Gibbs Free Energy

Calculate the heat of formation for the following reaction: 8 Al(s) + 3 Fe3O4(s) → 4 Al2O3(s) + 9 Fe(s) ΔH for Fe3O4(s) = -1120.9 kJ/mol ΔH for Al2O3(s) = -1669.8 kJ/mol (A) -3,316.5 kJ (B) -2,563.6 kJ (C) -1,657.2 kJ (D) -53.9 kJ

(A) -3316.5 kJ ΔH for a reaction is equal to the sum of the heats of formation of the product compounds minus the sum of the heats of formation of the reactant compounds: ΔH = Σ ΔHf products - Σ ΔHf reactants Omitting terms for the elements, the equation becomes: ΔH = 4 ΔHf Al2O3(s) - 3 ΔHf Fe3O4(s) ΔH = 4(-1669.8 kJ) - 3(-1120.9 kJ) ΔH = -3316.5 kJ

True or false? Hess's Law states that the energy change of a process is independent of the path that was taken to get there

True. Hess's Law states that the energy change of a process is independent of the path that was taken to get there

standard free energy change

Value of ΔG as calculated under standard conditions: under 1 atmosphere (atm) and 0 Kelvin.

changes in entropy, enthalphy, and free energy under standard conditions

standard entropy, standard enthalpy, and standard free energy changes - deltaG"naught"

bond dissociation energy

the energy required to break the bond between two covalently bonded atoms

standard enthalpy of reaction

the enthalpy of a reaction carried out under standard conditions

Hess's Law

the overall enthalpy change in a reaction is equal to the sum of enthalpy changes for the individual steps in the process

standard state

the state of a material at a defined set of conditions

equilibrium

when deltaG = 0 (where H = TS)

Hess's Law is true for variables such as enthalpy and entropy because these are _________ Variables. (A) Process (B) State (C) Fixed (D) Variable

(B) State Hess's Law is true for variables such as enthalpy and entropy because these are State Variables.

entropy

- measure of spontaneous dispersal of energy at a specific temperature - state function (independent of equilibrium pathway)

enthalpy

- the heat content of a system at constant pressure - state function (independent of pathway taken)

Suppose that a reaction has deltaH = -77 kJ and deltaS = -.48 kJ. At what temperature will it change from spontaneous to nonspontaneous? A. 47K B. 160K C. 243K D. 321K

B. 160K A change from spontaneous to nonspontaneous occurs with deltaG = 0.

A particular substance is dissolved in water and various energy changes are measured. It is known that heat is absorbed in this process and that the equilibrium constant is measured to be less than 1. What is the sign of the standard entropy change for this process? A. positive B. negative C. either positive or negative D. zero

C. either positive or negative We can connect Gibbs free energy with the equilibrium constant via the formula deltaG = -RTlnK, where K is the equilibrium constant. We know that K is < 1, so the sign of deltaG must be positive. We then know G = H - TS and we are given that H is positive (given that heat is absorbed during the process). Standard entropy change can be either positive or negative depending on the magnitude of the TS term.

Gibbs free energy, G

amount of energy available to do work - indicates spontaneity

thermodynamics vs kinetics

are separate!!! when a reaction is thermodynamically spontaneous, it has no bearing on how fast it goes. it only means that it will proceed eventually

actual free energy change

depends on the actual conditions of the reaction which depends on actual reactants and products

second law of thermodynamics

energy spontaneously disperses from being localized to becoming spread out if it is not hindered from doing so

Standard heat of combustion (ΔH°comb)

enthalpy change associated with combustion of a fuel

standard enthalpy of formation

enthalpy change if 1 mole of compound in standard state were formed directly from its elements (given on MCAT)