Kinetics And Equilibrium

Kinetic Collision Theory The main ideas of the collision-reaction theory are as follows: •

A chemical sample consists of particles (atoms, ions, or molecules) that are in constant random motion at various speeds, rebounding elastically from collisions with each other. The average kinetic energy of the particles is perceived as the temperature of the sample.

•

A chemical reaction must involve collisions of reactant particles.

•

An effective collision requires sufficient energy and correct orientation (positioning) of the colliding particles so that bonds can be broken and new bonds formed.

•

Ineffective collisions involve particles that rebound from the collision, essentially unchanged in nature.

•

The rate of a given reaction depends on two factors related to collisions: the frequency of collisions and the fraction of those collisions that are effective.

Factors Affecting Reaction Rates: The explanation of factors affection reaction rates is as follows:

•

A minimum activation energy (the energy of the activated complex) is required for a collision between particles to be effective, that is, energetic enough to cause the breaking and forming of bonds to product new particle structures

•

Many reactions occur as a sequence of elementary process, or two particle-collision steps, which make up the overall reaction mechanism.

•

The rate of any reaction depends on the nature of the chemical substances reacting because both the strength of bond(s) to be broken and the location of the bond(s) in the particle structure affect the likelihood that any given collision is effective.

•

An increase in reactant concentration or in reactant surface area increases the rate of a reaction because the total number of collisions possible per unit time is increased proportionately. Concentration – [reactant] ([ ] = concentration) , Rate , Surface Area – Surface Area , Rate 

•

An increase in temperature increases the rate of a reaction for two reasons. The total number of collisions possible per unit time is increased slightly, but more importantly, the fraction of collisions which are energetic enough to be effective is increased dramatically. Temperature – Temp , Rate  (and vice-versa)

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•

A catalyst increases the rate of a reaction by providing an alternative pathway to the same product formation that has a lower activation energy. A much larger fraction of collisions is effective following the changed reaction mechanism. Catalysts are involved in the reaction mechanism at some point, but are regenerated before the reaction is complete.

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Kinetic Collision Theory 1. Reaction Mechanism: individual steps during the process of a reaction.

2. Reaction Rate or Rate of Reaction: change in amounts of reactants or products over time.

3. Rate-Determining Step: the slowest reaction step in any reaction mechanism. It has a fairly high “activation energy” and the rate of the overall reaction is controlled by this step.

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Kinetic Collision Theory (Potential Energy Diagram) Exothermic Reaction Exothermic reactions are always negative.

Endothermic Reaction Endothermic reactions are always positive.

Things to Remember: •

Always start at zero.

•

Endothermic = +

•

Exothermic = -

•

Label the diagram!

• • •

Eact(rev)  +∆H + Eact(fwd) Eact(fwd)  -∆H + Eact(rev) ∆H  If stated, the other ∆H is either positive or negative.

•

Put all information that is given to you on your diagram.

•

If you are only given a chemical formula, then you must use Hess’s Law of Heat Summation (Formations) to determine the ∆H value. You will only need to draw a PE (potential energy) diagram for the value of ∆H and nothing else. For example: if ∆H is negative, then you draw a PE diagram for below zero (Exothermic), and vice-versa.

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Questions for the Kinetic Collision Theory (Potential Energy Diagram) 1. Carbon monoxide, CO, reacts with nitrogen dioxide, NO2. Carbon dioxide, CO2, and nitric oxide, NO, are formed. Draw a potential energy diagram to illustrate the progress of the reaction. Label the axis, transition state, and the activated complex. Indicate the activation energy for the forward reaction, Eact(fwd) = 134 Kj, as well as ∆H = -226 Kj. Calculate the activation energy for the reverse reaction, Eact(rev), and show it on the graph.

CO (g) + NO2 (g)  CO2 (g) + NO (g)

Eact(fwd) = 134 Kj ∆H = -226 Kj * Eact(rev) = ? (360 Kj) * ∆H = + 226 Kj

* Indicates that you are/have solving/solved for these variables.

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Kinetic Collision Theory (Reaction Intermediates) Reaction Intermediates are substances that are formed during the reaction, but immediately react again and are not present when the reaction is complete.

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Questions for the Kinetic Collision Theory (Reaction Intermediates) 1. A chemist proposes the following mechanism. Write the overall balanced equation. Identify any reaction intermediate.

Step 1 : NO2 (g) + F2 (g)  NO2F (s) + F (s) Step 2 : NO2 (g) + F (s)  NO2F (s) --------------------------------------------2NO2 (g) + F2 (g)  2 NO2F (s)

F (s) is the reaction intermediate. It is a product in the first step, it becomes a reactant in the second step, and it is not present in the overall formula.

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Kinetic Collision Theory (Catalysts) A catalyst is a substance that increases the rate of a chemical reaction without being consumed by the reaction. Catalysts are of tremendous importance in all facets of chemistry, from the laboratory to industry. A catalyst works by lowering the activation energy of a reaction so that a larger fraction of the reactants have sufficient energy to react. It lowers the activation energy by providing an alternative mechanism for the reaction.

Things to Remember: •

•

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The activation energy decreases and the ∆H remains the same when a catalyst is added to a reaction. A catalyst changes the activation energy of the reaction.

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Questions for the Kinetic Collision Theory (Catalysts) 1. Consider the following reaction:

C

2A + B2  D + E

A chemist proposes the following reaction mechanism:

Step 1 : A + B2  AB2 Step 2 : AB2 + C  AB2C Step 3 : AB2C + A  A2B2 + C Step 4 : A2B2  D + E

(a) What is the roll of AB2C and AB2?  AB2C and AB2 are both reaction intermediates.

(b) What is the role of C?  C is a catalyst, as it is a reactant first, then a product, and it is not in the final, overall formula.

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Reversible Reactions and Chemical Equilibrium 1. Dynamic Equilibrium: occurs when opposing changes are occurring at the same time and at the same rate. For instance: a jar that is sealed with water. The water evaporates and condenses at the same time, and at the same rate.

2. Homogenous Equilibrium: reactants and products are in the same phase (state of matter – solid, liquid, gas).

3. Heterogeneous Equilibrium: reactants and products are in a different phase.

The Four Conditions That Apply to all Equilibrium Systems: 1. Equilibrium is achieved in a reversible process when the rates of opposing changes are equal.

 H2O (l)

H2O (g)

 H2 (g) + Cl2 (g)

2HCl (g)

2. The observable properties of a system at equilibrium are constant (color, pressure, concentration, and pH level).

3. Equilibrium can only be achieved in a closed system – does not allow anything to escape or enter.

4. Equilibrium can be approached from either direction (left to right, or, right to left).

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The Equilibrium Constant The Law of Chemical Equilibrium states, at equilibrium, there is a constant ratio between the products and reactants in any change.

For any general equilibrium reaction: aP + bQ

Kc =

cR + dS

[ R ]c [ S ]d

products

[ P ]a [ Q ]b

reactants

Where “Kc” is the equilibrium at constant, “a, b, c, d” are the coefficients of “P, Q, R, S”, and [ ] is the concentration in mols/litre.

Things to Remember: • •

When solving for “Kc”, you do not have to use units, for instance: mol, litres, or mol/l. If a problem states “mol”, then you must divide the numbers of mol by the litres to obtain “mol/l”.

•

Pure liquids and solids are not included (this is in relation to the subscript of each chemical compound, (l) and (s)).

•

• •

If the coefficient (either a, b, c, d) is 1, do not write it as a superscript number in the general equilibrium reaction formula, as it is understood to already be there. If there is only 1 product or reactant when calculating “Kc”, replace with a 1. If a problem states that products or reactants are initially a number, do not use these numbers when determining “Kc” or a product or reactant; you must only use numbers when it states that they are at “equilibrium”.

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Questions for The Equilibrium Constant 1. The reaction between propane and oxygen to form carbon dioxide and water vapour:

C3H8 (g) + 5O2 (g)

3CO2 (g) + 4H20 (g)

[ R ]c [ S ]d

Kc =

[ P ]a [ Q ]b

[ CO2 (g) ]3 [ H2O (g) ]4

Kc =

[ C3H8 (g) ] [ O2 (g) ]5

2. A mixture of nitrogen and chlorine gases was kept at a certain temperature in a 5.0 litre reaction flask. When the equilibrium mixture was analyzed, it was found to contain 0.0070 mol of N2 (g), 0.0022 mol of Cl2 (g), and 0.95 mole of NCL3 (g). Calculate the equilibrium constant for this reaction.

N2 (g) + 3Cl2 (g)

2NCl3 (g)

[ N2 (g) ] = 0.0070 mols ÷ 5.0 l = 0.0014 mols/l [ Cl2 (g) ] = 0.0022 mols ÷ 5.0 l = 0.00044 mols/l [ NCl3 (g) ] = 0.95 mols ÷ 5.0 l = 0.19 mols/l

Kc =

[ C3H8 (g) ] [ Cl2(g) ]3

Kc =

Kc =

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[ N2 (g) ] [ Cl2 (g) ]3 [ CO2 (g) ]2 [ H2O (g) ]4

Kc =

UNIT

[ NCl3 (g) ]2

(0.19) 2 (0.0014)(0.00044) 3

0.036 (1.19 x 10 -13)

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Kc =

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3.03 x 10 11

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Le Châtelier’s Principle When a chemical system at equilibrium is disturbed by a change in a property of the system, the system adjusts to oppose the change.

Variables

Direction of Change

Concentration

Increase

Decrease Temperature

Volume (Overall pressure)

Shifts to consume some of the added reactant or product. Shifts to replace some of the removed reactant or product.

Increase

Shifts to consume some of the added heat.

Decrease

Shifts to replace some of the removed heat.

Increase (decrease in pressure)

Decrease (increase in pressure) Catalyst

Response of System

Added to the reaction

Shifts toward the side with the larger total amount of gaseous entities. Shifts toward the side with the smaller total amount of gaseous entities. No effect.

Things to Remember: •

If “energy” or “kj” is stated on the left side (reactants), it is considered to be endothermic; however, if it is stated on the right side (products), it is considered to be exothermic.

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Examples for Le Châtelier’s Principle Concentration: A few crystals of OH- are added. H20 (l)

H+ (aq) + OH- (aq)

 Effect: shift left

CO2 (g) is removed from the container. CaCO3 (s) + energy

CaO (s) + CO2 (g)

 Effect: shift right

Temperature: The container is heated. H2O (l) + energy H2O (g)  Effect: shift right

Energy is removed. 2SO2 (g) + O2 (g)

2SO3 (g) + energy

 Effect: shift right

Volume (Overall Pressure): The volume is decreased (increase in pressure). N2 (g) + 3H2 (g)

2NH3 (g)

 Effect: shift right

The volume is increased (decrease in pressure). 2NO (g) + O2 (g)

2NO2 (g)

 Effect: shift left

The volume is decreased (increase in pressure). H2 (g) + I2 (G)

2HI (g)

 Effect: no effect (the coefficients are equal)

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Finding Equilibrium Amounts for Initial Amounts and Kc (Quadratics) Problem: The following reaction has an equilibrium constant of 25.0 at 1100 K.

H2 (g) + I2 (G)

2HI (g)

2.00 mol of H2 (g) and 3.00 mol of I2 (g) are placed in a 1.00 litre reaction vessel at 1100 K. What is the equilibrium concentration of each gas?

What is Required? You need to find [ H2 (g) ], [ I2 (g) ], and [ HI (g) ].

What is Given? You have the balanced chemical equation. You know the equilibrium constant for this reaction, Kc = 25.0. You also know the concentrations of the reactants and product: [ H2 (g) ]i = 2.00 mol/l, [ I2 (g) ]i = 3.00 mol/l, and [ HI (g) ]i = 0.

Plan Your Strategy: Step 1: Set up an ICE table. Let the change in molar concentrations of the reactants be x. Use the stoichiometry of the chemical equation to write expressions for the equilibrium concentrations. Record these expressions in your ICE table.

Step 2: Write the equilibrium expression. Substitute the expressions for the equilibrium concentrations into the equilibrium expression. Rearrange the equilibrium expression into the form of a quadratic equation. Solve the quadratic equation for x.

Step 3: Substitute x into the equilibrium line of the ICE table to find the equilibrium concentrations.

Act on Your Strategy: Step 1: Set up an ICE table.

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Concentration (mol/l) Initial Change Equilibrium

H2 (g)

+

I2 (g)

2HI (g)

2.00

3.00

0

-x

-x

+2x

2.00 - x

3.00 - x

+2x

Finding Equilibrium Amounts for Initial Amounts and Kc (Quadratics) Continued… Step 2: Write and solve the equilibrium expression.

Kc =

25.0 =

[ HI (g) ]2 [ H2 (g) ] [ I2 (g) ] (2x) 2 (2.00 – x)(3.00 – x)

(2X)2

25.0 =

(6.00 – 2.00 x -3.00x + x2)

This equation does not involve a perfect square. It must be re-arranged into a quadratic equation.

25.0 =

4x2 (6.00 – 5.00x + x2)

6.00 – 5.00x + x2 = 4x2 ÷ 25.0 6.00 – 5.00x + x2 = 0.16 x2 6.00 – 5.00x + x2 – 0.16x2 = 0 0.840x2 – 5.00x + 6.00 = 0

x=

x=

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2a

-(-5.00) + √(-5.00)2 – 4(0.840)(6.00) 2(0.840)

x=

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-b + √b22 – 4ac

5.00 + √25 – 20.16 1.68

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x=

x=

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5.00 + √4.84 1.68

4.3 and 1.7

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Finding Equilibrium Amounts for Initial Amounts and Kc (Quadratics) Continued… Step 3: The value x = 4.3 is not physically possible. It would result in negative concentrations of H2 (g) and I2 (g) at equilibrium. The concentration of each substance at equilibrium is found by substituting x = 1.7 into the last line of the ICE table.

[ H2 (g) ] = 2.00 – x [ H2 (g) ] = 2.00 – 1.7 [ H2 (g) ] = 0.3 mol/l

[ I2 (g) ] = 3.00 – x [ I2 (g) ] = 3.00 – 1.7 [ I2 (g) ] = 1.3 mol/l

[ HI (g) ] = 2x [ HI (g) ] = 2(1.7) [ HI (g) ] = 3.4 mol/l

Check Your Solution: The coefficients in the chemical equation match the exponents in the equilibrium expression. To check your concentrations, substitute them back into the equilibrium expression.

25.0 =

25.0 =

25.0 =

25.0 =

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(3.4)2 (0.3)(1.3)

(3.4)2 0.39

11.56 0.39

29.64

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Solving the quadratic equation gives a value of x, correct to one decimal place. As a result, [ H2 (g) ] can only have one significant figure. The calculated value of “Kc” is equal to the given value, within the error introduced by rounding.

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Finding Equilibrium Amounts for Initial Amounts and Kc (Perfect Square) Problem: The following reaction increases the proportion of hydrogen gas for use as a fuel.

CO (g) + H2O (g)

H2 (g) + CO2 (g)

This reaction has been studied at different temperatures to find the optimum conditions. At 700 K, the equilibrium constant is 8.3. Suppose that you start with 1.0 mol of CO (g) and 1.0 mol of H2O (g) in a 5.0 litre container. What amount of each substance will be present in the container when the gases are at equilibrium, at 700 K?

What is Required? You need to find the amount (in mols) of CO (g), H2O (g), H2 (g), and CO2 (g) at equilibrium.

What is Given? You have the balanced chemical equation. You know the initial amount of each gas, the volume of the container, and the equilibrium constant.

Plan Your Strategy: Step 1: Calculate the initial concentrations, if needed.

[ CO (g) ] = 1.0 mol ÷ 5.0 L = 0.2 mol/l [ H2O (g) ] = 1.0 mol ÷ 5.0 L = 0.2 mol/l

Step 2: Set up an ICE (initial, change, and equilibrium) table. Record the initial concentrations you calculated in step 1 in your ICE table. Let the change in molar concentrations of the reactants be x. Use the stoichiometry of the chemical equation to write the record expressions for the equilibrium concentrations.

Concentration (mol/l)

5:

+

H2O (g)

H2 (g)

+

CO2 (g)

Initial

0.2

0.2

0

0

Change

-x

-x

+x

+x

0.2 - x

0.2 - x

0+x

0+x

Equilibrium

UNIT

CO (g)

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Finding Equilibrium Amounts for Initial Amounts and Kc (Perfect Square) Continued… Step 3: Write the equilibrium expression. Substitute the expressions for the equilibrium concentrations into the equilibrium expression. Solve the equilibrium expression for x.

[ H2 (g) ] [ CO2 (g)]

Kc =

[ CO (g) ] [ H2O (g) ] (x)(x)

8.3 =

(0.2 – x)(0.2 – x)

8.3 =

√8.3 =

+ 2.88 =

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√ (X)2 √ (0.2 – x)2

(x) (0.2 – x)

=x

+ 2.88 (0.2 - x)

=x

0.148

5:

(0.2 – x)2

+ 2.88 (0.2 - x)

0.576 ÷ 3.88

UNIT

(X)2

= 3.88x

=x

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- 2.88 (0.2 - x)

=x

Finding Equilibrium Amounts for Initial Amounts and Kc (Perfect Square) Continued… - 0.576 ÷ -1.88

0.306

= -1.88x

=x

Step 4: The value x = 0.306 is physically impossible because it would result in a negative concentration of both CO (g) and H2O (g) at equilibrium. Therefore: [ H2 (g) ] = 0.148 mol/l [ CO2 (g) ] = 0.148 mol/l [ CO (g) ] = 0.20 – x [ CO (g) ] = 0.20 – 0.148 [ CO (g) ] = 0.052 [ H2O (g) ] = 0.20 – x [ H2O (g) ] = 0.20 – 0.148 [ H2O (g) ] = 0.052 To find the amount of each gas, multiply the concentrations of each gas by the volume of the container (5.0 litre in this case). Amount of H2 (g) = CO2 (g) = 0.75 mol Amount of CO (g) = H2O (g) = 0.25 mol

Check Your Solution: The equilibrium expression has product terms in the numerator and reactant terms in the denominator. The amounts of chemicals at equilibrium are given in mols. Check “Kc”.

8.3 =

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(0.75)2 (0.25)2

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8.3

= 9.0

Finding Equilibrium Amounts for Initial Amounts and Kc (Perfect Square) Continued… This is close to the given value, Kc = 8.3. The difference is due to some mathematical rounding.

Things to Remember: •

UNIT

•

“x” will always be negative on the reactant side, while “x” will always be positive on the product side. Be sure to use the coefficient on the chemical compound. For example: 2HI(g).

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Questions for Finding Equilibrium Amounts for Initial Amounts and Kc 1. A chemist was studying the following reaction.

SO2 (g) + NO2 (g) (g)

NO (g) + SO3 (g)

In a 1.0 litre container, the chemist added 1.7 x 10-1 mol of SO2 (g) to 1.1 x 10-1 mol of NO2 (g). At equilibrium, the concentration of SO3 (g) was found to be 0.089 mol/l. What is the value of “Kc” for the reaction at this temperature?

[ SO2 (g)]i = 1.7 x 10-1 mol ÷ 1.0 l = 1.7 x 10-1 mol mol/l [ NO2 (g)]i = 1.1 x 10-1 mol ÷ 1.0 l = 1.1 x 10-1 mol/l

Concentration (mol/l) Initial Change Equilibrium

SO2 (g)

+

NO2 (g)

NO (g)

1.7 x 10-1

1.1 x 10-1

0

0

- 0.089

- 0.089

+ 0.089

+ 0.089

0.081

0.021

0.089

0.089

[ SO2 (g) ] [ NO2 (g) ]

Kc =

Kc =

Kc =

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SO3 (g)

[ NO (g) ] [ SO3 (g) ]

Kc =

UNIT

+

(0.089)(0.089) (0.081)(0.021)

0.007921 0.001701

4.7 (rounded)

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