Thermochemistry

  • The study of energy changes in chemical reactions

Nature of Energy

Two types of energy

1)Kinetic energy- energy of motion; atoms and molecules possess KE

KE = ½ mv2

Directly Proportional to Kelvin temperature

2)Potential energy- stored energy; found within the bonds of substances

**********potential energy can convert to kinetic energy***********

Energyis the ability to do work or produce heat

  • Work (w) -moving an object against a force
  • Heat (q)- energy transferred between object b/c of a temperature difference
  • Always moves from hot to cold
  • Heat is NOT temperature

State Function

  • A property of a substance that depends only on its present state
  • Independent of the pathway taken or how you get from point A to point B
  • Examples: internal energy, enthalpy, temperature, pressure
  • Non examples: distance traveled, work, heat

Think of the Universe being broken down into:

System Surroundings

part you are concerned with the rest

the experiment

Example: a reaction in a beaker

System: reactants(chemicals)

Surroundings: the beaker and everything beyond it

Types of Reactions

Endothermic Reactions

  • System absorbs E form the surroundings
  • Temperature of the surroundings lowers
  • products are generally less stable (weaker bonds) than reactants
  • ΔH is positive (+)
  • Energy acts as a reactant in a chemical equation
  • phase changes of melting, vaporization and sublimation are examples

Exothermic Reactions

  • System releases E into the surroundings
  • Temperature of the surroundings raises
  • products are generally more stable (stronger bonds) than reactants
  • ΔH is negative (-)
  • Energy acts as a product in a chemical equation
  • phase changes of freezing, condensation, & deposition are examples

Every Energy (E) measurement has 3 parts:

  1. A unit : Joules (J) or calories (cal))
  • SI unit is the Joule; 1 J = 1 kgm2/s2
  • Usually express in kilojoules
  • a calorie is equal to amount of heat needed to raise the temperature of 1 gram of water by 1°C; not a very common unit used anymore
  • Conversion factor between joules and calories 4.184J= l cal

*********1 nutritional Calorie (C) = 1000 cal or 1 kcal*********

  1. A magnitude
  2. A sign to tell direction

Three Laws of Thermodynamics

1. Law of Conservation of Energy (1st law of thermodynamics)

  • energy is can neither be created nor destroyed but converted from 1 form to another
  • energy lost by the system is gained by the surroundings and vice versa

∆E=q+w

∆E =internal energy- sum of all the PE and KE in a system ( Ef – Ei)

q = heat q is positive in endothermic reactions: heat added and

negative in exothermic reactions: heat removed

When dealing with gases, work is a function of volume

w = work w is positive if work is done on the system(compression)

negative if the system does work (expansion)

W = - P∆V

Work units: liter-atm ( L· atm)

1L· atm = 101.3 J

Example 1

If a gas expands from 46 L to 64 L at a constant pressure of 15 atm. How much work is done?

Example 2

What is the internal energy if 50.0 J of heat is added and 20.0 J of work are done on the system?

Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure

There are many ways to calculate this value: stoichiometry, calorimetry, Hess’s Law and bond energy

ΔHreaction = H products - Hreactants

  • ΔHrxn = (+)  endothermicΔHrxn = (-)  exothermic
  • Usually expressed as kilojoules per moleof reaction (kJ/mol)

Thermochemical Equations

  • Used to calculate the enthalpy released or absorbed in a chemical reaction
  • The stoichiometric coefficients always refer to the number of moles of a substance

Example: H2O (s) → H2O (l) ΔHrxn = 6.01 kJ/mol

  • If you reverse a reaction, the sign of ΔHreverses

Example: H2O (l) → H2O (s) ΔHrxn = -6.01 kJ/mol

  • If you multiply both sides of the equation by a factor n, then ΔH must change by the same factor n.

Example: 2H2O (s) → 2H2O (l) ΔH = 2 mol x 6.01 kJ/mol

•The physical states of all reactants and products must be specified in thermochemical equations.

Example: H2O (s) → H2O (l) ΔHrxn = 6.01 kJ/mol

H2O (l) → H2O (g) ΔHrxn = 44.0 kJ/mol

Example3

How much heat is evolved when 266 g of white phosphorus (P4) burn in air? P4(s) + 5 O2(g) P4O10(s) ΔHrxn = -3013 kJ/mol

Example4

How much heat is evolved when 4.03 g of hydrogen is reacted with an excess of oxygen?

2 H2(g) + O2(g) 2 H2O(l)ΔHrxn = -572 kJ/mol

Standard Enthalpies of Formation(Δ°Hf)

  • Is the change in enthalpy that accompanies the formation of 1 mole of a compound from its elements in their standard states
  • standard state for an element 1 atm for gases and 25°C for solids & liquids
  • There is a table of standard heats of formation Hf for reference in the back of your book
  • The standard enthalpy of formation of any element in its most stable form is zero.

Examples: ΔHf0 (O2) = 0ΔHf0 (O3) = 142 kJ/mol

ΔHf0 (Cgraphite) = 0ΔHf0 (Cdiamond) = 1.90 kJ/mol

To Calculate Standard Enthalpy

  • need to be able to write the a formation equation showing the formation of 1 mole of a compound from its elements in their standard states

Example 5

What is the equation for the formation of NO2?

Example 6

Write the equation for the formation of methanol, CH3OH.

The standard enthalpy change for any reaction can be found using this very important equation

ΣΔ°Hf products - ΣΔ°Hf reactants = °Hrxn

Example 7

Benzene (C6H6) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted?

2C6H6 (l) + 15O2 (g) → 12CO2 (g) + 6H2O (l)

ΔHf0 (CO2) = -393.5ΔHf0 (H2O) = -286. kJ/mol

ΔHf0 (C6H6) = 49.04

Example 8

Given the standard enthalpy of reaction , use the standard enthalpies of formation to calculate the standard enthalpy of formation of CuO

CuO(s) + H2(g)  Cu(s) + H2O(l) ΔH°reaction = -129.7 kJ

ΔHf0 (H2O) = -286 kJ/mol

Hess’s Law

  • Enthalpy is a state function- independent of the path
  • A reaction can be carried in a single step or in a series of steps.
  • If carried out in a series of steps, we can add equations to come up with the desired net equation and thus add ΔH for the overall reaction
  • One step:N2(g) + 2 O2(g) 2NO2(g) ΔHrxn = 68 kJ
  • Two step:N2(g) + O2(g) 2NO(g) ΔHrxn =180 kJ

2NO(g) + O2(g)  2NO2(g) ΔHrxn =-112 kJ

______

N2(g) + 2 O2(g) 2NO2(g) ΔHrxn = 68 kJ

Two Rules to Remember:

  • If the reaction is reversed, the sign of ΔH is reversed

N2(g) + 2 O2(g) 2NO2(g) ΔHrxn = 68 kJ/mol

2NO2(g)N2(g) + 2 O2(g)ΔHrxn = -68 kJ/mol

  • The magnitude of ΔH is directly proportional to the moles of reactants and products. If the coefficient of a reaction are multiplied by and integer, the ΔH is also multiplied by the integer

******when using Hess’s law, work by adding the equations to make it look like the answer, the other parts will cancel out******

Example 9 C(s) + 2 H2(g)CH4(g) ΔHrxn = ?

Eq 1 C(s) + O2(g)  CO2(g)ΔH1 = -393.5 kJ

Eq 2 H2(g) + ½ O2(g)  H2O(l)ΔH2 = -285.8 kJ

Eq 3 CH4(g) + 2O2(g)  CO2(g) + 2H2O(l)ΔH3 = -890.3 kJ

Example 10

Given O2(g) + H2(g)  2 OH(g) ΔºH = +77.9 kJ

O2(g) 2O(g) ΔºH = +495 kJ

H2(g) 2H(g)ΔºH = +435.9 kJ

Calculate theΔºHrxn for this reaction

O(g) + H(g)  OH(g)

Calorimetry

  • process that measures the transfer of heat
  • uses a calorimeter (an insulated container): usually the change in temperature of water is measured but often the heat capacity of the calorimeter is known since the calorimeter also absorbs some heat as

well as the water in the

device.

Heat Capacity (C)

  • Amount of heat energy needed to raise temperature of a substance by 1º Celsius

C = __q__

∆ T

Specific Heat Capacity (c) or (s)

  • amount of energy needed to raise 1 gram of a substance by 1º Celsius

c = _ q rearranges to q= mc∆T

m · ∆T

  • Units: J/g·°C or J/g·K or cal/g·ºC or cal/g· K
  • Each substance has its own specific heat

4.184 J/g·C for water 2.45 J/g·C for alcohol .452 J/g·C for iron

Example 11

The specific heat of graphite is .71 J/gºC. Calculate the energy needed to raise the temperature of 75 kg of graphite from 294 K to 348 K.

Molar Heat Capacity

  • amount of heat required to raise 1 mole of substance by 1°C

c = ___q rearranges to q = nc ∆T

n ·∆T

Coffee Cup Calorimetry (constant pressure)

  • used to determine ΔHrxn for reactions in solution
  • uses an insulated cup full of water as the calorimeter
  • use q= mc∆T to solve
  • Water has a specific heat of 4.184 J/g·C
  • Assume:

Heat released in a reaction = heat absorbed in water

Negative heatpositive heat

  • qrxn + qsoln = 0
  • qrxn = - qsoln if endothermic
  • qsoln = - qrxn if exothermic

Example 12

A 46.2 g sample of copper is heated to 95.4 ºC and then placed in a calorimeter containing 75.0 grams of water at 19.6 ºC. The final temperature of both the water and the copper is 21.8 C. What is the specific heat of copper?

Example 13 ΔH = q/mol of reactant

Suppose you place .500 g of magnesium chips into a coffee-cup calorimeter and then add 100.0 ml of 1.00 M HCl. The reaction occurs is

Mg(s) + 2HCl (aq) --> H2(g) + MgCl2(aq)

The temperature of the solution increases from 22.2 to 44.8º C. What is the enthalpy change per mole of magnesium. Assume specific heat capacity of the solution is 4.20 J/g·K and the density of the HCl solution is 1.00 g/ml.

Example 14 ΔH = q/mol of reactant

Assume you mix 200. ml of .400 M HCl with 200. ml of .400M NaOH in a coffee-cup calorimeter. The temperature of the solutions before mixing was 25.10 C; after mixing and allowing the reaction to occur, the temperature is 26.60C. What is the molar enthalpy of neutralization of the acid? Assume the densities of the solution are 1.00g/ml and their specific heat capacities are 4.20J/gK.

Bomb Calorimetry (constant volume)

  • Material is put into a chamber with pure oxygen. Wires start combustion; the container is put into a container of water
  • The heat capacity of the calorimeter is known and tested
  • Since volume cannot change, no work (-P ΔV) is done.

ΔE =q

  • The heat capacity of the calorimeter MUST be known

qbomb = (Ccal) (ΔT)

qbomb + qwater = - qreaction

Example 15

A bomb calorimeter has a heat capacity of 11.3kJ/ºC. When a 1.50 grams sample of methane(CH4) was burned in excess oxygen, the temperature increased by 7.3 ºC. When a 1.15 g sample of hydrogen gas was burned in excess oxygen, the temperature increased by 14.3 ºC. Calculate the Energy of combustion per grams of hydrogen and methane.

Example 16

A 1.00 g sample of table sugar (C12H22O11) is burned in a combustion calorimeter. The temperature of 1.50 x 103 grams of water in the calorimeter rises from 25.00C to 27.32 C. If the heat capacity of the bomb is 837 J/K and the heat capacity of the water is 4.184 J/gK, calculate the heat evolved per gram of sucrose and the heat evolved per mole of sucrose?

Bond Energies and Enthalpy

  • Bond energy is required to break a bond.
  • Since the breaking of a bond is an endothermic process, the bond energy is always a positive number.
  • When a bond is formed, energy equal to the bond energy is released.

Σ Bond energy of Reactants – Σ Bond energy of Products = Δ°Hrxn

Example:

Bond Bond Energy(kj/mol)

H – H436

O = O499

O – H463

Find the Δ°Hrxn for the following reaction

2 H2(g) + O2(g) 2H2O(g)

Physical Processes

Phase Changes melting vaporization

S  L  G

ENDOTHERMIC

Energy must be put into the substance to separate the attractive forces

freezing condensation

S  L  G

EXOTHERMIC

Energy must be removed from the substance to slow down molecules so the attractive forces can take effect

Making a Solution

Heat of Solution- heat released or absorbed when a certain amount of solute dissolves in solvent

ΔHsoln = ΔHsoln final –ΔHcomponents

ΔHsoln = H1 +H2 + H3

LE Hhydration

NaCl(s) Na+1(aq) + Cl1-(aq) ΔHsoln = ?

  1. Sodium and chloride ions are separated from each other. This requires energy(endothermic) LATTICE ENERGY
  2. Water molecules expand. This requires energy.(endothermic)
  3. Ions are hydrated by water molecules. New forces are formed. Usually exothermic HEAT OF HYDRATION

Measuring Heat during Phase Changes

Heat of Fusion/Solidification

Latent Heat of fusion (Hfus ) is the heat energy required to melt one gram of a solid at its melting point q = H fus x mass

For water, Hfus = 334 J/g

Latent Heat of solidification (Hsolid ) is the heat energy lost when one gram of a liquid freezes to a solid at its freezing point

q = Hsolid x mass

 For water, Hsolid = -334 J/g

Heat of Vaporization/Condensation

Latent Heat of vaporization (Hvap) is the heat to vaporize one gram of a liquid at its normal boiling point q = Hvap x mass

For water, Hvap= 2260 J/g

Latent Heat of condensation (Hcond ) is the heat energy released when one gram of a liquid forms from its vapor

 For water, Hcond = -2260 J/gq = Hcond x mass

Examples

  1. How much heat is needed to melt 500.0 g of ice at 0 C?
  1. How much heat is evolved when 1255 g of water condenses to a liquid at 100°C?