Chem II Chapter 5 Hess’s Law.

Slides:

Advertisements

Similar presentations

Calculating heat changes

Advertisements

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.

Hess’s law calculations. 2C(s) + 3H 2 (g) + ½O 2 (g) → C 2 H 5 OH(l) C(s) + O 2 (g) → CO 2 (g) ∆H = –393 kJ mol –1 H 2 (g) + ½O 2 (g) → H 2 O(l) ∆H =

Lecture 4: Hess’s Law Reading: Zumdahl 9.5 Outline: Definition of Hess’ Law Using Hess’ Law (examples)

Lecture 4: Hess’ Law Reading: Zumdahl 9.5 Outline –Definition of Hess’ Law –Using Hess’ Law (many examples)

Part 2.  Review…  Solve the following system by elimination:  x + 2y = 1 5x – 4y = -23  (2)x + (2)2y = 2(1)  2x + 4y = 2 5x – 4y = -23  7x = -21.

Solving a System of Equations using Multiplication

Algebra II w/ trig. Substitution Method: 1. Solve an equation for x or y 2. Substitute your result from step 1 into the other equation and solve for the.

The heat evolved or absorbed in a chemical process is the same whether the process takes place in one or several steps.

Hess’s Law Review  Q - What is the first Law of Thermodynamics?

By: Olivia Bohnhoff & Zach Feldker. What is kinetics?  The study of the speed of reactions.  It’s based on experiments.  We did it first semester,

Question of the Day: 1. A __ enthalpy and __ entropy are good indicators that a reaction is probably spontaneous. Day

Calculating Heats of Reaction

Solve by using the ELIMINATION method The goal is to eliminate one of the variables by performing multiplication on the equations, and then add the two.

Solve the following system using the elimination method.

Energy: Standard Enthalpy of Formation and Reaction – Direct & Hess’s Law Standard Enthalpy of Formation and Reaction.

Energy: Standard Enthalpy of Formation and Reaction – Direct & Hess’s Law Enthalpy Enthalpy (H): heat flow for a chemical reaction. q constant P.

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.

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.

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.

Solving Systems of Equations Using the Elimination Method.

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.

Ch. 16 Reaction Energy. Thermochemistry  Thermochemistry: the study of the transfers of energy as heat that accompany chemical reactions and physical.

Ch 5: Hess ’ s Law. Hess ’ s Law states.. “ the enthalpy change for a chem rxn is the same whether the rxn takes place in one step or several steps ”

Entry Task Monday, May 23 How many grams of HCl will you need to produce 514 kJ of energy using the following equation? CaCO 3 (s) + 2 HCl (aq) → CaCl.

16.1(b) Hess’s Law 1 2 POINT > Recall enthalpies of reaction, formation and combustion POINT > Define Hess’s Law POINT > Use Hess’s law to determine.

CHEM 100 Fall Page- 1 Instructor: Dr. Upali Siriwardane Office: CTH 311 Phone Office Hours: M,W, 8:00-9:00.

Fundamentals of Logic Design, 7 th editionRoth/Kinney © 2014 Cengage Learning Engineering. All Rights Reserved. 1 Boolean Algebra (continued) UNIT 3.

Calculating Heats of Reaction

Enthalpy and Hess’s Law

Thermodynamics.

10.3 Solving Linear Systems

Solving Systems of Equations

3.4 Solving Systems with 3 variables

Solving Systems of Equations

How much heat is released when 4

Thermodynamics Topic 7 Chemistry Semester 2

Solve Systems of Equations by Elimination

Warm up How many calories are in 535 kJ?

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.

Solve Systems of Equations by Elimination

Thermochemistry Chapter 6.

Unit 5: Thermochemistry

Hess’s Law Determine the enthalpy change of a reaction that is

2-4 Solving Multi-Step Equations

Hess' Law Learning Goals:

The Product Rule.

Calculating Heats of Reaction

Unit 10: Energy in Chemical Reactions

Solving Systems of Equations

Hess’s Law 17.4.

Solving Systems of Equations

Hess’s Law 17.4.

Hess’s Law Hess’s law allows you to determine the energy of chemical reaction without directly measuring it. The enthalpy change of a chemical process.

Solving Systems of Equations

THERMOCHEMISTRY Thermodynamics

Chapter 5 Thermochemistry Part B

Either way, you get to the finish.

Thermochemistry Lesson # 4: Hess’s Law.

__C3H8(g) + __O2(g) ⇌ __CO2(g) + __H2O(g)

Ch. 17: Reaction Energy and Reaction Kinetics

6.3 Elimination with Multiplication!

Given this NET reaction

Presentation transcript:

Chem II Chapter 5 Hess’s Law

N2 (g) + 2O2 (g) ------> 2NO2 (g) DH1 = 68kJ The equation below…. N2 (g) + 2O2 (g) ------> 2NO2 (g) DH1 = 68kJ ….is derived from a combination of two other equations…. N2 (g) + O2 (g) ---------> 2NO(g) DH2 = 180kJ 2NO(g) + O2 (g) -------------> 2NO2 (g) DH3 =-112kJ

Notice that when the two equations are added together…. N2 (g) + O2 (g) ---------> 2NO(g) DH2 = 180kJ + 2NO(g) + O2 (g) -------------> 2NO2 (g) DH3 =-112kJ ….that the NO molecules cancel.

…after eliminating the NO’s on both sides… N2 (g) + O2 (g) ---------> DH2 = 180kJ + O2 (g) -------------> 2NO2 (g) DH3 =-112kJ --------------------------------------------------------------- ...now, add up what is leftover….. N2 (g) + O2 (g) + O2 (g) ------> 2NO2 (g) …we get the original equation N2 (g) + 2O2 (g) ------> 2NO2 (g) DH1 = 68kJ

Let’s try another equation…… W(s) + C(graphite) --------> WC(s) DH=?? kJ Using the following three equations…. 2W(s) + 3O2 (g) ------> 2WO3(s) DH= -1680.6 kJ C(graphite) + O2 (g) ------> CO2 (g) DH= - 393.5 kJ 2WC(s) + 5O2 (g) --------> 2WO3 (s) + 2CO2(g) DH= - 2391.6 kJ How can these equations be converted derive the original equation???

Dealing with the first equation… 2W(s) + 3O2 (g) ------> 2WO3(s) DH= -1680.6 kJ …..need to keep W, but find a way to eliminate the 3O2 (g)+ 2WO3(s) . Take ½ of everything, including that of DH= -1680.6 kJ. ½ [2W(s) + 3O2 (g) ------> 2WO3(s) DH= -1680.6 kJ] This leaves… W(s) + 3/2O2 (g) ------> WO3(s) DH= -840.3 kJ

Now the 2nd equation… C(graphite) + O2 (g) ------> CO2 (g) DH= - 393.5 kJ …need to keep C, but not O2 (g) and CO2 (g) No need to alter the equation. The terms O2 (g) and CO2 (g) will be eliminated later.

The 3rd equation requires a lot of work. 2WC(s) + 5O2 (g) --------> 2WO3 (s) + 2CO2(g) DH= - 2391.6 kJ 1st, WC needs to become a product and everything else eliminated. Additionally, take ½ of everything, including DH= - 2391.6 kJ. Reverse the equation…. 2WO3 (s) + 2CO2(g) ---------> 2WC(s) + 5O2 (g) DH= + 2391.6 kJ …..then, take ½ of everything…. ½ [2WO3 (s) + 2CO2(g) --------> 2WC(s) + 5O2 (g) DH= + 2391.6 kJ] This leaves… WO3 (s) + CO2(g) ---------> WC(s) + 5/2 O2 (g) DH= + 1195.8 kJ

Now, put the three equations along with their Enthalpy values. Notice all of the terms that cancel out. W(s) + 3/2O2 (g) ------> WO3(s) DH= -840.3 kJ C(graphite) + O2 (g) ------> CO2 (g) DH= - 393.5 kJ WO3 (s) + CO2(g) ---------> WC(s) + 5/2 O2 (g) DH= + 1195.8 kJ (2 ½ O2 molecules = 5/2 O2 molecules ) ----------------------------------------------------------- W(s) + C(graphite) --------> WC(s) DH= -38.0 kJ