SB 15.4-15.5

Calculate the entropy change of the surroundings in J/mol⋅K when 30 kJ of heat is released by the system at 27°C.

+100 Reason: ΔSsurr = - ΔHsys/T = -30,000J/300K = +100 J/K

Which of the following conditions are specified by standard-state conditions?

- A solution will have a concentration of 1 M. - Gases are at 1 atm pressure. - The standard state of an element is its most stable allotropic form at standard-state temperature and pressure.

Select all the options that describe the behavior of a reversible reaction for which both ΔH and ΔS < 0.

- At low temperatures, ΔGrxn < 0. - The reaction will become less spontaneous as the temperature increases. - The formation of the reactants will become increasingly favored as the temperature rises.

Describe the free-energy change and spontaneity of the reaction CuO (s) → Cu (s) + 1212O2 (g) at 375 K if ΔH = 168 kJ/mol and ΔS = -9.63 J/K⋅mol.

- ΔG = +172 kJ - The reaction is not spontaneous at this temperature.

Which of the following statements correctly describe the standard free-energy of formation (ΔG°f) for a substance?

- ΔG°f is the free energy change for the synthesis of 1 mole of a compound from its constituent elements in their standard states. - ΔG°f for any element in its most stable allotropic form at 1 atm is equal to zero.

A reaction is found to be spontaneous only at high temperatures. Which of the following must be true?

- ΔHrxn > 0 Reason: If the reaction is spontaneous only at high temperatures, it must be endothermic (ΔHrxn > 0) and ΔSrxn > 0. - ΔSrxn > 0 Reason: A reaction that is spontaneous only at high temperatures is endothermic (ΔHo > 0) with an increase in entropy (ΔSo > 0).

When heat, q, is transferred from the system to the surroundings, the amount of energy that is dispersed is greater when the temperature is low. Which of the following options reflect this relationship (at constant pressure)?

- ΔSsurr= - ΔHsys/T - ΔSsurr= - qsys/T

Determine the change in entropy for the deposition of water vapor whose molar heat of sublimation is 54.2 kJ/mol at 1.0°C.

-198 J/K⋅mol Reason: At the phase change when both phases are present, ΔG is zero. Therefore, 0 = ΔH − TΔS and ΔS = ΔH/T= -54.2kJ/mol)/(274K)= -198 J/K⋅mol

A reaction taking place at 100°C has an entropy change of 200 J/K and an enthalpy change of -20 kJ. Which of the following expressions would be used to calculate ΔG under these conditions?

-20,000 - (373)(200) Reason: ΔG = -20 kJ × 1000J/1kJ - [200 J/K × 373 K]

Use the information below to solve for ΔG°rxn for the simple reaction A → 2B. A = -10 B = -20

-30 kJ Reason: ΔG°rxn = 2(-20) - (-10) = -30

The combustion of glucose (shown below) is exothermic. C6H12O6 (s) + 6O2 (g) → 6CO2 (g) + 6H2O (g) Does this reaction cause an increase or a decrease in entropy for the system and surroundings?

-Increase in the entropy of the system Reason: There is an overall increase in the number of gas molecules present, so ΔSsys > 0 -Increase in the entropy of the surroundings Reason: Since the reaction is exothermic, there is a transfer of heat from the system to the surroundings, resulting in an increase in the entropy of the surroundings.

Calculate ΔSuniv and identify the following process as spontaneous, nonspontaneous, or at equilibrium. Use the values provided in the table below. CO2 (g) → CO2 (aq) at 25°C CO2 (g) S°(J/K⋅mol) 213.6 ΔH°f(kJ/mol) -393.5 CO2 (aq) S°(J/K⋅mol) 121.3 ΔH°f(kJ/mol) -412.9

-The reaction is nonspontaneous at this temperature. Reason: Since ΔSuniv < 0, the process is nonspontaneous -ΔSuniv = -27.2 J/mol⋅K

Which of the following correctly reflect the relationships between ΔSsurr, ΔHsys, and T? Select all that apply.

-ΔSsurr is inversely proportional to temperature. -If ΔHsys is positive, ΔSsurr will be negative.

Solutions standard state

1 M

Gases standard state

1 atm pressure

Select all the options that correctly reflect the steps necessary to calculate the temperature at which a reaction changes from being nonspontaneous to spontaneous if ΔH = 171 kJ/mol and ΔS = 161 J/K⋅mol.

1) T = ΔH/ΔS 2) Set ΔG = 0 3) T = (171kJ/mol)/(0.161kJ/mol⋅K)

If we know that the entropy change for the surroundings is x J/K, and the system absorbed y kJ of heat during that reaction, what is the temperature at which this occurred (in K)?

1000y/x -Reason: This equation allows all units to cancel except for K.

A chemist finds that the sublimation of a compound has ΔG = -2.5 kJ at 350oC, and ΔG = +0.5 kJ at 140oC. From this information, what is a reasonable value for the temperature of sublimation?

175 degrees C Reason: 175 degrees C is between 140 degrees and 350 degrees C, and could be the temperature at which ΔG = 0.

Which equation would be used to calculate ΔG°rxn for the reaction 2A + B → 3C + D? A=10 B=-5 C=40 D=-20

3(40) - 20 - 2(10) + 5 Reason: ΔG°rxn = ΣmΔG°products - ΣnΔG°reactants = [3(40)+(-20)]-[2(10)+(-5)] = [3(40)-20]-[2(10)-5] = 3(40)-20-2(10)+5

A given reaction has ΔH = -380.1 kJ/mol and ΔS = -95.00 J/K⋅mol. Calculate the temperature (in °C) below which the reaction is spontaneous

3728 Reason: T=(−380.1kJ/mol)/(−0.095kJ/mol⋅K) = 4001 K T(°C) = 4001 -273 = 3728°C

For a particular reaction ΔH° = -5. kJ and ΔS° = -100 J/K. What is the temperature (in K) at which this reaction favors the products? Multiple choice question.

50 Reason: T = ΔHΔSΔHΔS when ΔG = 0. The energy units of ΔH and ΔS must match.

Determine the change in entropy for the melting of a solid whose molar heat of fusion is 22.3 kJ/mol at 27°C. Multiple choice question.

74.3 J/K⋅mol Reason: At the phase change when both phases are present, ΔG is zero. Therefore, 0 = ΔH − TΔS and ΔS = ΔH/T = (22.3kJ/mol)/(300K) = 74.3 J/K⋅mol

Select the statement that correctly describes the relationship between the entropy and spontaneity of a process.

A reaction/process in which ΔSsys > 0 and ΔSsurr > 0 will be spontaneous. Reason: If both ΔSsys and ΔSsys are positive, ΔSuniv must be positive, and the process will be spontaneous.

Which conditions will result in a spontaneous reaction? A) ΔSsys = -9 J/K; ΔSsurr = -2 J/K B) ΔSsys = 5 J/K; ΔSsurr = -7 J/K C) ΔSsys = -20 J/K; ΔSsurr = 25 J/K D) ΔSsys = 30 J/K; ΔSsurr = 10 J/K

C) Reason: ΔSuniv = ΔSsys + ΔSsurr must be > 0 for a spontaneous reaction. D) Reason: ΔSuniv = ΔSsys + ΔSsurr must be > 0 for a spontaneous reaction.

Which of the following is an application of the second law of thermodynamics? A) A nuclear fission reaction results in the conversion of mass to energy. B) The heat and motion induced by an engine is equal to the energy of combustion. C) The combined masses of sulfur and iron are equal to the mass of iron(II) sulfide formed through a chemical reaction. D) A gas expands because matter tends to spread out.

D) The expanded gas has a higher entropy

Since entropy is typically tabulated in units of _____ / K and standard enthalpies are typically tabulated in units of _____ / mol, it is usually necessary to perform a unit conversion when calculating Gibbs free energy.

J (joules); kJ (kilojoules)

thermodynamic quantity of ΔG° < 0

Products are favored at equilibrium.

thermodynamic quantity of ΔG° > 0

Reactants are favored at equilibrium.

thermodynamic quantity of ΔG > 0

Reaction will be nonspontaneous.

thermodynamic quantity of ΔG < 0

Reaction will occur spontaneously.

What is a correct statement of the third law of thermodynamics?

The entropy of a perfect crystalline substance is 0 at 0 K.

Elements standard state

The most stable form at 1 atm and 25 degrees celsius

True or false: A reaction with a positive ΔSuniv may occur quickly or slowly.

True

The importance of the third law of thermodynamics is that it allows us to experimentally determine _____.

absolute entropies

The third law of thermodynamics states that the entropy of a perfect crystal at 0 K is 0. This allows for the calculation of _____ entropies unlike standard _____ of formation, which are derived using an arbitrary reference.

absolute; enthalpies

The third law of thermodynamics states that a perfect _____ of a pure solid substance has _____ entropy at 0 K.

crystalline structure; zero

The second law of thermodynamics states that the total _____ of the universe will _____ for any spontaneous process.

entropy; increase

A system is in a state of _____ when ΔSuniv = 0.

equilibrium

The Gibbs _____-energy change (symbolized by Δ_____) is a measure of the spontaneity of a process, and of the useful energy available from it.

free; ΔG

If a given reversible reaction has positive values for both ΔH and ΔS, the value of ΔG will become _____ negative as temperature increases, and the formation of the _____ will be increasingly favored.

more; products

In order for the products to be favored in a reaction, the standard free-energy change (ΔG°) must have a ______ sign.

negative Reason: In order for products to be favored, the sum of the standard free energies of the products must be lower than the sum of the standard free energies of the reactants.

ΔG > 0

nonspontaneous process

ΔG = 0

process at equilibrium

Solids standard state

pure

ΔG < 0

spontaneous process

When the Gibbs free-energy change for a reaction is less than zero (negative), that reaction is ______ and the entropy change (ΔS) for the universe is ______.

spontaneous, positive

The ΔG for a reaction that occurs under standard-state conditions is called the _____ Gibbs free energy of reaction. The term "standard state" implies a temperature of _____ C or _____ K for the reaction.

standard; 25; 298

Gibbs free energy is used most prevalently to determine the spontaneity of a reaction because it is only dependent on the ______, making it much more convenient to measure experimentally.

system

In order to determine the spontaneity of a reaction or process, the entropy changes in both the _____ and the _____ must be considered.

system and surroundings

Thermodynamics tells us nothing about the _____ of a reaction.

the rate

The sign of which quantity indicates whether a reaction or process will occur spontaneously?

ΔG

Which of the following is the correct form of the change in the Gibbs equation for a process occurring at constant temperature? Multiple choice question.

ΔGsys = ΔHsys - TΔSsys Reason: Note that all of the values are written in terms of the system.

The reaction is spontaneous only at low T if

ΔH < 0 and ΔS < 0.

A reaction is always spontaneous if...

ΔH < 0 and ΔS > 0.

The reaction is always nonspontaneous if

ΔH > 0 and ΔS < 0.

The reaction is spontaneous only at high T if

ΔH > 0 and ΔS > 0.

Which of the following values must be known in order to calculate the change in Gibbs free energy using the Gibbs equation?

ΔHsys, ΔSsys, and T

The reaction 4Fe (s) + 3O2 (g) + 6H2O (l) → 4Fe(OH)3 (s) is spontaneous at temperatures below 1950°C. If ΔH = -1582 kJ, calculate the entropy change for this reaction at 1950°C. Multiple choice question.

ΔS = -712 J/K Reason: At 1950°C = 2223 K ΔS = ΔH/T = −1582/2223= -0.712 kJ/K = -712 J/K.

At a certain temperature, the change in entropy of the system is calculated to be ΔSsys. If the system is at equilibrium, what is the value of ΔSsurr under these conditions? Multiple choice question.

ΔSsurr = -ΔSsys Reason: The total entropy is equal to zero; the entropy change of the surroundings cancels the change in the system.

A particular process results in a decrease in the entropy of the system. If this process is spontaneous, what must be true about the entropy change of the surroundings?

ΔSsurr > -ΔSsys -Reason: The increase in the entropy of the surroundings must more than compensate for the decrease in the entropy of the system in order for the process to be spontaneous.