

Increase volumeMore molesFewer moles

Volume Increase and EquilibriumA volume increaseShifts the equilibrium towards the side with the greater number of moles.Shifts towards reactants B and C in this example.


Heat and Endothermic ReactionsFor an endothermic reaction at equilibrium,A decrease in temperature removes heat and the equilibrium shifts towards the reactants.An increase in temperature adds heat and the equilibrium shifts towards the products.

CaCO3 (s) + 133 kcal CaO(s) + CO2(g)

Decrease TIncrease T

Temperature Change and EquilibriumCopyright 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings

TABLE 9.5

Heat and Exothermic ReactionsFor an exothermic reaction at equilibrium,A decrease in temperature removes heat and the equilibrium shifts towards the products.An increase in temperature adds heat and the equilibrium shifts towards the reactants.

N2(g) + 3H2(g) 2NH3(g) + 22 kcal

Decrease TIncrease T

Temperature Change and EquilibriumCopyright 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings

TABLE 9.6

Changes and EquilibriumCopyright 2007 by Pearson Education, Inc. Publishing as Benjamin Cummings

TABLE 9.7

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Chemical Equilibrium