Chapter 15

Chapter 15Thermochemistry

What is energy? Energy is the ability to do work or produce

heat. The Law of Conservation of Energy:

◦ This law states that can not be created or destroyed only transferred.

Two types of energy:

Energy

Heat is energy transferred from a warmer object to a cooler object.

Heat is represented mathematically as q.

Heat

A breakfast of cereal, orange juice, and milk contains 230 Calories. Convert this amount of energy in to Joules.

Problem

Glucose is a simple sugar found in fruit. Burning 1.00 g of glucose releases 15.6 kJ of energy. How many Calories are released?

3.73 Calories

An fruit and oatmeal bar contains 142 Calories. Convert this energy to Joules

5.94 x 105

A chemical reaction releases 86.5 kJ of heat. How many Calories are released?

20.7 Calories

The specific heat of any substance is the amount of heat required to raise the temperature of one gram of that substance by 1 degree Celsius.

Specific Heat

The specific heat of a substance can be used to calculate the heat energy absorbed or given off when that substance changes temperature.

Using Specific Heat

Thermochemistry is the study of heat changes during chemical reactions or phase changes.

When studying thermochemistry we look at two things:

System◦ The system is the specific part of the universe

that we are studying. Surroundings

◦ The surroundings are everything else in the universe.

Thermochemistry

Enthalpy is defined as the heat content of a system at constant pressure.

The change in enthalpy for a reaction is called the enthalpy (heat) of reaction.

ΔHrxn ΔHrxn = Hproducts – Hreactnats

Enthalpy and Enthalpy Change

A thermochemical equation is a balanced equation that includes the physical states of all reactants and products and the enthalpy change.

4 Fe(s) + 3 O2(g) 2 Fe2O3(s) ΔH = -1625 kJ

NH4NO3(s) NH4+(aq) + NO3

- ΔH = 27 kJ

Thermochemical Equations

Hess’s Law states that if you can add two or more equations to produce a final equation for a reaction than the sum of the enthalpy changes of the individual reactions is the enthalpy change of the overall reaction.

Hess’s Law

Calculate ΔH for the reaction 2 H2O2(l) 2 H2O(l) + O2(g)

2 H2(g) + O2(g) 2 H2O(l) ΔH = -572 kJ

H2(g) + O2(g) H2O2(l) ΔH = -188 kJ

Use equations (a) and (b) to determine ΔH for the following reaction:

2 CO(g) + 2 NO(g) 2 CO2(g) + N2(g)

a) 2 CO(g) + O2(g) 2 CO2(g) ΔH = -566.0 kJ

b) N2(g) + O2(g) 2NO(g) ΔH = -180.6 kJ

ΔH for the following reaction is -1789 kJ. Use equation (a) to determine ΔH for reaction (b).

4 Al(s) + 3 MnO2(s) 2 Al2O3(s) + 3 Mn(s)

a) 4 Al(s) + 3 O2 2 Al2O3(s) ΔH = -3352 kJ

b) Mn(s) + O2(g) MnO2 ΔH = ?

Formation Reaction:

S(s) + 3 F2(g) SF6 ΔHof = -1220 kJ

Sometimes we need to use fractional coefficients.

Enthalpy of Formation

We can use the enthalpy of formation for components of a reaction to calculate the total enthalpy change of the reaction (ΔHrxn).

H2S(g) + 4 F2(g) 2 HF(g) + SF6(g)

a) ½ H2(g) + ½ F2(g) HF ΔHof = -273 kJ

b) S(s) + 3 F2(g) SF6 ΔHof = -1220 kJ

c) H2(g) + S(s) H2S(g) ΔHof = -21 kJ

Using Enthalpy of Formation

Determine ΔH for CH4(g) + 2 O2(g) CO2(g) + 2 H2O(l)

Using: ΔHo

f(CO2) = -394 kJ ΔHo

f(H2O) = -286 kJ ΔHo

f(CH4) = -75 kJ ΔHo

f(O2) = 0 kJ

When things rust the reaction taking place is:

4 Fe(s) + 3 O2(g) 2 Fe2O3(s) ΔH = -1625 kJ

Any physical or chemical that occurs with no outside intervention is a spontaneous process.

Reaction Spontaneity

Entropy is a measure of the number of possible ways a system can be configured.

If we have a piece of paper cut into 8 different sections there would be 56 different ways we could arrange them. (8 x 7)

If we cut the paper into 16 different pieces there would be 240 different ways we could arrange them. (16 x 15)

We have increased the papers entropy.

Entropy

The second law of thermodynamics states that a spontaneous reaction will always occur in such a way that entropy increases.

Remember that the change in enthalpy (ΔH) is defined as:

Hproducts – Hreactants

Similarly: ΔS = Sproducts – Sreactants

If ΔS is positive the entropy of the system is increasing. If ΔS is negative the entropy of the system is

decreasing.

The Second Law Of Thermodynamics

• Phase Changes:◦ When a phase change occurs from a more

ordered state to a less ordered state ΔS will be positive.

◦ Solid Liquid ΔS > 0

◦ When a phase change occurs from a less ordered state to a more ordered state ΔS will be negative.

◦ Gas Liquid ΔS < 0• Dissolving a gas into a solvent always

results in a decrease in entropy.

Predicting Entropy Changes

Assuming no change in physical state, entropy increases when the number of moles of products is greater than the number of moles of reactants.◦ 2 SO3(g) 2 SO2(g) + O2(g)

Entropy increases when a solute dissolves in a solvent.◦ NaCl(s) Na+(aq) + Cl-(aq)

Entropy increases as temperature increases.

Predict the sign of ΔS for each of the following chemical of physical processes

ClF(g) + F2(g) ClF3(g) ΔS =

NH3(g) NH3(aq) ΔS =

Entropy has the units Joules/Kelvin

Named after physicist J. Willard Gibbs, free energy is the maximum amount of energy available during a chemical reaction.

Gibbs Free Energy Equation: ΔG = ΔH – TΔS

When a reaction occurs at standard conditions (298 K and 1atm)

ΔGo = ΔHo - ΔSo

Gibbs Free Energy

ΔG = ΔH – TΔS

A reaction where ΔH is negative and ΔS is positive will always be spontaneous.

N2(g) + 3 H2(g) 2 NH3(g) @ 298 K ΔH = -91.8 kJ ΔS = -197 J/k

ΔG = ΔG = -33.1 kJ

ΔH ΔS ΔG Reaction Spontaneity

ΔG = ΔH - TΔS

For a process ΔH is 145 kJ and ΔS is 322 J/K. Calculate ΔG for this reaction at 298 K. Is it spontaneous?