Thermodynamics. Heat and Temperature Thermochemistry is the study of the transfers of energy as heat that accompany chemical reactions and physical changes

ThermodynamicsThermodynamics

Heat and TemperatureHeat and Temperature

ThermochemistryThermochemistry is the study of the is the study of the transfers of energy as heat that transfers of energy as heat that accompany chemical reactions and accompany chemical reactions and physical changesphysical changes

CalorimetersCalorimeters measure the energy measure the energy that is absorbed or releasedthat is absorbed or released


TemperatureTemperature is a measure of the is a measure of the average kinetic energy of the average kinetic energy of the particles in a sample of matterparticles in a sample of matter

The The joule (J)joule (J) is the SI unit of heat and is the SI unit of heat and all other forms of energyall other forms of energy


Heat Heat is the energy transferred is the energy transferred between substances because of their between substances because of their temperature differencestemperature differences

Heat always moves from matter at a Heat always moves from matter at a higher temperature to matter at a higher temperature to matter at a lower temperaturelower temperature

Specific HeatSpecific Heat

Specific HeatSpecific Heat is the energy required is the energy required to raise the temperature of one gram to raise the temperature of one gram of substance by one degree Celsiusof substance by one degree Celsius

The energy (heat) gained or lost with The energy (heat) gained or lost with a change in temperature can be a change in temperature can be calculated using calculated using q = C x m x q = C x m x ΔΔTT– q is heat (energy), C is specific heat, m q is heat (energy), C is specific heat, m

is mass, and is mass, and ΔΔT is temperature changeT is temperature change

EnthalpyEnthalpy

The energy absorbed in a reaction is The energy absorbed in a reaction is represented by represented by ΔΔH. H is the symbol H. H is the symbol for for enthalpyenthalpy

An An enthalpy changeenthalpy change is the amount of is the amount of energy absorbed by a system as heat energy absorbed by a system as heat during a process at constant during a process at constant pressurepressure

EnthalpyEnthalpy

ΔΔH = HH = Hproductsproducts – H – Hreactantsreactants

The The enthalpy of reactionenthalpy of reaction is the is the quantity transferred as heat during a quantity transferred as heat during a chemical reactionchemical reaction

Its units are joules/moleIts units are joules/mole

EnthalpyEnthalpy

A A thermochemical equationthermochemical equation includes includes the energy absorbed or released the energy absorbed or released during the given reactionduring the given reaction– 2H2H22 + O + O22 2H 2H22O + 483.6 kJO + 483.6 kJ

If If ΔΔH is negative (products side), the H is negative (products side), the reaction is exothermic. If it is positive reaction is exothermic. If it is positive (reactants side), it is endothermic(reactants side), it is endothermic

EnthalpyEnthalpy

The enthalpy change for reaction The enthalpy change for reaction relates to the stoichiometric relates to the stoichiometric proportions for that reactionproportions for that reaction

Therefore, stoichiometry can be used Therefore, stoichiometry can be used to calculate enthalpy changes for a to calculate enthalpy changes for a given sample amount for a particular given sample amount for a particular reactionreaction

EnthalpyEnthalpy

Example:Example:– How much heat will be released if 1.0 g of How much heat will be released if 1.0 g of

HH22OO22 decomposes according to the decomposes according to the following reaction (2Hfollowing reaction (2H22OO22 2H 2H22O + OO + O2 2 ΔΔH H = -190kJ= -190kJ))

– Calculate moles of sample (use formula wt)Calculate moles of sample (use formula wt) 1.0 g H1.0 g H22OO22 x (1 mol/34.0 g H x (1 mol/34.0 g H22OO22) = 0.029 mol) = 0.029 mol

– Multiply by enthalpy change Multiply by enthalpy change 0.029 mol H0.029 mol H22OO22 x (-190 kJ/2 mol H x (-190 kJ/2 mol H22OO22) = -2.8 kJ) = -2.8 kJ

[the 2 mol is the coefficient in the balanced eqn][the 2 mol is the coefficient in the balanced eqn]

Hess’ LawHess’ Law

Hess’ LawHess’ Law – The overall enthalpy – The overall enthalpy change for a reaction is equal to the change for a reaction is equal to the sum of enthalpy changes for the sum of enthalpy changes for the individual steps of the processindividual steps of the process


Smog is caused by NOSmog is caused by NO22, a gas formed , a gas formed from Nfrom N22 and O and O2 2 in a two-step reaction. in a two-step reaction. The net reaction is as follows: The net reaction is as follows: NN22 + O + O22 2 NO 2 NO ΔΔH = 181 kJH = 181 kJ

+ 2NO + O+ 2NO + O22 2NO 2NO22 ΔΔH = -113 kJH = -113 kJ

NN22 + 2O + 2O22 + 2NO + 2NO 2NO + 2 NO 2NO + 2 NO22

ΔΔHHnetnet = = ΔΔHH11 + + ΔΔHH22


The net equation is:The net equation is:

NN22 + 2O + 2O22 2NO 2NO22

According to Hess’ Law, the enthalpy According to Hess’ Law, the enthalpy change for the net reaction will be change for the net reaction will be sum of the enthalpy changes for the sum of the enthalpy changes for the individual steps [181 kJ + (-113 kJ) = individual steps [181 kJ + (-113 kJ) = 68 kJ]68 kJ]


Reactions may also be manipulated Reactions may also be manipulated as algebra problems are using the as algebra problems are using the following rules:following rules:1.If the coefficients of an equation are 1.If the coefficients of an equation are

multiplied by a factor, the enthalpy multiplied by a factor, the enthalpy change is multiplied by that factor. This change is multiplied by that factor. This is because the heat absorbed/released is because the heat absorbed/released by a reaction depends on the quantities by a reaction depends on the quantities of reactants & productsof reactants & products


2. If an equation is reversed, the sign 2. If an equation is reversed, the sign of of ΔΔH changes also. This is because H changes also. This is because if a reaction releases heat in one if a reaction releases heat in one direction, in the reverse reaction it direction, in the reverse reaction it will absorb heat (which changes the will absorb heat (which changes the sign of the enthalpy change)sign of the enthalpy change)

CalorimetryCalorimetry

In a reaction carried out in a In a reaction carried out in a calorimeter, the heat of the reaction calorimeter, the heat of the reaction (q(qrxnrxn) is transferred to the surroundings ) is transferred to the surroundings (q(qsurrsurr). Therefore, ). Therefore, qqrxnrxn = -q = -qsurrsurr

Remember that you can calculate qRemember that you can calculate qsurrsurr using using q = m x C x q = m x C x ΔΔT (T (the temperature the temperature change is the final temp minus the change is the final temp minus the initial temp)initial temp)


Once qOnce qsurrsurr is determined, q is determined, qrxnrxn is is determined by changing the sign of determined by changing the sign of qqsurrsurr

ΔΔH is equal to the qH is equal to the qrxnrxn divided by the divided by the moles of the sample used in the moles of the sample used in the reaction multiplied by the reaction multiplied by the stoichiometric proportion of moles stoichiometric proportion of moles needed for the reactionneeded for the reaction


Example: Example:

When a 4.25 g sample of solid When a 4.25 g sample of solid NHNH44NONO33 dissolves in 60.0 g H dissolves in 60.0 g H22O in a O in a calorimeter, the temperature drops calorimeter, the temperature drops from 21.0 C to 16.9 C. Calculate from 21.0 C to 16.9 C. Calculate ΔΔH H for the solution process.for the solution process.

NHNH44NONO33 NH NH44++ + NO + NO33

--


Step 1: Calculate qStep 1: Calculate qsurrsurr

qqsurrsurr = m x C x = m x C x ΔΔTT

qqsurrsurr = (60.0 g)(4.184 J/g C)(16.9 C – 21.0 = (60.0 g)(4.184 J/g C)(16.9 C – 21.0 C)C)

qqsurrsurr = -1.03 x 10 = -1.03 x 1033 kJ kJ

Step 2: Determine qStep 2: Determine qrxnrxn

qqrxnrxn = -q = -qsurrsurr = -(-1.03 x 10 = -(-1.03 x 1033 kJ) = 1.03 x 10 kJ) = 1.03 x 1033 kJkJ


Step 3: Calculate Step 3: Calculate ΔΔHHFirst calculate moles of your sampleFirst calculate moles of your sample

4.25 g NH4.25 g NH44NONO33 = 0.0531 mol NH = 0.0531 mol NH44NONO33

ΔΔH = (qH = (qrxnrxn/mol of sample) x mol from /mol of sample) x mol from equationequation

ΔΔH = (1.03 x 10H = (1.03 x 1033 kJ/0.0531 mol) x (1 mol) kJ/0.0531 mol) x (1 mol)

ΔΔH = 19.4 kJH = 19.4 kJ

EntropyEntropy

Most reactions are exothermic, and Most reactions are exothermic, and tend to proceed spontaneously, tend to proceed spontaneously, making products that are more making products that are more stable than the reactantsstable than the reactants

Some endothermic reactions are Some endothermic reactions are spontaneous, which means spontaneous, which means something other than something other than ΔΔH determines H determines whether a reaction will occur.whether a reaction will occur.

EntropyEntropy

The 2The 2ndnd Law of Thermodynamics Law of Thermodynamics states that the entropy (disorder) of states that the entropy (disorder) of a system will increase over timea system will increase over time

Entropy (S)Entropy (S) is a measure of the is a measure of the degree of randomness of the degree of randomness of the particles in a systemparticles in a system

EntropyEntropy

The entropy of a pure crystalline The entropy of a pure crystalline solid is zero at absolute zero (3solid is zero at absolute zero (3rdrd law law of thermodynamics)of thermodynamics)

Entropy change (Entropy change (ΔΔS)S) is the difference is the difference between the entropy of the products between the entropy of the products and the entropy of the reactants. and the entropy of the reactants. The more random the system, the The more random the system, the more positive this value ismore positive this value is

Gibbs Free EnergyGibbs Free Energy

Processes in nature are driven in two Processes in nature are driven in two directions: toward least enthalpy and directions: toward least enthalpy and greatest entropygreatest entropy

(Gibbs) Free Energy (G)(Gibbs) Free Energy (G) is a relation of is a relation of the enthalpy and entropy factors at a the enthalpy and entropy factors at a given temperature and pressure. given temperature and pressure. Natural processes tend in the Natural processes tend in the direction that lowers the free energy direction that lowers the free energy of the systemof the system


Only the change in free energy can be Only the change in free energy can be measured using the following equation:measured using the following equation:ΔΔG = G = ΔΔH – TH – TΔΔSS

(change in free energy = change in enthalpy (change in free energy = change in enthalpy minus absolute temperature times change minus absolute temperature times change in entropy)in entropy)

If If ΔΔG is less than zero, the reaction is G is less than zero, the reaction is spontaneousspontaneous


ΔΔH and H and ΔΔS can be positive or negative, S can be positive or negative, which leads to the following possibilities:which leads to the following possibilities:

ΔΔHH ΔΔSS ΔΔGG Spontaneous?Spontaneous?

- Value - Value (exothermic)(exothermic)

+ value + value (more (more random)random)

Always Always negativenegative

YesYes

- Value - Value (exothermic)(exothermic)

- Value (less - Value (less random)random)

Negative at Negative at low low temperaturestemperatures

At low tempsAt low temps

+ value + value (endothermic)(endothermic)

+ value + value (more (more random)random)

Negative at Negative at high tempshigh temps

At high tempsAt high temps

+ value + value (endothermic)(endothermic)

- Value (less - Value (less random)random)

Never Never negativenegative

NoNo

Gibbs Free EnergyGibbs Free Energy Example calculation of free energyExample calculation of free energy

For the reaction NHFor the reaction NH44Cl (s) Cl (s) NH NH33 (g) + HCl (g) at 298.15 K, (g) + HCl (g) at 298.15 K, ΔΔH = 176 kJ/mol and H = 176 kJ/mol and ΔΔS = 0.285 kJ/(mol ∙ K). Calculate S = 0.285 kJ/(mol ∙ K). Calculate ΔΔG, G, and tell whether the reaction is spontaneous in the forward and tell whether the reaction is spontaneous in the forward reaction at this temperature.reaction at this temperature.

Use Use ΔΔG = G = ΔΔH – TH – TΔΔSS

ΔΔG = (176 kJ/mol) – (298.15 K)[0.285 kJ/(mol ∙ K)]G = (176 kJ/mol) – (298.15 K)[0.285 kJ/(mol ∙ K)]

ΔΔG = (176 kJ/mol) – (84.9 kJ/mol)G = (176 kJ/mol) – (84.9 kJ/mol)

ΔΔG = 91 kJ/molG = 91 kJ/mol

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Thermodynamics. Heat and Temperature Thermochemistry is the study of the transfers of energy as heat that accompany chemical reactions and physical changes