EnergeticsUnit 7

CDO Chemistry 2015

1st Law of Thermodynamics1st Law of Thermodynamics1st Law – energy cannot be created or destroyed it can just change formsEnergy can be changed from Potential to Kinetic or vice versa

System Vs. SurroundingsSystem Vs. Surroundings

Two Parts to the UniverseSystem – is the part of the universe being studied

Surroundings – the rest of the universe that interacts with the system

Directionality of HeatHeat – (q)

Flow from the surrounding to the system is positive; q > 0

Flow from the system into the surroundings is negative; q < 0

Endothermic Processes Endothermic

process (endo = in) is one that absorbs heat from the surrounding (it feels cold).

Exothermic Processes Exothermic

process (exo = out) is one that transfers heat to the surrounding (it feels hot) All combustion

and neutralization reactions

Heat and TemperatureTemperature – is the measure of

the average kinetic energy of the particles

Heat energy does not necessarily change the temperature of 2 objects in the same way.

Increase in temperature depends on Mass of the objectThe heat added Nature of the substance

Heat Capacity and Specific HeatThe amount of energy required to raise

the temperature of a quantity of a substance by 1 K (1C) is its heat capacity.

We define specific heat capacity (or simply specific heat - C) as the amount of energy required to raise the temperature of 1 g of a substance by 1 K.

Specific Heat of water = 4.184 J/g oC

Finding Heat (q)

q = m C DTq = quantity of heat in Joulesm = mass in gramsC = specific heat capacity = J/(g oC)DT = change in Temp

Example 1How much heat is released when 10.0 g of

Cu with a specific heat capacity of 0.385 J/g oC is cooled from 85.0 oC to 25.0 oC?

Example 2If 500. J of heat is added to 100.0 g samples of water (specific heat = 4.184 J/goC) and Silver (specific heat = 0.237 J/goC) which substance will have the biggest change in Temperature?

CalorimetryWe measure the transfer of heat (at a

constant pressure) by a technique called calorimetry.

In calorimetry ...the heat released by the system is equal to the heat absorbed by its surroundings.

the heat absorbed by the system is equal to the heat released by its surroundings.

The total heat of the system and the surroundings remains constant.

CalorimetryWe use an insulated device called a

calorimeter to measure this heat transfer.

A typical device is a “coffee cup calorimeter.”

The calorimeter is often filled with water and the change in temperature of the water is measured

Calorimetry-q substance = +q water

So…….-mCΔT = +mCΔT

ExampleIf a 25.5 g sample of a metal at 95.0oC is placed into a 100.0 g of water at 25.0oC and the temperature of the water is raised to 32.6oC, what is the specific heat of the metal?

ExampleAn insulated cup contains 75.0g of

water at 24.00oC. A 26.00g sample of metal at 82.25oC is added. The final temperature of the water and metal is 28.34oC. What is the specific heat of the metal?

Enthalpy When a reaction is carried out a

constant pressure, the heat (q) that is transferred in the reaction is given a special name “Enthalpy”, ΔH

Enthalpy is measured in kilojoules per mole (KJ/mol)

∆HEndothermic Reactions - Heat is transferred

from the surrounding to the system, so the change in enthalpy is positive (ΔH > 0)

Exothermic Reactions – Heat transferred from the system to the surrounding, so the change in enthalpy is negative (ΔH < 0)

Methods of Finding DHMeasure it using a coffee cup calorimeterCalculate it using Hess’s LawCalculate it using average bond energies

Coffee Cup CalorimetryThe ΔH for the reaction

is equal in magnitude but opposite in sign to the q for the calorimeter

q is heat lost or gainedn is moles of substance

1000n

q- H rxn

Important AssumptionsAssume the volume of the water is equal to

its mass in grams.If water is present use its mass in the q

=mcΔTAlways use the specific heat of water 4.184

J/goC as the c valueIf two solutions are being reacted add

volumes together and assume it is equal to mass in grams.

Use the substance identified in the problem to determine n.

ExampleCalculate the enthalpy change of the

combustion of ethanol (C2H5OH) from the following data. Assume all of the heat from the reaction is absorbed by the water..

Mass of water in calorimeter

200.00 g

Temperature increase in water

13.00 oC

Mass of ethanol burned 0.45 g

Practice1.56 grams of methanol (CH3OH) is burned in a

calorimeter. This causes the 150.Og of water to increase its temperature from 20.0oC to 38.6oC. What is enthalpy of combustion for methanol?

Practice100.0 mL of water was measured out and

poured into a polystyrene cup with an initial temp of 18.3 oC. 5.20 g of NH4Cl was added to the cup and it dissolved. The minimum temp. was recorded to be 15.1 oC. Calculate the enthalpy of solution for NH4Cl per mole.

Practice200.0 mL of water was measured out and

poured into a polystyrene cup with an initial temp of 20.0 oC. 8.25 g of CaCl2 was added to the cup and it dissolved. The maximum temp. was recorded to be 35.0 oC. Calculate the enthalpy of solution for CaCl2.

ExampleWhen 1.00 L of 1.00M Ba(NO3)2 solution at 25.0 °C

is mixed with 1.00 L of 1.00 M Na2SO4 solution at 25 °C in a calorimeter, the white solid BaSO4 forms and the temperature of the mixture increases to 28.1 °C . What is the enthalpy of reaction?

PracticeWhen 1.00 L of 1.50 M HCl solution at 30.0 °C is

mixed with 1.00 L of 1.50 M NaOH solution at 30.0 °C in a calorimeter, and the temperature of the mixture increases to 41.5 °C . What is the enthalpy of neutralization?

Hess’s LawIf a chemical equation is the sum of

multiples of other equations, the DH of this equation equals a similar sum of multiples of DH's for the other equations.

Enthalpy RelationshipsEnthalpy RelationshipsGiven: H2(g) + 1/2 O2(g) H2O(g)

∆H˚ = -242 kJ

If multiplied:

2 H2(g) + O2(g) 2 H2O(g)

∆H˚ = -484 kJ

If Reversed the sign on the enthalpy value switches:

H2O(g) H2(g) + 1/2 O2(g)

∆H˚ = +242 kJ

If a different Phase the enthalpy value changes:

H2(g) + 1/2 O2(g) H2O(liquid)

∆H˚ = -286 kJ

Example 8CO2(g) CO(g) + 1/2O2(g) DH = +283.0 KJ

C(s) + O2(g) CO2(g) DH = -393.5 KJ

_________________________________________

C(s) + 1/2O2(g) CO(g) DH = ?

Example 9SO2(g) S(s) + O2 (g) DH = +296.8

KJ

2SO3(g) 2S(s) + 3O2 (g) DH = +791.4 KJ________________________________2SO2 (g) + O2 (g) 2SO3(g) DH = ?

Average Bond EnthalpiesThis table lists the

average bond enthalpies for many different types of bonds.

Average bond enthalpies are positive, because bond breaking is an endothermic process.

© 2009, Prentice-Hall, Inc.

Enthalpies of ReactionDHrxn = (bond enthalpies of bonds broken) -

(bond enthalpies of bonds formed)

Prior to using this formula the Lewis Structure for reactant and product must be drawn

ExampleEstimate the enthalpy change for the

chlorination of ethylene:CH2CH2(g) + Cl2(g) CH2ClCH2Cl

Example2H3COH + 3O2 2CO2 + 4H2O