11:11 PM Unit 1 Kinetics and Equilibrium Chemistry 3202

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Unit 1Kinetics and

EquilibriumChemistry 3202

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Part 1: Reaction Kinetics (Chp. 12) Reaction Kinetics is the study of the

rate of a chemical reaction Qualitative:

Reactions may be described as being FAST or SLOW Fast – burning, explosions, precipitation Slow – rusting, fermentation

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Reaction Kinetics

Quantitative:

The rate of a reaction measures how fast products are formed or how fast reactants are consumed

Rate = Change in quantity

Change in time

POSSIBLE UNITS ??

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Measuring Reaction Rate The method used to determine

reaction rate will depend on the reaction being studied. (p. 466)

Methods:1. monitor pH if there is an acid or

base in the equation2. record gas volume or changes in

pressure if there is a gas in the reaction

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Measuring Reaction Rate

Methods: (cont’d)3. record changes in mass if solids are

present4. monitor absorption of light if there is a

color change5. changes in electrical conductivity

indicate changes in ion concentration

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MC: What could we use to measure the rate of this reaction?

Cu(s) + 2 AgNO3(aq) 2 Ag(s) + Cu(NO3)2(aq)

a) pressure c) gas volume

b) pH d) mass

Answer: d) because a solid is present

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MC: What could we use to measure the rate of this reaction?

SO3(g) + H2O(l) H2SO4(aq)

a) pressure c) gas volumeb) pH d) massAnswers: a) and c) because a gas is present b) because an acid is being produced

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What determines RATE??

All chemical reactions are bond breaking/bond forming events

The rate of a reaction depends on how quickly bonds are broken and how rapidly new bonds form.

KMT and Collision Theory are used to explain reaction rates.

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Kinetic Molecular Theory (KMT)

Matter is made of particles (atoms, ions, or molecules) in continuous motion

An increase in temperature: increases the speed of particles reduces the forces of attraction between

particles

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Kinetic Molecular Theory (KMT) KMT is supported by:

Diffusion – particles of a gas spread to fill their container (‘perfume in a room’)

- solids dissolve uniformly in liquids over time.

Pressure – a balloon remains inflated because gas particles are continuously hitting the sides of the balloon

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Energy Distribution of Particles

# of

par

ticle

s

25 °C

200 °C

Kinetic Energy

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Collision Theory

reactant particles must collide with one another for a chemical reaction to occur

particles must collide with proper orientation

collisions must have enough intensity to break old bonds and allow new bonds to form

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Collision Theory

to increase reaction rate you must increase the number of successful collisions between reactant molecules

VIDEO (VHS): Reaction Rates LASERDISK: 3 VIDEO CLIPS

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Factors Affecting Reaction Rate

1. Concentration

– an increase in the concentration of a reactant usually increases the rate of a chemical reaction

- the rate increases because there are:

- more particles resulting in

- more collisions between particles &

- more successful collisions.

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Factors Affecting Reaction Rate

2. Temperature

- an increase in the temperature increases the rate of a chemical reaction

- the higher temperature results in:

- more collisions between particles

- more intense collisions

NOTE: A temperature increase of 10 ºC usually causes reaction rate to double.

more successful collisions

faster rate

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Factors Affecting Reaction Rate3. Nature of Reactants– compounds with fewer bonds to break

will react more rapidly than compounds with many bonds

eg. propane (C3H8) burns faster than candlewax (C25H52) because it has fewer bonds

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Factors Affecting Reaction Rate3. Nature of Reactants

- compounds with weak bonds react more rapidly than compounds with strong bonds

– ions will react more rapidly than atoms and molecules

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Factors Affecting Reaction Rate4. Surface Area- crushing a solid to produce a powder, or

changing a substance to the gas phase, exposes more particles for collision

- if more particles are available for collision there will be:- more collisions- more successful collisions

faster ratefaster rate

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Factors Affecting Reaction Rate5. Catalysts

- a catalyst increases the reaction rate by providing a different reaction pathway or mechanism with a lower activation energy

- a catalyst IS NOT consumed by a chemical reaction.

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# of

par

ticle

s

Kinetic Energy

Ea withoutcatalyst

Ea withcatalyst

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Potential Energy Diagrams (p. 473)

PE diagrams show changes in potential energy (stored chemical energy) during chemical reactions

Exothermic reactions release more energy than they absorb (eg. burning)

Endothermic reactions absorb more energy than they release

(eg. photosynthesis)

NEW SLIDE

ΔH written in the equation and outside the equation (Thermochemical Equation)

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Potential Energy Diagrams

∆H represents the heat of reaction or enthalpy of reaction

∆H is the difference between the PE of the reactants and the PE of the products

the minimum energy needed for a chemical reaction to occur is the activation energy

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Potential Energy Diagrams

the activated complex for a reaction is a temporary, unstable, intermediate species that quickly decomposes to products

eg. H2 + I2 → H2I2 → 2 HI

activated complex

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Reaction Progress

PE

Reactants

Products

∆H

ENDOTHERMIC

(positive)

site of the activated complex

activation energy (Ea forward)

Eareverse

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Reaction Progress

PE Reactants

Products

∆H

EXOTHERMIC

(negative)Ea forward

site of AC

Ea reverse

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Formula: (OPTIONAL)

Eaforward - Eareverse = ΔH

This formula is NOT necessary if you prefer using the PE diagram.

Animation

Ea fwd Ea rev ΔH Endothermic or

Exothermic

25 -30

50 20

150 250

65 28

Sketch a PE diagram for each reaction

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Reaction Progress

PE

∆HEafwd

CO2 + H2O

C6H12O6 + O2

PhotosynthesisEarev

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Reaction Progress

PE

∆H

Eafwd

CO2 + H2O

C6H12O6 + O2

Respiration

Earev

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p. 474

∆H

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∆H

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∆H

p. 475

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Reaction Progress

PE

no catalyst

catalyzed

Effect of a catalyst

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Reaction Progress

PE

no catalystEXOTHERMIC

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Sample problem: p. 475 Questions:

p. 476; #’s 1 – 4

p. 484; #’s 1 – 4

p. 486; #’s 1,2, 4, 6, & 7

p. 468; # 4

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Reaction Mechanisms (pp. 477 – 485)reaction mechanism – the steps that

occur in a chemical reactionelementary reaction - each step in a reaction mechanismreaction intermediate – a molecule, atom or ion formed in one step and consumed in a later step NOTE: reaction intermediates are NOT included in the overall equation

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Reaction Mechanisms

eg. #1

Step #1 NO(g) + O2(g) NO3(g)

Step #2 NO3(g) + NO(g) 2 NO2(g)

Overall Equation:

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

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HBr + O2 → HOOBrfast

HOOBr + HBr → 2 HOBrslow

2 HOBr + 2 HBr → 2 H2O + 2 Br2

fast

p. 478 #’s 5 – 8

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Reaction Mechanisms

rate-determining step (RDS)

- the slowest step in a reaction mechanism

- to increase the rate of a reaction you must speed up the RDS

- increasing the concentration of a reactant will increase the rate ONLY IF the reactant is in the RDS

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Reaction Mechanisms

PE diagrams

- every step in a reaction mechanism has an activation energy which can be drawn on a PE diagram

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Reaction Progress

PE

Reaction Mechanisms

3-step mechanism

#1

#2

#3RDS ??

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Reaction Mechanisms

eg:

Step #1 H2CO2 + H+ H2CO2H+ fast

Step#2 H2CO2H+ HCO+ + H2O slow

Step #3 HCO+ CO + H+ fast

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Reaction Mechanisms

eg:

Overall H2CO2 H2O + CO

Omit H+ - catalyst

Omit H2CO2H+ & HCO+ - reaction intermediates

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Reaction Mechanisms

Reaction Progress

PE

H2CO2 + H+CO + H+

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p. 476; #’s 1 – 4

p. 484; #’s 1 – 4

p. 486; #’s 1,2, 4, 6, & 7

p. 468; # 4

p. 478; #’s 5 - 8 p. 484 #’s 5 – 9 p. 485 #’s 10, 12 p. 486 #’s 8, 10, 11 p. 487 #’s 14, 17 p. 829 #’s 128,129, 131, 132

p. 829 # 128

Step 1 H2(g) + NO(g) → H2O(g) +

N(g)

Step 2

Step 3 H2(g) + O(g) → H2O(g)

2H2(g) + 2NO(g) → N2(g) + 2H2O(g)

Part 2: Chemical EquilibriumEquilibrium

A balancing Act!Text Ch 13: p 488 - 541

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Part 2: Chemical Equilibrium

All reactions we have done have shown reactants being converted 100% to products

Many reactions are reversible with some products being converted back to reactants

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Part 2: Chemical Equilibrium

Dynamic equilibrium occurs when 2 opposing processes occur at the same rate

A chemical equilibrium occurs when two opposing chemical reactions occur at equal rates.

Demo: p. 491

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Types of Equilibria

1. Phase EquilibriaAn equilibrium may be established between different phases of a compound in a sealed container

eg.

H2O(l) in a sealed container

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Types of Equilibria

Initially: H2O(l) changes to H2O(g)

H2O(l) → H2O(g)

Gradually: H2O(g) changes to H2O(l)

H2O(l) ← H2O(g)

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Types of Equilibria

Using equilibrium notation:

H2O(l) ⇌H2O(g)

Temperature changes?

Closed vs. open system?

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Types of Equilibria

2. Solubility Equilibria occur in saturated solutions when NaCl(s) is placed in water, the

initial rate of dissolving is fast

NaCl(s) NaCl(aq)

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Types of Equilibria

as more solid dissolves, the rate of dissolving slows and recrystallization begins.

eg. NaCl(s) NaCl(aq)

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Types of Equilibria

when the solution is saturated there are NO VISIBLE CHANGES

At equilibrium, the RATE of dissolving and the RATE of recrystallization are EQUAL.

eg. NaCl(s) ⇌ NaCl(aq)

equilibrium

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Types of Equilibria

Temperature change?? Open vs. Closed ??

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Types of Equilibria

3. Chemical Equilibrium Chemical reactions that are reversible

usually result in chemical equilibrium

eg. NO2 gas changing to N2O4

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Types of Equilibria

Initially the forward rate is high

eg. 2 NO2(g) N2O4(g)

as more product forms, the reverse reaction begins and increases in rate.

eg. 2 NO2(g) N2O4(g)

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Types of Equilibria eventually the forward rate slows and

the reverse rate increases such that the FORWARD AND REVERSE RATES ARE EQUAL

eg. 2 NO2(g) ⇌ N2O4(g)

http://www.chm.davidson.edu/ronutt/che115/EquKin/EquKin.htm

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Conditions for Equilibrium (p. 492)

1. Macroscopic properties are constant

ie. NO OBSERVABLE CHANGE

2. Forward and reverse rates must be equal

3. A CLOSED SYSTEM is required for equilibrium

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Conditions for Equilibrium4. Equilibrium may occur from either

direction

eg. 2 NO2(g) ⇌ N2O4(g)

OR

N2O4(g) ⇌ 2 NO2(g)

p. 493; #’s 1 – 6Kinetics & Equilibrium #4

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Shifts in Equilibrium

Equilibrium occurs when the forward rate equals the reverse rate.

Changes in concentration, temperature and pressure/volume can cause the forward or reverse reaction rate to change.

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Shifts in Equilibrium

Eventually, a new equilibrium will be established with different reactant and product concentrations

Le Châtelier’s Principle is used to predict changes in concentrations when a stress is applied to a system at equilibrium.

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Le Châtelier’s Principle (p.520)When a stress is applied to a system at equilibrium, the system will adjust or shift to relieve the stress.

A change(stress) is applied to a system at equilibrium

Forward or reverse rate will change New equilibrium established

WHAT???

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Le Châtelier’s Principle1. Changes in concentration

An increase in concentration on one side of an equation favors or drives the reaction to the opposite side.

eg. What will happen if CO is added to this system at equilibrium?

CO(g) + 2 H2(g) ⇌ CH3OH(g)

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Le Châtelier’s Principle- the system will ‘shift’ to the right to use

the CO and produce more CH3OH

- some H2 will be used

[CH3OH] will increase

[H2] will decrease[CO] ‘spikes’ and then drops to a value higher than it was before the change

?? possible graph ??

forward rate increases;reverse rate catches up

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mol/L

time

CO

CH3OH

H2

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Le Châtelier’s Principle

An equilibrium shifts away from a substance that increases in concentration or toward a

substance that decreases in concentration.

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Le Châtelier’s PrincipleWHY???

- rates at equilibrium ???

- increasing the [CO] & forward rate ??

- reverse rate??

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Le Châtelier’s Principle- [H2] ??

- new equilibrium concentrations ??

** NO CHANGE in Keq **

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Le Châtelier’s Principle

decrease [CO]

increase [H2]

decrease [CH3OH]

CO(g) + 2 H2(g) ⇌ CH3OH(g)

What happens if we:

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Le Châtelier’s Principle

IMPORTANT NOTE:

- adding a solid or liquid does not change molar concentration

- changing the amount of a solid or liquid in an equilibrium will NOT cause a shift

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Le Châtelier’s Principle

eg.

CaCO3(s) ⇌ CaO(s) + CO2(g)

add CaCO3(s) ??

add CO2(g) ??

What happens if we:

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Le Châtelier’s Principle (ΔH & equations)2. Temperature

- Raising the temperature of an exothermic equilibrium favors the formation of reactants.

eg. CO(g) + 2 H2(g) ⇌ CH3OH(g) + 65 kJ

What happens if we increase temperature in this equilibrium?

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Le Châtelier’s Principle- Raising the temperature of an endothermic equilibrium favors formation of products.

eg. CaCO3(s) + heat ⇌ CaO(s) + CO2(g)

What happens if we increase temperature in this equilibrium?

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Le Châtelier’s PrincipleOR

- Raising the temperature shifts the equilibrium away from the energy term.

- Decreasing the temperature shifts the equilibrium toward the energy term.

CO(g) + 2 H2(g) ⇌ CH3OH(g) + 65 kJ

CaCO3(s) + heat ⇌ CaO(s) + CO2(g)

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Le Châtelier’s Principleeg. How would an increase in

temperature affect these equilibria?

N2(g) + 3 H2(g) ⇌ 2 NH3(g) + heat

2 SO3(g) ⇌ 2 SO2(g) + O2(g) ΔH = +197 kJ

2 SO3(g) + 197 kJ ⇌ 2 SO2(g) + O2(g)

**A change in temperature changes Keq**

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Le Châtelier’s Principle

3. Pressure/Volume An increase in pressure of a system at

equilibrium has the same effect as a decrease in the volume of the system.

(inverse relationship)

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Le Châtelier’s Principle Increasing the pressure of a system at

equilibrium by reducing volume causes the equilibrium to shift in the direction that reduces pressure

ie. shift to the side with fewer molecules of GAS!!

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Le Châtelier’s Principleeg. How would an increase in pressure -

caused by a decrease in volume - affect these equilibria?

N2(g) + 3 H2(g) ⇌ 2 NH3(g)

4 NH3(g) + 5 O2(g) ⇌ 4 NO(g) + 6 H2O(g)

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Le Châtelier’s Principle

eg. Why would a change in pressure NOT affect the following equilibria?

H2(g) + I2(g) ⇌ 2 HI(g)

2 Ag(s) + Zn2+(aq) ⇌ 2 Ag+

(aq) + Zn(s)

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Le Châtelier’s Principle

pp. 529, 530 #’s 33 – 37

(omit 36 for now)

p. 533 #’s 1 – 3, 5

Answers on p. 537

Lab: Perturbing Equilibrium

Animation

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Le Châtelier’s Principle

4. Catalyst- Does NOT cause a shift in equilibrium- Increases BOTH rates equally so

equilibrium is reached faster

5. Surface Area - same as a catalyst

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Equilibrium Constant (Keq)For any system at equilibrium, there is a

mathematical relationship between reactant and product concentrations

(Guldberg and Waage, 1864)

P. 494

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Equilibrium Constant (K)

See p. 495 eg. N2O4(g) ⇌ 2 NO2(g)

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Equilibrium Constant (K)

For the general equilibrium below:

aP + bQ ⇌ cR + dS

Keq = [R]c [S]d

[P]a [Q]b

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Equilibrium Constant (K)

- a, b, c, & d are coefficients used to balance the equation

- P, Q, R, & S are the reactants and products

- Kc is sometimes used instead of Keq when units are molar concentration

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Equilibrium Constant (K)

eg. Write the expression for Keq for:

2 SO2(g) + O2(g) ⇌ 2 SO3(g)

2 HCl(g) ⇌ H2(g) + Cl2(g)

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Equilibrium Constant (K)

NOTE!!

Solids or liquids ARE NOT included in the Keq or Kc expression because their concentration is constant

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Equilibrium Constant (K)eg. Write the expression for Kc for:

2 NaCl(s) + H2SO4(aq) ⇌ 2 HCl(g) + Na2SO4(aq)

CaCO3(s) ⇌ CaO(s) + CO2(g)

p. 497; #’s 1 - 5

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Interpreting K

Large vs small K values?

K much larger than 1 - products favoured

K much smaller than 1 - reactants favoured

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Changing K (p. 497)

For a given system at equilibrium, the value of the equilibrium constant depends only on temperature

ie. For any equilibrium, the only way to change the actual value of K is to change the temperature

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Calculations with K

Types of K calculations:

1. Given equilibrium concentrations, find K

2. Find a missing concentration given K and other concentrations.

3. Given initial concentrations and equilibrium data, find K (ICE tables)

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1.Given equilibrium concentrations, find K

eg. Calculate Kc for this equilibrium using the equilibrium concentrations given:

H2(g) + I2(g) ⇌ 2 HI(g)

[H2] = 0.22 mol/L

[I2] = 0.30 mol/L

[HI] = 1.56 mol/L

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Given the equilibrium concentrations;

[CO(g)] = 0.105 mol/L,

[H2(g)] = 0.250 mol/L

[CH3OH(g)] = 0.00261 mol/L,

What is the value of Keq for the equilibrium below?

CO(g) + 2 H2(g) ⇌ CH3OH(g)

(A) 0.0994

(B) 0.398

(C) 2.51

(D) 10.0

Be Careful!!The other answers are possible if you mess

up the calculation

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Given the equilibrium concentrations below, what is the value of Keq for

N2(g) + O2(g) ⇌ 2 NO(g) ?

[N2(g)] = 0.10 mol/L

[O2(g)] = 0.20 mol/L

[NO(g)] = 0.0030 mol/L

(A) 2.2×10−4

(B) 4.5×10−4

(C) 1.5×10−1

(D) 3.0×10−1

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Find Keq for the equilibrium below:

4 NH3(g) + 5 O2(g) ⇌ 4 NO(g) + 6 H2O(g)

[NH3(g) ] = 0.100 mol/L

[O2(g) ] = 0.200 mol/L

[ NO(g) ] = 0.300 mol/L

[ H2O(g) ] = 0.250 mol/L

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2. Find a missing concentration . . .

eg. Find the [HI] in the equilibrium below if Kc = 36.9, [H2] = 0.125 mol/L and [I2] = 2.56 mol/L.

H2(g) + I2(g) ⇌ 2 HI(g)

p. 499; #’s 6 – 9

p. 538; # 6 Which reaction is faster - forward

or reverse?

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3. ICE tables

Keq may be calculated given initial concentrations and at least one equilibrium concentration

Using the Initial concentration and the Change in concentration we can find the missing Equilibrium concentrations and calculate Keq

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ICE tables

eg. 4.30 mol of NH3 was placed in a 1.00 L closed container to establish this equilibrium:

2 NH3(g) ⇌N2(g) + 3 H2(g)

Calculate Kc if the equilibrium concentration of H2(g) = 0.500 mol/L

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ICE tables

2 [NH3] [N2] 3 [H2]

I

C

E

4.30 mol/L 0 0

-2x +x +3x

4.30 - 2x x 3x

0.500 mol/L

2x x 3x

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[H2] = 0.500 mol/L3x = 0.500 x = 0.167

[N2] = x

= 0.167 mol/L

[NH3] = 4.30 - 2x

= 3.97 mol/L

K = (0.500)3 x (0.167)

(3.97)2

= 0.00132

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ICE tableseg. 2.00 mol of N2, 4.00 mol of H2 and

3.00 mol of NH3 were allowed to come to equilibrium in a 1.00 L container

2 NH3(g) ⇌N2(g) + 3 H2(g)

Calculate Kc if the equilibrium concentration of NH3(g) = 3.50 mol/L

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ICE tables

2 [NH3] [N2] 3 [H2]

I

C

E

3.00 2.00 4.00

2x x 3x

3.00 + 2x 2.00 - x 4.00 - 3x

+2x -x -3x

= 3.50

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[H2] = 4.00 – 3x

= 3.25

[N2] = 2.00 - x

= 1.75 mol/L

[NH3] = 3.50 mol/L

3.00 + 2x = 3.50

x = 0.25

K = (3.25)3 x (1.75)

(3.50)2

= 4.90

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ICE tableseg. The oxidation of ammonia occurs

according to the following expression:

4 NH3(g) + 5 O2(g) ⇌4 NO(g) + 6 H2O(g)

0.800 mol of each chemical were placed in a 1.00 L container and there was 0.450 mol of NH3 at equilibrium. Calculate the equilibrium concentrations and Kc

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4 NH3(g) 5 O2(g) 4 NO(g) 6 H2O(g)

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[NH3] = 0.450 mol/L

0.800 – 4x = 0.450

x = 0.0875

[O2] = 0.3625 mol/L

[NO] = 1.15 mol/L

[H2O] = 1.325 mol/L

K = 37000

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ICE tables

eg. 2.30 mol of NH3 was placed in a 2.00 L closed container to establish this equilibrium:

2 NH3(g) ⇌N2(g) + 3 H2(g)

Calculate Kc if 25 % of the NH3(g) reacts.

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[NH3] = 2.30 mol/2.00 L = 1.15 mol/L

2 [NH3] [N2] 3 [H2]

I

C

E

1.15 0 0

2x x 3x

1.15 - 2x x 3x

-2x +x +3x

= (0.75 x 1.15) = 0.8625 mol/L

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[NH3] = 0.8625

1.15 - 2x = 0.8625

x = 0.1438

[N2] = x

= 0.1438 mol/L

[H2] = 3x

= 0.4314 mol/L

K = 0.0155

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ICE tables

eg. 0.800 mol of NH3 was placed in a 1.00 L closed container to establish this equilibrium:

2 NH3(g) ⇌N2(g) + 3 H2(g)

Calculate Kc if 70% of the NH3 reacts.

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Le Châtelier’s Principle

Chapter Review

p. 535 #’s 2, 3, 5-7

p. 536 #’s 13, 15

p. 538 #’s 7, 8, & 10 (MC)

p. 539 #’s 15, 17, 19, 20, 22

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ICE tables

Assignment

test

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ΔH & equations The energy term may be included in a

chemical equation, or written as ΔH to the right of the equation.

eg. CO(g) + 2 H2(g) ⇌ CH3OH(g) + 65 kJ

OR

CO(g) + 2 H2(g) ⇌ CH3OH(g) ΔH = - 65 kJ

EXOTHERMIC

Energy is PRODUCED

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ΔH & equations

eg. N2(g) + O2(g) + 90 kJ ⇌ 2 NO(g)

OR

N2(g) + O2(g) ⇌ 2 NO(g) ΔH = + 90

kJ

Energy is REQUIRED ENDOTHERMIC

Back to Temperature

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ΔH & equations The energy term may be included in a

chemical equation, or written as ΔH to the right of the equation.

eg. CO(g) + 2 H2(g) ⇌ CH3OH(g) + 65 kJ

OR

CO(g) + 2 H2(g) ⇌ CH3OH(g) ΔH = - 65 kJ