Causes of Change

Changes in Enthalpy DuringChemical Reactions

Changes in Enthalpy Accompany Reactions

• Changes in enthalpy occur during reactions• Changes depend on many variables– Temperature is the most important

• For simplicity, data is usually obtained at 25.00ºC or 298.15K.– This is known as the standard thermodynamic

temperature

Thermodynamic Values for a Chemical Reaction

• Chemists usually present a thermodynamic value for a chemical reaction by using the chemical equation:

½ H2(g) + ½ Br2(l) HBr(g) ΔH = -36.4 kJ

• This represents the following:– 0.5 mol of H2 reacts with 0.5 mol of Br2 to produce

1 mol HBr and all have a temp of 298.15K, the enthalpy decreases by 36.4 kJ.

Signs of Enthalpies

• Looking at the last reaction:

½ H2(g) + ½ Br2(l) HBr(g) ΔH = -36.4 kJ

• Is this exothermic or endothermic? ___________

• How do you know? ________________________

_____________________________________

Chemical Calorimetry

• So what happens to the energy lost in the ½ H2(g) + ½ Br2(l) HBr(g) reaction?

– It does not disappear, so where is it?

• It is released as heat by the system• If the reaction was endothermic, energy in the

form of heat would be absorbed by the system.

Calorimetry

• The experimental measurement of an enthalpy change for a reactions is called calorimetry.

Calorimetry – the measurement of heat-related constants, such as specific heat or latent heat.

• Combustions reactions are ALWAYS exothermic

Combustion Reactions

• The enthalpy changes of combustion reactions are determined using a bomb calorimeter.

• A sample is burned in a very well insulated chamber fueled by high pressure oxygen. The released energy is absorbed by the surroundings (water) where we can calculate this change in temp (of the H2O) and deter-mine the energy gained by the water (energy released from the combustion).

Specific Heats

• Calorimeter – A device used to measure the heat absorbed or released in a chemical or physical change.

• The water, or surroundings, have known specific heats and known masses, allowing us to calculate the change in energy (q) from the observed change in temperature.

Calorimeters

Organic Matter

• Nutritionists use bomb calorimetry to measure the energy in a sample.

• Sample sizes are selected to make the oxygen the excess reactant.– Why is this? _____________________________

• These combustion reactions burn to completion and produce ____________ and ____________.

Balanced Reactions

• The coefficients in a reaction indicate the number of moles reacting to produce the number of moles of products.

• Calorimetry is quite precise, most thermodynamic quantities are known to many significant figures.

Adiabatic Calorimetry

• Instead of using a water bath to absorb the energy generated by a chemical reaction, adiabatic calorimetry using an insulating vessel.

• “Adiabatic” means “not allowing to pass through.”

• No energy can enter or escape this vessel.

Adiabatic Calorimetry

• As a result of this vessel, the reaction mixture increases in temperature if the reaction is _______________ and decreases in temperature if the reaction is _____________.

• If the system’s Cp is known, the rxn’s enthalpy can be calculated.

• Adiabatic calorimetry is used for rxns that are not ignited, such as reactions in aqueous solutions.

Hess’s Law

• Any two processes that both start with the same reactants in the same start and finish with the same products in the same state will have the same ΔH or __________________.– Regardless of the number of steps to get there.

• Hess’s Law – The law that states that the amount of heat released or absorbed in a chemical reaction does not depend on the number of steps in the reaction.

• States that the overall enthalpy change in a rxn is equal to the sum of the enthalpy changes for the individual steps in the process.

Hess’s Law

• Consider the synthesis of PCl5:

P4(s) + 10Cl2(g) 4PCl5(g) ΔH = -1596 kJ

• Phosphorous pentachloride can also be synthesized in the following two steps:

P4(s) + 6Cl2(g) 4PCl3(g) ΔH = -1224 kJ

PCl3(g) + Cl2(g) PCl5(g) ΔH = -93 kJ

• The second step must occur FOUR times to produce the same amount of PCl5 (4 x -93kJ)

Homework:Vocab sheets

due

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THE WORKED OUT

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