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Chapter 16 Reaction Energy & Kinetics. 16-1 Thermochemistry. Thermochemistry. The study of the transfers of energy as heat that accompany chemical reactions and physical changes This heat can be measured in a calorimeter. Units.
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Chapter 16Reaction Energy & Kinetics 16-1 Thermochemistry
Thermochemistry • The study of the transfers of energy as heat that accompany chemical reactions and physical changes • This heat can be measured in a calorimeter
Units • Temperature units may be in Kelvin (K) or degrees Celsius (°C) • Energy units are the joule (J) which is the SI unit for energy. The kJ maybe used as well.
Heat Transfer • Think of heat transfer as the difference between two objects’ temperatures. The energy will flow from areas of high temperature to areas of low temperature.
Heat Capacity & Specific Heat • Different materials can absorb energy differently. • Specific Heat is the amount of energy required to raise the temperature of one gram of a substance by one degree Celsius or Kelvin. • See page 533 Table 16-1
Heat Capacity & Specific Heat • Cp = q / (m x ∆T) or q = Cp x m x ∆T Where: Cp = specific heat q = energy lost or gained m = mass of sample ∆T = difference between initial & final temperatures
Heat of Reaction • The quantity of heat released or absorbed during a chemical rxn. • When a rxn includes this information, we call it a thermochemical equation: 2H2(g) + O2(g)→ 2H2O (g) + 483.6 kJ If you double the amount of reactants, you double the amount of energy. Sometimes fractional coefficients are used.
Thermochemistry • Enthalpy – The heat energy of reaction ∆H Enthalpy can only be measured by a change, not directly. ∆H is the amount of energy absorbed or lost by a rxn at constant pressure
∆H will be negative since energy has left the system Reaction Pathways The products have less energy than the reactants. The rxn released energy (heat) = exothermic
∆H will be positive since energy has been added to the system Reaction Pathways The products have more energy than the reactants. The rxn absorbed energy (heat) = endothermic
Tips • Coefficients = moles • Physical states must be indicated as ΔH changes for different states of matter • ΔH is directly proportional to the molar coefficients, as they change, ΔH changes • Temperature usually does not effect ΔH
Heat of Formation, ΔHf • The energy released or absorbed when one mole of a substance is formed from its elements (in terms of product) • Standard Heats of Formation, ΔHf0are values when the reactants and products are at their standard states of matter, at atmospheric pressure and room temperature (298 K)
Heat of Formation, ΔHf • Compounds with a high negative heat of formation, largely exothermic, are very stable and occur readily. • An element’s ΔHf0 = zero • Compounds with a positive, or small negative heat of formation values are unstable and will decompose into their constituents.
Heat of Combustion, ΔHc • The energy released when one mole of substance undergoes complete combustion (in terms of reactant) • All reactants are in their standard state. C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O (g) ΔHc = -2219.2
Calculating Heats of Reactions: Hess’s Law • The overall enthalpy change in a rxn is equal to the sum of enthalpy changes for the individual steps in the process. • Rules: • If a rxn is reversed, the sign on ΔH is reversed • Multiply the coefficients, if necessary, on rxns so the end rxn has the desired quantities
These values are only for the formation of the compound Heat of Formation, ΔHf NaOH + HC2H3O2 NaC2H3O2 + H2O Na + O2 + H2 NaOH ∆Hf = - 469.15 kJ O2 + H2 + C HC2H3O2∆Hf = - 484.5 kJ Na + O2 + H2+ C NaC2H3O2∆Hf = 298.15 O2 + H2 H2O ∆Hf = -285.83 kJ
NaOH + HC2H3O2 NaC2H3O2 + H2O Reverse these signs These are not being formed, rather they are being used Na + O2 + H2 NaOH ∆Hf = - 469.15 kJ O2 + H2 + C HC2H3O2∆Hf = - 484.5 kJ Na + O2 + H2+ C NaC2H3O2∆Hf = 298.15 O2 + H2 H2O ∆Hf = - 285.83 kJ Keep these the same since these compounds are being formed in the rxn
NaOH + HC2H3O2 NaC2H3O2 + H2O NaOH Na + ½O2 + ½H2 ∆Hf = 469.15 kJ HC2H3O2 O2 + 2H2 + 2C ∆Hf = 484.5 kJ Add them Na + O2 + 3/2H2+ 2C NaC2H3O2∆Hf = 298.15 kJ ½O2 + H2 H2O ∆Hf = -285.83 kJ The value is positive, so the rxn is endothermic ∆H = 965.97 kJ Hess’s Law
Practice Problem! Calculate ∆H for the following reaction: 2N2 (g) + 5O2 (g) 2N2O5 (g) Using the following data: ½O2 (g) + H2 (g) H2O (l) ∆Hf° = -285.8 kJ N2O5 (g) + H2O (l) 2HNO3 (l) ∆H° = -76.6 kJ 3/2O2(g)+½H2(g)+½N2(g)HNO3 (l) ∆Hf° = -174.1 kJ ∆ H = 28.4 kJ
Chapter 16Reaction Energy & Kinetics 16-2 Driving Force of Reactions
Enthalpy & Reaction Tendency • Most rxns tend to favor production of more stable products from less stable reactants. • That means exothermic rxns are usually favored in nature. • However, some endothermic rxns do occur spontaneously.
Entropy, S • A measure of the degree of randomness of the particles in a system; tendency toward disorder. • There is a tendency for systems in nature to favor the disordered state. Gas Solid Liquid Increase in Entropy
Entropy • A positive value for ∆S means an increase in entropy (disorder) • A negative value for ∆S means a decrease in entropy (disorder) • Rxns tend to favor high entropy and low enthalpy
Free Energy, G • A combined enthalpy-entropy function • Rxns tend to favor a lower free energy system • Free Energy cannot be measured, only the change can be measured, ∆G ∆G0 = ∆H0 - T∆S0 @ Standard States
Free Energy • If ∆G is positive, the reaction is nonspontaneous and will not readily occur. • If ∆G is negative, the reaction is spontaneous and will occur.
Relationship between Enthalpy, Entropy & Free Energy Exothermic Disordering Always Ordering Exothermic @ low temps Endothermic Disordering @ high temps Endothermic Ordering Always
Practice • Predict whether the value of ∆S will be greater than, less than or equal to zero: • 3H2 (g) + N2 (g) 2NH3 (g) • 2Mg (s) + O2 (g) 2MgO (s) • C6H12O6 (s) + 6O2 (g) 6CO2 (g)+ 6H2O (g) • KNO3 (s) K+ (aq) + NO3-(aq) ∆ S < 0 ∆ S < 0 ∆ S > 0 ∆ S > 0
Practice • Calculate the value of ∆G0 for the reaction below. Will the reaction be spontaneous at 298 K? Cu2S (s) + S(s) 2CuS(s) ∆H0 = -26.7 kJ ∆S0 = -19.7 J/(mol∙K) ∆ G0 = -20.8 kJ/mol Yes, it is spontaneous