Hess’s Law. Several reactions in chemistry occur in a series of steps, rather than just one step. For example, the following reaction explains the combustion.

Slides:



Advertisements
Similar presentations
5.1 Exothermic and endothermic reactions If a reaction produces heat (increases the temperature of the surroundings) it is exothermic. If a reaction produces.
Advertisements

Thermochemistry Exothermic reactions release heat to the surroundings. Fe 2 O Al  2 Fe + Al 2 O kJ Potassium Permanganate Reaction Demo.
Thermochemistry 2 Hess’s Law Heat of Formation Heat of Combustion Bond Enthalpy.
Hess’s Law EQ: Why is Hess’s Law a useful tool in solving for ∆Hrxn?
Lecture 4: Hess’s Law Reading: Zumdahl 9.5 Outline: Definition of Hess’ Law Using Hess’ Law (examples)
EXAMPLE: How much heat is required to heat 10.0 g of ice at o C to steam at o C? q overall = q ice + q fusion + q water + q boil + q steam.
Using Standard Molar Enthalpies of Formation SCH4U0.
Standard Enthalpy Changes =  H o P = 1 bar (0.997 atm) T = 298K, unless otherwise specified n = 1 mole for key compound.
Chapter 6B Notes Thermochemistry West Valley High School AP Chemistry Mr. Mata.
Chemistry Chapter 7 – Hess’s Law Teacher: H. Michael Hayes.
It has been suggested that hydrogen gas obtained by the decomposition of water might be a substitute for natural gas (principally methane). To compare.
Hess’s Law Review  Q - What is the first Law of Thermodynamics?
Chapter : Chemical Reactions That Involve Heat Suggested Reading: Pages
Industrial Chemistry Hess’s law. Index Hess’s Law and its experimental verification Hess’s Law calculations, 4 examples. Hess’s Law.
Standard Enthalpies of Formation Learning Goal: You will be able to write formation equations, find the enthalpies of formation and use them & Hess’ Law.
More Heat Calculations What have we done?. We can figure out heat values and then put them into kJ / mole.
Dr Ku Syahidah Ku Ismail CHAPTER 5 STANDARD THERMODYNAMIC FUNCTIONS OF REACTION.
ERT 108/3 PHYSICAL CHEMISTRY FIRST LAW OF THERMODYNAMICS Prepared by: Pn. Hairul Nazirah Abdul Halim.
Thermochemistry Chapter 6. Thermochemistry is the study of heat change in chemical reactions.
It is impossible to measure enthalpy directly
The basis for calculating enthalpies of reaction is known as Hess’s law: the overall enthalpy change in a reaction is equal to the sum of enthalpy changes.
Thermochem Hess’s Law and Enthalpy of Formation Sections 5.6 and 5.7.
ENERGY EXCHANGES IN CHEMICAL REACTIONS
1 Hess suggested that the sum of the enthalpies (ΔH) of the steps of a reaction will equal the enthalpy of the overall reaction. Hess’s Law.
Hess’ Law. Many reactions can occur by many alternative routes. Hess' Law states: The enthalpy change for a reaction depends only on the energy of the.
JOHNMAR S. DELIGERO Chemistry/Biology 12 (Nova Scotia Curriculum) Sino-Canadian Program Henan Experimental High School Zhengzhou Henan, China
 Certain reactions cannot be measured by calorimetry ◦ Ex: slow reactions, complex reactions, hazardous chemicals…  We can substitute in other reactions.
Thermochemistry. Thermodynamics  Study of the changes in energy and transfers of energy that accompany chemical and physical processes.  address 3 fundamental.
Topic 5.1 and 5.2 Hess’s Law and Bond Enthalpies.
Thermochemisty (Enthalpy) and Hess’s Law Chapter 10 Sections
IIIIII Chapter 16 Hess’s Law. HESS’S LAW n If a series of reactions are added together, the enthalpy change for the net reaction will be the sum of the.
Stoichiometry Chemistry 6.0. The Mathematics of Chemical Reactions: STOICHIOMETRY I. Balanced Chemical Equations A. Provide qualitative and quantitative.
Hess’s Law. Enthalpy is a State Variable ‘State variable’ just means something that doesn’t change depending on the path you take to get from A to B.
Hess’s Law “In going from a particular set of reactants to a particular set of products, the change in enthalpy is the same whether the reaction takes.
Topic 5.3 and 5.4 Hess’s Law and Bond Enthalpies.
CHEMISTRY 122 Calculating the Heats of Reaction. Hess's Law  It is sometimes difficult to measure the change in enthalpy in a reaction because the reaction.
HESS’S LAW what is it ? how is it used ? AS Chemistry.
CHAPTER Every chemical reaction involves a change in energy (usually heat).
Hess’ Law Practice Problems 1)-79.6 kJ 2)-155 kJ 3)17.4 kJ 4)-3.70 kJ 5)-142 kJ 6) 88.1 kJ 7) kJ 8) kJ or kJ 9) -256 kJ, exo 10)
1.4 Energetics Practical 1.6 – Finding an enthalpy change that cannot be measured directly e. recall Hess’s Law and apply it to calculating enthalpy.
Hess’s Law 5.3 Energetics.
Section 4: Calculating Enthalpy Change
Hess’s Law.
Industrial Chemistry Hess’s law.
PROBLEMS 3 BATAA EL GAFAARY.
FLOW OF ENERGY Heat, Enthalpy, & Thermochemical Equations
How much heat is released when 4
In going from a particular set of reactants to a particular set of products, the change in enthalpy is the same whether the reaction takes place in one.
Hess’ Law.
Hess’s Law & Standard Enthalpies of Formation
Hess’s Law.
Hess's Law.
Unit 5: Thermochemistry
Hess’s Law Germain Henri Hess.
Stoichiometry Calculations involving Enthalpy
Standard Enthalpies of Formation
List of enthalpies for several kinds of reactions.
Hess’s Law.
Hess’s Law and Standard Enthalpies of Formation Unit 10 Lesson 4
AP Chem Get Heat HW stamped off Today: Enthalpy Cont., Hess’ Law
Enthalpy (∆H).
Hess’s Law and Standard Enthalpies of Formation Unit 10 Lesson 4
Chapter 16 Preview Objectives Thermochemistry Heat and Temperature
Hess’s Law.
Hess’s Law Germain Henri Hess.
1.2.5 Hess’s Law- the equation
Thermochemistry Lesson # 4: Hess’s Law.
__C3H8(g) + __O2(g) ⇌ __CO2(g) + __H2O(g)
Hess’s Law and Standard Enthalpies of Formation
1.2.5 Hess’s Law- the equation
Presentation transcript:

Hess’s Law

Several reactions in chemistry occur in a series of steps, rather than just one step. For example, the following reaction explains the combustion (burning) of carbon: C (s) + O 2 (g) → CO 2(g) ΔH = kJ However, sometimes the series below actually takes place in order to burn the carbon: Step 1: C (s) + ½ O 2(g) → CO (g) ΔH = kJ Step 2: CO (g) + ½ O 2(g) → CO 2 (g) ΔH = kJ

Hess’s Law If we add these two reactions together, namely recognize that CO in step 2 can be replaced by the reaction that first produced it, we get: Step 2: CO (g) + ½ O 2(g) → CO 2 (g) Then: C (s) + ½ O 2(g) + ½ O 2(g) → CO 2(g) Which we can simplify to: C (s) + O 2(g) → CO 2(g)

Hess’s Law An easier way to show this addition is by putting all the steps in a row and doing the following: C (s) + ½ O2(g) → CO (g) CO (g) + ½ O2(g) → CO2 (g) We can add the ½ O2 molecules together to make 1 mole of O2. In summary: 1. Cancel off like terms on opposite sides of the equations. 2. Add like terms on same side of equations.

Hess’s Law C (s) + ½ O 2(g) → CO (g) ΔH = kJ CO (g) + ½ O 2(g) → CO 2 (g) ΔH = kJ C (s) + O 2(g) → CO 2 (g)

Hess’s Law Now, if we add the heat of the first reaction to the heat of the second reaction, we should get the heat of reaction for the overall reaction. ΔH 1 + ΔH 2 = ΔHtotal kJ + ( kJ) = kJ Therefore, the end result does not matter which path it takes. The carbon could combust right away, or it could take the path we just investigated. Either way, the enthalpy change in the end was the same.

Hess’s Law C (s) + ½ O 2(g) → CO (g) ΔH = kJ CO (g) + ½ O 2(g) → CO 2 (g) ΔH = kJ C (s) + O 2(g) → CO 2 (g) ΔH = kJ

Hess’s Law Hess’s Law states that the enthalpy change for any reaction only depends on the energy states of the final products and initial reactants and is independent on the number of steps (or the pathway) in between.

Hess’s Law Ex) Given the intermediate steps in the production of tetraphosphorus decaoxide, P 4 O 10, calculate ΔH°f for P 4 O 10 given the following reactions: 1. 4 P + 3 O 2 → P 4 O 6 ΔH = kJ 2. P 4 O O 2 → P 4 O 10 ΔH = kJ

Tips Tips for using Hess’s Law: 1. If you need to flip an equation, the energy term must change sign, or it must be on the opposite side of the equation. i.e. 4 P + 3 O 2 → P 4 O 6 ΔH = kJ P 4 O 6 → 4 P + 3 O 2 ΔH = kJ

Tips 2. If you change the balancing coefficients by multiplying by a number, multiply the energy value by the same number. i.e 4 P + 3 O 2 → P 4 O 6 ΔH = kJ if you wanted to produce 2 moles of P4O6 then: 8 P + 6 O2 → 2P4O6 ΔH = kJ

Example The combination of coke and steam produces a mixture called coal gas, which can be used as a fuel or as a starting material for other reactions. If we assume coke can be represented by graphite, the equation for the production of coal gas is: 2C (s) + 2H 2 O (g)  CH 4 (g) + CO 2 (g) Determine the overall enthalpy change for the above reaction based on the following intermediate steps: C (s) + H 2 O (g)  CO (g) + H 2 (g)∆H = 131.3kJ CO (g) + H 2 O (g)  CO 2 (g) + H 2 (g) ∆H = kJ CH 4 (g) + H 2 O (g)  3H 2 (g) + CO (g) ∆H = kJ