Unit4Day3-LaBrake Monday, November 11, 2013 1:15 PM Vanden - - PDF document

CH301 Vanden Bout/LaBrake Fall 2013

Vanden Bout/LaBrake/Crawford CH301 THERMODYNAMICS Quantifying Heat Flow – Chemical Change UNIT 4 Day 3

CH302 Vanden Bout/LaBrake Spring 2012

Important Information

HW10 DUE T 9AM

Unit4Day3-LaBrake

Monday, November 11, 2013 1:15 PM Unit4Day3-LaBrake Page 1

CH301 Vanden Bout/LaBrake Fall 2013

What are we going to learn today?

Use calorimetry to calculate ΔHrxn Use different methods to calculate ΔHrxn Define Heats of Formation, Hess’s Law, and Bond Energies

CH301 Vanden Bout/LaBrake Fall 2013

A bomb calorimeter measures heat at constant volume, which is equivalent to a) ΔU b) ΔH c) Work

QUIZ: iClicker Question 1

CH301 Vanden Bout/LaBrake Fall 2013

System and State

A system is the part of the universe on which we want to focus our attention. The surroundings are everything else The universe is the system and the surroundings Universe = system + surroundings We also describe chemical changes with beginning and end states A change in a chemical reaction is described as Δ State = Stateend – Statebeginning

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CH301 Vanden Bout/LaBrake Fall 2013

We also describe chemical changes with beginning and end states A change in a chemical reaction is described as Δ State = Stateend – Statebeginning

CH301 Vanden Bout/LaBrake Fall 2013

ENERGY REACTION PATH 2 CH3OH + 3 O2  2 CO2 + 4 H2O + heat

Enthalpy for a Chemical Change

CH301 Vanden Bout/LaBrake Fall 2013

Thermochemical equation A chemical reaction written with the corresponding enthalpy change CH4(g) + 2O2(g)  CO2(g) + 2H2O(l) ΔH = -890 kJ mol rxn-1 2CH4(g) + 4O2(g)  2CO2(g) + 4H2O(l) ΔH = -1780 kJ mol rxn-1 CO2(g) + 2H2O(l)  CH4(g) + 2O2(g) ΔH = +890 kJ mol rxn-1

Enthalpy for a Chemical Change

How much heat is released when 10 g of CH4 is combusted? CH4(g) + 2O2(g)  CO2(g) + 2H2O(l) ΔH = -890 kJ mol rxn-1

• A. -890 kJ

B. 890 kJ

• C. -556 kJ

D. 556 kJ

POLL: iClicker Question 2

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CH301 Vanden Bout/LaBrake Fall 2013

CH4(g) + 2O2(g)  CO2(g) + 2H2O(l) ΔH = -890 kJ mol rxn-1

• A. -890 kJ

B. 890 kJ

• C. -556 kJ

D. 556 kJ

CH301 Vanden Bout/LaBrake Fall 2013

Reaction enthalpies are based on all reactants and products in their standard state (1 bar pressure). Tabulated data can be assumed to be at 25°C. Standard conditions are represented by

Standard Enthalpies, ΔHrxn

CH301 Vanden Bout/LaBrake Fall 2013

It is possible to calculate the enthalpy of a chemical change using tabulated data. We can do this because Enthalpy is a State Function These type of calculations allow us to Estimate reaction enthalpy Run an experiment that is typically too expensive Predict the spontaneity of a reaction

Enthalpy Calculations

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CH301 Vanden Bout/LaBrake Fall 2013

There are three strategies for calculating Enthalpy

• 1. Hess’s Law
• 2. Hess’s Law using Heats of Formation
• 3. Bond Energies

Enthalpy Calculations

CH301 Vanden Bout/LaBrake Fall 2013

Activity

Please open your course pack to page 107. Thermo Unit – Reaction Enthalpies

CH301 Vanden Bout/LaBrake Fall 2013

Calculate the reaction enthalpy for the combustion of methanol.

• A. -2233 kJ mol-1
• B. -1516 kJ mol-1
• C. -3793 kJ mol-1
• D. -1277 kJ mol-1

POLL: iClicker Question 4

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CH301 Vanden Bout/LaBrake Fall 2013 CH301 Vanden Bout/LaBrake Fall 2013

Hess’s Law The combination of a series of chemical reactions to estimate the change in enthalpy for an overall net reaction What is the standard change in enthalpy for the following reaction? C(s) + 0.5 O2(g) CO(g) GIVEN: CO2(g )  CO(g ) + ½O2(g ) ΔHº = + 283 kJ mol rxn-1 C(s) + O2(g)  CO2(g) ΔHº = - 393 kJ mol rxn-1

Hess’s Law

CH301 Vanden Bout/LaBrake Fall 2013

GOAL: C(s) + 0.5 O2(g) CO(g)

CO2(g )  CO(g ) + 0.5 O2(g ) ΔHº = + 283 kJ mol rxn-1 C(s) + O2(g)  CO2(g) ΔHº = - 393 kJ mol rxn-1

Hess’s Law

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CH301 Vanden Bout/LaBrake Fall 2013

Suppose the controlled reaction of the oxygen in air with Methane could produce Methanol, which is a clean burning liquid fuel source. Find the standard reaction enthalpy for the formation of 1 mole of CH3OH(l) from methane and oxygen, given the following information: CH4(g) + H2O(g)  CO(g) + 3 H2(g) ΔHº = +206.10 kJ 2 H2(g) + CO(g)  CH3OH(l) ΔHº = -128.33 kJ 2H2(g) + O2(g)  2H2O(g) ΔHº = -483.64 kJ

Hess’s Law

CH301 Vanden Bout/LaBrake Fall 2013

GOAL: 2 CH4(g) + O2(g)  2 CH3OH(l)

CH4(g) + H2O(g)  CO(g) + 3 H2(g) ΔHº = +206.10 kJ 2 H2(g) + CO(g)  CH3OH(l) ΔHº = -128.33 kJ 2 H2(g) + O2(g)  2H2O(g) ΔHº = -483.64 kJ

Hess’s Law

CH301 Vanden Bout/LaBrake Fall 2013

POLL: iClicker Question 5

Use the enthalpies of formation from the table on page 108 to calculate the enthalpy for the same reaction.

• A. -2233 kJ mol-1
• B. -1516 kJ mol-1
• C. -3793 kJ mol-1
• D. -1277 kJ mol-1

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CH301 Vanden Bout/LaBrake Fall 2013

Standard Enthalpy of Formation ΔH for the formation of 1 mole of a compound from its elements in their most stable form at standard conditions 2 C(gr) + 3 H2(g) + 0.5 O2(g)  1 C2H5OH(l) ΔHf

º = -277.67 kJ mol-1

Standard Enthalpy of Formation, ΔHf

CH301 Vanden Bout/LaBrake Fall 2013

What is the standard enthalpy of formation for O2(g)?

• A. -277.67 kJ mol-1

B. 277.67 kJ mol-1

• C. -6.02 kJ mol-1

D. 6.02 kJ mol-1 E. 0 kJ mol-1

POLL: iClicker Question 6

We can use the standard enthalpy of formation (ΔHf

°) to calculate the standard

enthalpy of reaction (ΔHf

°)

This is possible because ΔH is a state function (independent of pathway), where ΔH = ΔHfinal – ΔHinitial

ΔHr

°= ΣnΔHf ° products - ΣnΔHf ° reactants

Standard Enthalpy of Formation, ΔHf

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CH301 Vanden Bout/LaBrake Fall 2013

Δ Δ This is possible because ΔH is a state function (independent of pathway), where ΔH = ΔHfinal – ΔHinitial

ΔHr

°= ΣnΔHf ° products - ΣnΔHf ° reactants

Δ

CH301 Vanden Bout/LaBrake Fall 2013

Calculate the standard enthalpy of combustion of methanol from the provided data. ΔHf, CH3OH

° = -239 kJ mol-1

ΔHf, CO2

°

= -394 kJ mol-1 ΔHf, H2O

°

= -286 kJ mol-1

Example

CH301 Vanden Bout/LaBrake Fall 2013

POLL: iClicker Question 7

Use the bond energy table on page 108 to calculate the enthalpy for this same reaction.

• A. Approximately -1277 kJ mol-1 x 0
• B. Approximately -1277 kJ mol-1 x 1
• C. Approximately -1277 kJ mol-1 x 2
• D. Approximately -1277 kJ mol-1 x 3

Bond Enthalpy The heat required to break a mole of bonds at constant pressure. ΔHr

°= ΣBEreactants - ΣBEproducts

Bond Enthalpies

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CH301 Vanden Bout/LaBrake Fall 2013

Bond Enthalpy The heat required to break a mole of bonds at constant pressure. ΔHr

°= ΣBEreactants - ΣBEproducts

Bond Enthalpies

CH301 Vanden Bout/LaBrake Fall 2013

Estimate the ΔHr

° for

CCl3CHCl2(g) + 2HF(g)  CCl3CHF2(g) + 2HCl (g) BEC-Cl= 338 kJ mol-1 BEH-F = 567 kJ mol-1 BEC-F = 484 kJ mol-1 BEH-Cl = 43 kJ mol-1

Example

CH302 Vanden Bout/LaBrake Fall 2012

The transfer of heat energy into or out of a system at constant pressure is a state function called Enthalpy. The change in Enthalpy can be determined experimentally using a coffee cup calorimeter at constant pressure. Then change in Enthalpy can be calculated based on a variety of tabulated data: Heats of formation/Other Heats of Reaction/Bond Energies

What have we learned today?

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CH301 Vanden Bout/LaBrake Fall 2013

Write a formation chemical equation for a compound Calculate change in enthalpy for a reaction based on calorimetry data Calculate change in enthalpy for a reaction based on tabulated data (Hess’s law, formation data, bond energy data).

Learning Outcomes

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