Thermodynamics Notes

This topic overlaps with thermochemistry so you will recognize some of the material

Thermodynamics- the study of energy and its transformations

Spontaneous- refers to a whether a reaction will occur (favor product formation), NOT how quickly.

A reaction can be spontaneous at certain conditions, but is so slow that it appears not to occur.

System - the part of the universe being observed.

Surroundings - the part of the universe not being observed, i.e. everything outside the system.

Heat- energy transferred between a system and its surroundings due to a temperature difference.

q = m C ΔT m = mass in grams C = specific heat in J/g/°C ΔT = °C

q = + endothermic (heat absorbed) q = - exothermic (heat released)

at constant pressure, qp = ΔH

Specific heat capacity, C - the energy required to raise the temperature of one gram of a substance

by one degree Celsius. UNITS: J/g/°C

Water liquid: 4.184 J/g/°C

Molar Heat capacity- the energy required to raise the temperature of one mole of substance by one

degree Celsius. Units are J/mol /°C

Temperature Change, ΔT = Tf - Ti

Notice that the temperature change is the same for Celsius as it is for Kelvin!!

All units for a temperature change can be expressed as °C or K.

Ex. specific heat of water liquid = 4.184 J/g/°C OR 4.184 J/g/K

Heat capacity- Cp, a measure of how much the temperature of an object is raised when it absorbs heat.

Let your UNITS guide you to heat!

Cp = ΔH / ΔT ; J/°C

qp = Cp • ΔT Use this equation to obtain heat when given heat capacity.

o Consider your units heat energy in joules = (J/°C) (°C)

Calorimetry – the science of measuring heat.

Concept: heat lost by system = - (heat gained by surroundings)

For placing a hot metal in cool water: metal loses heat and the water absorbs this heat

q (metal) = - q (water ) assuming no heat is lost to surroundings (as with a coffee cup calorimeter)

Constant pressure calorimetry is used in determining the heat of reactions for solutions. Why?

Example:

50.0 mL of 1.0 M HCl is mixed with 50.0 mL of 1.0 M NaOH in a coffee cup calorimeter. The

temperature of each solution is measured before mixing and found to be 24.0⁰C . After mixing the

highest temperature recorded is 30.5⁰C. Find the enthalpy for this reaction.

(a) Write the net ionic equation for the reaction. _______________________________

(b) This reaction is called a ___________________________________________.

(c) Write the equation you will use to solve this problem.

_________________________________________________________

(d) What assumption did you make?

_________________________________________________________

(e) Show work and box answer WITH UNITS.

(f) For reactants that are at the same initial temperature before mixing, if the resulting solution

gets warmer the reaction is _____________________________, cooler the reaction is

_______________________________________.

(g) How much energy would have been released if solutions doubled in concentrations?

You Try It:

When 1.00 L of 1.00 M Ba(NO3)2 solution at 25.0⁰C is mixed with 1.00 L of 1.00 M Na2SO4 solution at

25⁰C in a calorimeter, the white solid BaSO4 forms and the temperature of the mixture increases to

28.1⁰C. Assuming that the calorimeter absorbs only a negligible quantity of heat, that the specific heat

capacity of the solution is 4.18 J/g•⁰C, and that the density of the final solution is 1.0 g/mL, calculate

the enthalpy change per mole of BaSO4 formed.

PAUSE FOR THOUGHT:

If two reactants at the same temperature are mixed and the resulting solution gets WARMER,

this means that the reaction lost heat (EXOthermic reaction), to the surrounding’s causing the

temperature of the mixture to increase.

If two reactants at the same temperature are mixed and the resulting solution gets COLDER,

this means that the reaction gained heat (ENDOthermic reaction), from the surrounding’s causing

temperature of the mixture to decrease.

# 52 page 268

in a coffee-cup calorimeter, 1.60 g of ammonium nitrate is mixed with 75.0 g of water at an initial

temperature of 25.00 ⁰C. After dissolution of the salt, the final temperature of the calorimeter

contents is 23.34⁰C. Assuming the solution has a heat capacity of 4.18 J/g•⁰C and assuming no heat loss

to the calorimeter, calculate the enthalpy change for the dissolution of ammonium nitrate in units of

kJ/mol.

For dissolving an ionic solid in water there is no chemical reaction.

o The system is the process of dissolving (dissolution).

o The surroundings are the water-solute solution and the insulating cup.

o q (process) = - q (water-solute mixture)

o Measure the temperature change dealing with the solution and measure the total mass of

the solution.

Assumptions: 1. mass of solute compared to solvent mass is negligible

2. specific heat of the solution (solute plus solvent) is same water’s, 4.18 J/g•⁰C

# 54 page 268

The reaction of barium hydroxide with hydrochloric acid releases 118kJ per mole of barium hydroxide.

Calculate the heat when 100.0 mL of 0.500 M HCl is mixed with 300.0 mL of 0.100 M Ba(OH)2. Assuming

that the temperature of both solutions was initially 25.0⁰C and that the final mixture has a mass of

400.0 g and a specific heat capacity of 4.18 J/g•⁰C, calculate the final temperature of the mixture.

Constant Volume Calorimetry:

Work is defined as pressure-volume W = - PΔV

From this formula it is clear that at constant volume no work is done.

A bomb calorimeter is used to study systems in which the volume of the system does not change.

Weighed reactants are placed inside a rigid steel container, the “bomb”, and ignited. (Combustion

reactions) The energy change is determined by measuring the temperature change of the water

and other calorimeter parts. The reaction is the system, and the water and other calorimeter

parts are the surroundings.

q (reaction) = - q (water + bomb calorimeter)

Example:

It has been suggested that hydrogen gas obtained by the decomposition of water might be a

substitute for natural gas (principally methane). To compare the energies of combustion of these

fuels, the following experiment was carried out using a bomb calorimeter with a heat capacity of

11.3 kJ/⁰C. When a 1.50 g sample of methane gas was burned with excess oxygen in the

calorimeter, the temperature increased by 7.3 ⁰C. When a 1.15-g sample of hydrogen gas was

burned with excess oxygen, the temperature increase was 14.3 ⁰C. Calculate the energy of

combustion (per gram) for hydrogen and methane. ANS. 55kJ/g CH4 and 141 kJ/g H2

1st Law of thermodynamics- is really just the law of conservation of energy

ΔE = q + w ΔE = internal energy q = heat w = work

Work- force through a distance. For gases @constant pressure (chemistry) it is expansion/compression

work and w = - PΔV

Work = (-) work is done by the system Big Idea: Gas expansion, work is done at the expense of

Internal Energy. It’s going to cost the system some energy to do work!

Work = (+) work is done on the system Big Idea: Gas compression increases the internal energy of

the system.

*If internal energy increases then it becomes more positive.

State Function – refers to a property of the system that depends only on its present state.

State Functions are pathway independent. You can hike up the mountain

along the red path indicated in the photo, or you can take a helicopter from BC (base) to the summit. The

change in state is the same for both scenarios.

State Functions: ΔE, ΔH, ΔS Δ implies final state minus initial state.

Change in Internal Energy, 𝚫𝑬 = 𝑬𝒇 − 𝑬𝒊

Change in Enthalpy, 𝚫𝑯 = 𝑯 𝒑𝒓𝒐𝒅𝒖𝒄𝒕𝒔 − 𝑯𝒓𝒆𝒂𝒄𝒕𝒂𝒏𝒕𝒔

Change in Entropy, 𝚫𝑺 = 𝑺 𝒑𝒓𝒐𝒅𝒖𝒄𝒕𝒔 − 𝑺𝒓𝒆𝒂𝒄𝒕𝒂𝒏𝒕𝒔

Change in Gibbs Free Energy, 𝚫𝑮 = 𝑮 𝒑𝒓𝒐𝒅𝒖𝒄𝒕𝒔 − 𝑮𝒓𝒆𝒂𝒄𝒕𝒂𝒏𝒕𝒔

NOT State Functions: Heat (q) and Work (w), ( also mass, volume, etc.)

State functions are PATH INDEPENDENT and so a Hess’ Law approach can be used for entropy

and free energy just like we did with enthalpy!

ΔH⁰ rxn = ∑ ∆𝑯𝒇° (𝒑𝒓𝒐𝒅𝒖𝒄𝒕𝒔) − ∑ ∆𝑯𝒇

° (𝒓𝒆𝒂𝒄𝒕𝒂𝒏𝒕𝒔)

ΔS⁰ rxn = ∑ 𝑺𝒇° (𝒑𝒓𝒐𝒅𝒖𝒄𝒕𝒔) − ∑ 𝑺𝒇

° (𝒓𝒆𝒂𝒄𝒕𝒂𝒏𝒕𝒔)

CAUTION: Entropy units are in joules/mol•K Must convert to kJ when comparing to ΔH, ΔG

ΔG⁰ rxn = ∑ ∆𝑮𝒇° (𝒑𝒓𝒐𝒅𝒖𝒄𝒕𝒔) − ∑ ∆𝑮𝒇

° (𝒓𝒆𝒂𝒄𝒕𝒂𝒏𝒕𝒔)

Enthalpy- the measure of the energy released or absorbed by a substance when bonds are broken and

formed during a reaction.

CAUTION: ΔH is given for the reaction as written; that is, for the mole ratios in the given reaction.

ΔH = H products – H reactants. ΔH = - exothermic* ΔH = + endothermic**

*If products have stronger bonds than reactants, products have lower enthalpy than the reactants

and are more stable; energy is released and the reaction is exothermic.

**If products have weaker bonds than reactants, the products have higher enthalpy than the

reactants and are less stable; energy is absorbed by the reaction (endothermic).

ΔH rxn = Σ bonds broken – Σ bonds made

Key words used for enthalpy;ΔH -the heat of reaction, enthalpy change for a reaction, the heat of

solution, enthalpy of combustion, enthalpy of formation, heat of fusion(or crystallization), heat of

vaporization(or condensation). An older term is latent heat of fusion, latent heat of vaporization

Standard Enthalpy of formation, ΔHf⁰- the enthalpy change for a reaction in which 1 mol of a

compound is formed from its elements in their standard states. KNOW YOUR DEFINITIONS!

Standard Enthalpy of combustion, ΔHc°- the enthalpy change for combustion (hydrocarbon). Always

use the context of the problem to relate the amount of heat released to the amount of substance

burned.

How to calculate Enthalpy of a reaction:

Use table of standard enthalpies in appendices in your textbook

o ΔH°f = 0 for elements in their standard states

Use table of Bond energies

Use Hess’s Law to calculate enthalpy of a reaction Use ΔH for given reactions

For a review of these type problems look at the end of these notes.

2nd Law of Thermodynamics- the entropy of the universe is increasing; Entropy is NOT conserved

Entropy, S- measure of the randomness or disorder of the system. Units = J/mol/K

All substances have a positive(non-zero) entropy at standard conditions SEE 3rd Law of Thermo

o Standard conditions: 1atm for gases, 1M for solutions, 25°C (298K); symbolized by °

What you should know about entropy …

o increases as you go from solid liquid gas;

o particles in solution have higher entropies than solids;

NaCl(s) Na+ (aq) + Cl – (aq)

o compounds have more entropy than elements(comparing same phase of each);

o for elements in the same phase, the element with the larger molar mass has more entropy;

o two moles of a substance have higher entropy than one mole;

o the higher the temperature the greater the randomness, entropy (temperature α KEavg);

o in a reaction the side with greater moles of gases has higher entropy (Δng = + );

2B(s) + 3H2(g) → B2H6(g)

Entropy is decreasing in the above reaction because reactants have three moles of gases and products

have only one mole of gases. The number of probable positions decreases (randomness decreases).

Remember: Δng = moles gaseous products – moles of gaseous reactants

was used in converting Kc to Kp and vice versa.

For reversible phase changes: ΔS = - 𝜟𝑯

𝑻

QUANTITATIVELY: USE THE BIG MAMMA EQTN FOR 𝜟𝑺⁰

ΔS⁰ rxn = ∑ 𝑺𝒇° (𝒑𝒓𝒐𝒅𝒖𝒄𝒕𝒔) − ∑ 𝑺𝒇

° (𝒓𝒆𝒂𝒄𝒕𝒂𝒏𝒕𝒔)

CAUTION: From the 3RD Law of Thermodynamics, all entropy values (S⁰) are positive and nonzero:

𝑆𝑓0 ≠ 0

Q: Predict the sign of ΔS⁰ for each of the following changes.

(a) 2H2 (g) + O2 (g) 2H2O (l) (b) H2O(l) H2O(g) (c) HCl (g) H+ (aq) + Cl – (aq)

Q: Calculate the entropy change for the reactions above.

(a) 2H2 (g) + O2 (g) 2H2O (l) (b) H2O(l) H2O(g) (c) HCl (g) H+ (aq) + Cl – (aq)

3rd Law of Thermodynamics- S = zero for a perfect crystal at absolute zero(0 Kelvin)

For standard conditions then S⁰ > 0

Do not get S⁰ confused with ΔS⁰; ΔS⁰ CAN be ( + ) randomness increasing OR ( - ) randomness decreasing

Gibbs Free Energy, G- is a measure of the spontaneity of a process

ΔG = + reaction is non-spontaneous

ΔG = - reaction is spontaneous

ΔG = 0 reaction is at equilibrium

Change in free energy, cannot be measured directly by experimentation.

ΔG°f = 0 for elements

o Big Mamma ΔG° = Σ ΔG°f (products) - Σ ΔG°f (reactants)

o ΔG° = ΔH° - T ΔS°

o ΔG = ΔG° + RT (lnQ) ΔG is not at standard conditions

o @equilibrium ΔG =0 and Q = K, the new relation is ΔG° = - RT (lnK)

To solve for K you must use the inverse log function (ex); K = e-ΔG°/RT

K>1 product favored, ΔG° = neg K<1 react fav, ΔG° = pos K=1 ΔG° = 0 at energy minimum

Spontaneity and ΔG = ΔH - T ΔS:

ΔH ΔS T Spontaneous (ΔG = neg)

- + All temperatures ALWAYS

- - LOW YES*

HIGH NO*

+ - All temperatures NEVER

+ + LOW NO*

HIGH YES*

*Notice that when ΔH and ΔS have the SAME SIGN, there is a temperature at which the reaction

becomes spontaneous (sign of ΔG changes).

Transition Temperature between spontaneous and non-spontaneous reaction: T = ΔH/ΔS

This particular temperature is sometimes referred to as T reversible.

Energy Diagrams: Read graphs from left to right

EXOTHERMIC ENDOTHERMIC

a) activation energy, Ea

b) activation energy for the reverse rxn

c) Enthalpy of reaction, ΔH (negative)

Hess’s Law- The change in enthalpy for a rxn is the same whether the rxn takes place in one step or a

series of steps.

If a reaction can be described as a series of steps, then ΔH for the overall reaction is simply the sum of

the ΔH values for all the steps.

1. If write rxn REVERSED : CHANGE SIGN OF ΔH: +to- or –to+

2. If MULTIPLY by a constant: MULT ΔH BY THAT # TOO

3. If reactions add up to the specified reaction:

ADD all ΔH’s to get net ΔHrxn

CAUTION: Rules for manipulating ΔH are different from Keq !!

Q: Calculate the heat of rxn for nitrogen gas reacting with oxygen to produce two moles of

nitrogen dioxide using the given enthalpy changes.

N2(g) + O2(g) → 2NO(g) ΔH = 180 kJ

2NO2(g) → 2NO(g) + O2(g) ΔH = 112 kJ

Look for a substance that is in the net equation but only appears in one step.

Q: Diborane (B2H6) is a highly reactive boron hydride that was once considered as a possible rocket fuel

for the U.S. space program. Calculate ΔH for the synthesis of diborane from its elements, according to

the equation

2B(s) + 3H2(g) → B2H6(g) ΔH = ?

using the following data:

2B(s) + 3/2 O2(g) → B2O3(s) ΔH = -1273 kJ

B2H6(g) + 3O2(g) → B2O3(s) + 3H2O(g) ΔH = -2035 kJ

H2(g) + ½O2(g) → H2O(l) ΔH = -286 kJ

H2O(l) → H2O(g) ΔH = 44 kJ

Q: Using Bond Energies below, calculate the enthalpy of reaction for the given equation.

Table of Bond Energies

Single

bonds

Bond

energy

Single

bonds

Bond

energy

Single

Bonds

Bond

energy

Multiple

bonds

Bond

energy

H-H 436 C-H 413 C-Br 285 C=C 614

C-C 348 C-O 358 N-H 389 O=O 498

Cl-Cl 243 O-H 464 C-N 305 C=O 741

Br-Br 192 C-Cl 327 H-Cl 431 C≡C 836

(a) CH4 + 2Cl2 → CH2Cl2 + 2HCl

(b) CH3CH2CH3 + 5 O2 → 3 CO2 + 4 H2O

Q: Calculate the heat of reaction for combustion of 60-g ethane using the enthalpy of formation values from

table in textbook.