17.1-17.4.ppt

Thermochemistry

Thermochemistry

Study of the transfer of energy in

Study of the transfer of energy in

Heat and Temperature

Heat and Temperature

What’s the difference??

What’s the difference??

Temperature (C0): measures the average kinetic

energy of particles.

Heat

Heat

Heat (Q): Q= m ∆t Cp

(Total Thermal Energy of a system) Instrument: calorimeter p.519

Heat of Reaction (∆H)

Heat of Reaction (∆H)

The net energy released or absorbed in a

The net energy released or absorbed in a

chemical reaction.

chemical reaction.



Thermochemical equation

Thermochemical equation

: includes the

: includes the

amount of energy released/absorbed.

amount of energy released/absorbed.

Write the thermochemical equation of the

Write the thermochemical equation of the

formation of water if 483.6 kJ of energy are

formation of water if 483.6 kJ of energy are

produced from 2 moles of hydrogen.

produced from 2 moles of hydrogen.

2H

2H₂₂ (g) + O (g) + O₂₂ (g) (g)  2H 2H₂₂O (l) + 486.3 kJ (exo)O (l) + 486.3 kJ (exo) Always include states of matter!

Enthalpy (heat) Changes (

Enthalpy (heat) Changes (

Δ

Δ

H)

H)

The amount of energy absorbed or lost by a system.

The amount of energy absorbed or lost by a system.



How to calculate

How to calculate

Δ

Δ

H:

H:



A)

A)

from a graph

from a graph

:

:

Heat

Heat

– Heat

– Heat

If

If ΔΔH = negative: exothermic (Energy Lost)H = negative: exothermic (Energy Lost) If

B)

∆H can be calculated using:

Hess’s Law :The overall ΔH is equal to the sum of the ΔH for individual steps.

Molar heat of formation: ∆H_f is the net heat released or absorbed when one mole of a compound is formed by the combination of its elements. (synthesis)

∆H_f values (Appendix A-14) are based on one mole of product. -∆H_f indicates a stable compound.

Molar heat of combustion: ∆H_c is the heat released by the complete combustion of one mole of a substance.

Calculate the

Calculate the

Δ

Δ

H

H

for:

for:

2SO

2SO

+ O

+ O





2SO

2SO



Equation must be balanced.

Equation must be balanced.



Table A-14 (

Table A-14 (

Δ

Δ

H

H

)

)



Δ

Δ

H

H

SO

SO

= (-296.8)kJ/mol

= (-296.8)kJ/mol



Δ

Δ

H

H

SO

SO

= (-395.7)kJ/mol

= (-395.7)kJ/mol



∆

∆

H

H

= (∆H

= (∆H

products – (∆H

products – (∆H

reactants)

reactants)



(-791.4) - (-593.6)

(-791.4) - (-593.6)

Multiply both SO₂ and SO₃ by 2 because we have 2 moles of each.

-593.6 kJ -791.4 kJ

Sample problems(continued)

Sample problems(continued)

 2) Calculate the heat of reaction for:2) Calculate the heat of reaction for:

 2 H2 H₂₂OO₂₂ 2H 2H₂₂O + OO + O₂₂ꜛꜛ (must be balanced) (must be balanced)  Find ∆HFind ∆H_ff table A-14 table A-14

 ∆∆HH_ff H H₂₂OO₂₂ (-187.8 kJ/mol) (2) = (-187.8 kJ/mol) (2) = -375.6 kJ-375.6 kJ  ∆∆HH_ff O O₂₂ (0.00 kJ/mol) (0.00 kJ/mol)

 ∆∆HH_ff H H₂₂O (-285.8 kJ/mol) (2) = O (-285.8 kJ/mol) (2) = -571.6 kJ-571.6 kJ  ∆∆HH_rxnrxn = ∆H = ∆H_ff products - ∆H products - ∆H_{f f} reactantsreactants

 Calculating ∆H from Calculating ∆H from ∆H∆H_cc values.values.

 1. Calculate the heat of formation for:1. Calculate the heat of formation for:

 CC_(s)(s) + 2H + 2H_2(g)2(g) CH CH_{4(g) 4(g)}

 Make sure the equation is balanced.Make sure the equation is balanced.

 Use Use ∆H∆H_{c c} values of C, Hvalues of C, H₂₂ and methane (tbl. A-5) and methane (tbl. A-5)  NOTE:NOTE: ∆H∆H_{c c} values are based on one mole ofvalues are based on one mole of

REACTANTREACTANT. . Therefore:Therefore: ∆

∆HH_ff = ∆H = ∆H_cc of of REACTANTSREACTANTS – (∆H – (∆H_cc PRODUCTS) PRODUCTS)

C = -393.5 kJ/mol CH₄ = -890.8 kJ/mol

H = (-285.8 kJ) (2)

17.2 Two

17.2 Two Driving Forces Driving Forces of Reactions…Allow one to of Reactions…Allow one to predict if a reaction will occur spontaneously.

predict if a reaction will occur spontaneously.

 Enthalpy Enthalpy ((ΔΔH)H):: exothermic reactions favored (- exothermic reactions favored

(-∆H) (products at

∆H) (products at lowerlower energy) _energy)

 Entropy Entropy ((ΔΔS)S):: measures disorder of a system. measures disorder of a system.

(+∆S)

(+∆S)High disorderHigh disorder favored. _favored.

 What does high disorder look like? What does high disorder look like? Increase Increase

in randomness (disorder): in randomness (disorder):

 Solid to liquid to gas_{Solid to liquid to gas}

 Lesser # particles to greater #particles_{Lesser # particles to greater #particles}

Possible reactions

Possible reactions: RESULTS:_{: RESULTS:} 1. Exothermic and Entropy increase GO

2. Endothermic and Entropy decrease NO GO 3. Exothermic and Entropy decrease ?

4. Endothermic and Entropy increase ?

#3 & #4 may or may not “Go” depending on which of the two driving forces is dominant .

#3 example: water freezes. In this case, enthalpy(∆H) is the dominant factor.

Gibb’s Free Energy (G)

Gibb’s Free Energy (G)

Predicts whether ∆H or ∆S will dominate in a reaction

Predicts whether ∆H or ∆S will dominate in a reaction



Natural processes proceed in the direction

Natural processes proceed in the direction

that lowers the free energy (G) of a system.

that lowers the free energy (G) of a system.



Only

Only

Δ

Δ

G can be measured:

G can be measured:

∆G=∆H-(T∆S)

∆G=∆H-(T∆S)

…

… Predicts whether a reaction will be Predicts whether a reaction will be spontaneous (at constant T & P).

spontaneous (at constant T & P).

Negative ΔG = spontaneous reaction

Negative ΔG = spontaneous reaction 

Positive ΔG = nonspontaneous reaction

Sample problem: Sample problem:

Calculate the free energy change for the following reaction at

250 C:

Ca_(s) + 2H₂O_(l) Ca(OH)₂ + H₂

Data: ∆H = -411.6 kJ ∆S = 31.8 J/mol.K

∆G = ∆H – (T∆S)

-411.6 - (298 x .0318 kJ/mol.K)

Summary

Summary

Favorable conditions for a spontaneous reaction:

 Free Energy (ΔG) = Negative

 Enthalpy (ΔH) = Negative

 Entropy (ΔS) = Positive

 _{When enthalpy and entropy are at odds,}

17.3 The Reaction Process

17.3 The Reaction Process

Collision Theory

Collision Theory

: Explanation of reactions

: Explanation of reactions

as a result of collisions.

as a result of collisions.

– Particles must collideParticles must collide

– Particles must collide in the Particles must collide in the correct spatial orientation

correct spatial orientation

– Collision must be energetic Collision must be energetic enough to disrupt bonds

Reaction Pathways

Reaction Pathways



Activation Energy

Activation Energy

: minimum energy

: minimum energy

required to transform reactants

required to transform reactants





activated

activated

complex.

complex.



Activated Complex

Activated Complex

: transitional structure

: transitional structure

formed from an effective collision.

formed from an effective collision.

A= Energy of Reactants B= Activation Energy

C= Energy of Activated Complex D= Energy of Products

17.4 Reaction Rates and Kinetics

17.4 Reaction Rates and Kinetics

Rate-Influencing Factors

Rate-Influencing Factors

: Anything that will

: Anything that will

influence collision frequency and/or efficiency.

influence collision frequency and/or efficiency.

– Nature of ReactantsNature of Reactants

– Surface Area (increased surface area)Surface Area (increased surface area)

– Temperature (increased temperature)_{Temperature (increased temperature)}

– Concentration (increased conc. of reactants)Concentration (increased conc. of reactants)

Rate Laws

Rate Laws

…Relate reaction rate to the [concentration] of

…Relate reaction rate to the [concentration] of

reactants, at a given temperature.

reactants, at a given temperature.

As concentration increases, rate increases.

As concentration increases, rate increases.

1)

1)Elementary Reactions Elementary Reactions (single step)(single step)

2A + 3B A2A + 3B A₂₂BB₃₃

Rate Law: R = k[A]_{Rate Law: R = k[A]}22 [B] [B]3 3

3D E + F3D E + F

2)

2) Multi-step ReactionsMulti-step Reactions

Rate is determined from the slowest step. example:

Net equation: NO₂ + CO NO + CO₂

step 1: 2NO₂ NO₃ + NO (slow)

step 2: NO₃ + CO NO₂ + CO₂ (fast)

Rate Law?

Rate law determination

Rate law determination

by experimental data

by experimental data

:

:

In a one step reaction:

•Doubling [A]: the rate increases by 2X •_{Doubling [B]: the rate increases by 4X}

•Reduce [B] to 1/3: the rate decreases to 1/9 What is the rate law?