- 28 Views
- Uploaded on
- Presentation posted in: General

Energy Changes in Chemical Reactions -- Chapter 17

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

E = internal energy = sum of all potential and kinetic energy of the system:

DE = Efinal - Einitial

1. First Law of Thermodynamics (Conservation of energy)

DE = q + w

where, q = heat absorbed by system

w = work done on system

Surroundings

SYSTEM

Sign Conventions:

Q(+) -->

System absorbs

heat

W (+) --> work

done on system

Heat in

Work out

W < 0

Q > 0

Heat out

Work in

Q < 0

W > 0

E is a “state function” -- independent of pathway,

but, q and w are not! (they do depend on pathway)

Two factors affect any change or reaction

(a) Enthalpy (H) (Recall that DH = DE + PDV)

exothermic processes (negative DH) tend to be spontaneous, but not always

(b) Entropy (S) – degree of disorder or randomness

Change in entropy: DS = Sfinal - Sinitial

For a reaction: DS = Sproducts - Sreactants

Positive DS means:

an increase in disorder as reaction proceeds

products more disordered (random) than reactants

For a reaction, positive DS favors spontaneity

(usually small)

Entropy (S) increases with the number of energetically equivalent ways to arrange the components of a system to achieve a particular state.

Trends:

- In general: Sgas >> Sliquid > Ssolid
- larger mass => more entropy
- more complex structure => more entropy
- more atoms in molecule => more entropy
Second Law of Thermodynamics

“…for any spontaneous process, the overall entropy of the universe increases…”

A spontaneous process can have a negative DS for the system only if the surroundings have a larger positive DS.

DSuniv = DSsys + DSsurr

where DSsurr =

–DHsys

T

Third Law of Thermodynamics

“…the entropy of a pure crystalline substance equals zero at absolute zero…”

S = 0 at T = 0 K

(a) Standard Entropy (at 25 ºC) = Sº [Table 17.2, p789]

(note the units!)

Entropy change for a reaction: (e.g. J/mole K)

DSº = S Sº(products) - S Sº(reactants)

(b) Gibbs Free Energy = G

Defined as: G = H - TS

- A combination of enthalpy and entropy effects
- Related to maximum useful work that system can do
For a process (e.g. a reaction),

DG = DH - TDS

For any spontaneous change, DG is negative!!!

(i.e. the free energy must decrease)

DGº (at 1 atm) generally used to decide if a reaction is spontaneous (Yes, if negative)

Three ways to obtain DGº for a reaction

(a) from DHº and DSº

requires DHºf and Sº data for all reactants and products

DGº = DHº - TDSº

DHº = SDHºf(products) - SDHºf(reactants)

DSº = SSº(products) - SSº(reactants)

Three ways to obtain DGº for a reaction….

(b) from standard “Free Energies of Formation” DGºf

DGº = SDGºf(products) - SDGºf(reactants)

where DGºf is the free energy change for the formation of one mole of the compound from its elements, e.g.

DGºf for Al2(SO3)3(s) equals DGº for the following reaction

2 Al(s) + 3 S(s) + 9/2 O2(g) --> Al2(SO3)3(s)

If reaction is reversed, change sign of DG°.

If reaction is multiplied or divided by a factor, apply same factor to DG°.

DG° for overall reaction = sum of DG° values for individual reactions.

Problem

Given the following thermochemical reactions:

(eq 1) C2H4(g) + 3 O2(g) --> 2 CO2(g) + 2 H2O(l)DG° = -1098.0 kJ/mol

(eq 2) C2H5OH(l) + 3 O2(g)--> 2 CO2(g)+ 3 H2O(l)DG° = -975.2kJ/mol

Calculate DG° for the following reaction:

C2H4(g) + H2O(l) --> C2H5OH(l)

Reverse 2nd reaction to put C2H5OH on product side then rewrite 1st equation and add them together.

(eq 2) 2 CO2(g) + 3 H2O(l) --> C2H5OH(l) + 3 O2(g)

DH° = + 975.2 kJ/mol (note the sign change!!!)

(eq 1) C2H4(g) + 3 O2(g) --> 2 CO2(g) + 2 H2O(l)

DH° = -1098.0 kJ/mol

Net: C2H4(g) + H2O(l) --> C2H5OH(l)

{note: 3 O2, 2 CO2, and 2 H2O cancel out}

DH° = DH°1 + DH°2 = 975.2 + (-1098.0) = -122.8 kJ/mol

(a) Effect of Temperature on DG

DG depends on DH and DS: DG = DH – TDS

but DH and DS are relatively independent of temperature, so DG at some temperature T can be estimated.

DGºT≈DHº298 - TDSº298

(b) for a system at equilibrium:

Gproducts = Greactants and DG = 0

since DG = DH - TDS = 0

DH = TDS or T = DH/DS

Example Problem

Given the following, determine the normal boiling point of mercury (Hg).

DHvaporization of Hg = 60.7 kJ/mole

entropies: liquid Hg: Sº = 76.1 J/mole K

gaseous Hg: Sº = 175.0 J/mole K

Equilibrium: Hgliq HggasDG = 0

so, DH - TDS = 0 or T = DH/DS

T = [60.7 x 103 J/mole]/[(175.0 - 76.1) J/mol K]

= 614 K = 341 ºC

(c) Relationship between DGº and Equilibrium Constant (K)

For any chemical system:

DG = DGº + (RT) ln Q

If DG is not zero, then the system is not at equilibrium. It will spontaneously shift toward the equilibrium state.

At Equilibrium: DG = 0 and Q = K

∴DGº = -RT ln K

for gaseous reactions: K = Kp

for solution reactions: K = Kc

{units of DG must match those of R value}

DGº = – RT ln K

K values can be determined from thermodynamic data !

when K > 1 DGº is negative

∴ spontaneous reactions have

large K and negative DGº values

1. Consider the gas-phase reaction: N2O5(g) + H2O(g) --> 2 HNO3(g) and the following thermodynamic data.

(a) Decide whether or not the above reaction is spontaneous at 25 ºC by calculating the value of the appropriate thermodynamic quantity.

(b) Calculate the temperature (in ºC) at which the above reaction should have an equilibrium constant (Kp) equal to 1.00.

For a certain reaction, DH° = -95.2 kJ and DS° = -157 J/K. Determine DSsurr and DSuniv (in J/K) for this reaction at 850 K. Based on these results, is the reaction spontaneous at this temperature?