Unit4Day3-LaBrake - University of Texas at Austinctp.cm.utexas.edu/courses/fall2013/ch301/slides/Unit4Day... · 2013. 11. 14. · Tabulated data can be assumed to be at 25°C. Standard

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-LaBrakeMonday, November 11, 20131:15 PM

Unit4Day3-LaBrake Page 1


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


A bomb calorimeter measures heat at constantvolume, which is equivalent toa) ΔUb) ΔHc) Work

QUIZ: iClicker Question 1


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 surroundingsUniverse = 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



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 surroundingsUniverse = 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


ENER

GY

REACTION PATH

2 CH3OH + 3 O2 2 CO2 + 4 H2O + heat

Enthalpy for a Chemical Change


Thermochemical equationA 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?


A. -890 kJ

B. 890 kJ

C. -556 kJ

D. 556 kJ

POLL: iClicker Question 2



How much heat is released when 10 g of CH4 is combusted?


A. -890 kJ

B. 890 kJ

C. -556 kJ

D. 556 kJ



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


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 enthalpyRun an experiment that is typically too expensivePredict the spontaneity of a reaction

Enthalpy Calculations



There are three strategies for calculating Enthalpy

1. Hess’s Law2. Hess’s Law using Heats of Formation3. Bond Energies

Enthalpy Calculations


Activity

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


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




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



Hess’s LawThe 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


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



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 kJ2 H2(g) + CO(g) CH3OH(l) ΔHº = -128.33 kJ2H2(g) + O2(g) 2H2O(g) ΔHº = -483.64 kJ

Hess’s Law


GOAL: 2 CH4(g) + O2(g) 2 CH3OH(l)

CH4(g) + H2O(g) CO(g) + 3 H2(g) ΔHº = +206.10 kJ2 H2(g) + CO(g) CH3OH(l) ΔHº = -128.33 kJ2 H2(g) + O2(g) 2H2O(g) ΔHº = -483.64 kJ

Hess’s Law



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



Standard Enthalpy of FormationΔH 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) ΔHfº = -277.67 kJ mol-1

Standard Enthalpy of Formation, ΔHf


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



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




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



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



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

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


Bond EnthalpyThe heat required to break a mole of bonds at constant pressure.

ΔHr° = ΣBEreactants - ΣBEproducts

Bond Enthalpies



Bond EnthalpyThe heat required to break a mole of bonds at constant pressure.

ΔHr° = ΣBEreactants - ΣBEproducts

Bond Enthalpies


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


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?



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


Documents

Unit4Day3-LaBrake - University of Texas at Austinctp.cm.utexas.edu/courses/fall2013/ch301/slides/Unit4Day... · 2013. 11. 14. · Tabulated data can be assumed to be at 25°C. Standard