Chapter 19

Classify each of the following reactions as one of the four possible types: 1. spontaneous at all temperatures; 2. nonspontaneous at all temperatures; 3. spontaneous at low T; nonspontaneous at high T; 4. spontaneous at high T; nonspontaneous at low T. (a) N2(g)+3F2(g)→2NF3(g) ΔH∘=−249kJ;ΔS∘=−278J/K (b)N2(g)+3Cl2(g)→2NCl3(g) ΔH∘=460kJ;ΔS∘=−275J/K (c) N2F4(g)→2NF2(g) ΔH∘=85kJ;ΔS∘=198J/K

(a) 3. spontaneous at low T; nonspontaneous at high T; (b) 2. nonspontaneous at all temperatures; (c) 4. spontaneous at high T; nonspontaneous at low T.

Calculate the standard free energy (ΔGrxn0) and determine if the reaction is spontaneous for:HCl(g)+NH3(g)→NH4Cl(s)Where: ΔGf0 [NH4Cl(s)]= -203 kJ/mol ΔGf0 [NH3(g)]= -17 kJ/mol ΔGf0 [HCl(g)]= -95 kJ/mol +91 kJ/mol spontaneous reaction -91 kJ/mol, nonspontaneous reaction -91 kJ/mol, spontaneous reaction -9.1 kJ/mol nonspontaneous reaction

-91 kJ/mol, spontaneous reaction

What is the entropy of a pure crystalline substance at absolute zero

0

Determine the equilibrium constant, K, at 25°C for a reaction in which ΔGo = −20.5 kJ/mol. 1.88 × 10^8 3.92 × 10^3 6.82 × 10^4 21.01

3.92 × 10^3 ΔGo and K are related by the equation ΔGo = -RT ln K

Which of the following statements correctly describes a spontaneous process? All spontaneous processes are exothermic. The reverse process of a spontaneous process is always spontaneous. All spontaneous processes proceed without outside intervention. All spontaneous processes are fast.

All spontaneous processes proceed without outside intervention. An example of a spontaneous process is when an egg is dropped above a hard surface and it breaks upon impact.

___________ is amount of heat at constant pressure

Enthalpy

________________is a measure of randomness or disorder in a system due to energy dispersion in the system

Entropy

Spontaneous processes are reversible. T or F

False Irreversible

For each of the following pairs, predict which substance has the higher entropy per mole at a given temperature. He(l) or He(g) He(g) at 5 atm pressure or He(g) at 1.8 atm pressure. 1 mol of Ne(g) in 16.0 L or 1 mol of Ne(g) in 1.60 L. CO2(g) or CO2(s)

He(g) He(g) at 1.8 atm pressure 1 mol of Ne(g) in 16.0 L CO2(g)

What are the criteria for spontaneity in terms of free energy? In any spontaneous process operating at constant temperature, the free energy of the system remains the same. In any spontaneous process operating at constant temperature, the free energy of the system decreases. In any spontaneous process operating at constant temperature, the free energy of the system increases.

In any spontaneous process operating at constant temperature, the free energy of the system decreases.

What are the criteria for spontaneity in terms of entropy? In any spontaneous process the entropy of the universe decreases. In any spontaneous process the entropy of the universe remains the same. In any spontaneous process the entropy of the universe increases.

In any spontaneous process the entropy of the universe increases.

At constant temperature and pressure the sign of the free energy relates to spontaneity of process. If a process is spontaneous in the forward direction, then the sign of ΔG is what? Negative ΔG>0 Positive Neither positive or negative but equal to zero

Negative The use ΔH and ΔS in the formulation of the state function ΔG (Gibb's free energy) to predict whether a given reaction occurring at constant temperature and pressure will be spontaneous .

Predict the sign of the entropy change, ΔS∘, for each of the reaction below. 2NH3(g)→N2(g)+3H2(g) CO2(s)→CO2(g) Ba2+(aq)+SO42−(aq)→BaSO4(s) C4H8(g)+6O2(g)→4CO2(g)+4H2O(g) 2H2(g)+O2(g)→2H2O(l) Ca(OH)2(s)→CaO(s)+H2O(g)

Positive Positive Negative Positive Negative As a state function, entropy change, ΔS, depends only on initial and final states. ΔS has a positive value when disorder increases and a negative value when disorder decreases. The following conditions usually result in an increase in entropy: a change of phase: solid→liquid→gas an increase in the number of gas molecules, or a solid dissolving to form a solution. Although the sign of the entropy change can be predicted as described above, the actual value of ΔS∘ must be calculated from the absolute entropy values, S∘, of the reactants and products: ΔS∘=S∘(products)−S∘(reactants)

For the decomposition of calcium carbonate, consider the following thermodynamic data (Due to variations in thermodynamic values for different sources, be sure to use the given values in calculating your answer.): ΔH∘rxn 178.5kJ/mol ΔS∘rxn 161.0J/(mol⋅K) Calculate the temperature in kelvins above which this reaction is spontaneous. Calculate the equilibrium constant for the following reaction at room temperature, 25 ∘C: CaCO3(s)→CaO(s)+CO2(g) When adjusted for any changes in ΔH and ΔS with temperature, the standard free energy change ΔG∘T at 500 K is equal to 9.8×104J/mol . Calculate the equilibrium constant at 500 K.

T =1109 K The spontaneity of a reaction can be determined from the free energy change for the reaction, ΔG∘. A reaction is spontaneous when the free energy change is less than zero. A reaction is nonspontaneous when the free energy change is greater than zero. A reaction is in equilibrium when the free energy change is equal to zero. K =1.09×10−23 This very tiny value for the equilibrium constant indicates that calcium oxide is not produced at room temperature. K =5.78×10−11 An increase in temperature from 298 to 500 K causes a significant increase in the equilibrium-constant value. Therefore, some amount of calcium oxide can be produced at 500 K even though the reaction is not spontaneous.

The third law of thermodynamics states that? The entropy of a pure, perfect crystalline substance at absolute zero is not zero. The entropy of a pure, perfect crystalline substance at absolute zero is negative. The entropy of a pure, perfect crystalline substance at absolute zero is positive. The entropy of a pure, perfect crystalline substance at absolute zero is zero.

The entropy of a pure, perfect crystalline substance at absolute zero is zero. The entropy of a system at the point of a single microstate with zero energy is examined at the temperature of absolute zero.

The second law of thermodynamics states that? The entropy of the universe increases for any spontaneous process. The entropy of the universe decreases for any spontaneous process. The entropy of the universe increases for any nonspontaneous process. The entropy of the universe remains unchanged for any spontaneous process.

The entropy of the universe increases for any spontaneous process.

At molecular level, there are three parameters that can change entropy. What are they?

Volume Temperature Motion of particles

Which of the following processes is not spontaneous? Washing the dishes Automobile rusting Alka-Seltzer tablet dissolving in water Kool-Aid dye mix mixing in water

Washing the dishes

Standard molar enthalpy for gases are generally (greater or lesser) than liquids and solids. Standard entropies (increase or decrease) with molar mass Standard entropies (increase or decrease) with number of atoms in a formula

greater increase increase

The normal boiling point of Br2(l) is 58.8 ∘C, and its molar enthalpy of vaporization is ΔHvap = 29.6 kJ/mol When Br2(l) boils at its normal boiling point, does its entropy increase or decrease? Calculate the value of ΔS when 5.00 mol of Br2(l) is vaporized at 58.8 ∘C.

increase The entropy of a system increases from the liquid phase to the gaseous phase. The gaseous phase has the most disorder, the greatest number of possible microstates, and the greatest freedom of movement of all the phases. ΔS =446J/K Use the formula that relates entropy to temperature to determine the value of ΔS when 5.00 mol of Br2(l) is vaporized at 58.8 ∘C. Note that there are 5.00 moles that are vaporized, the temperature must be converted to K, and the answer must have units of joules per Kelvin. ΔS=ΔH/T (29.6 kJ/mol Br2(l))×(5.00 mol Br2(l))×(1/(273.15+58.8)) K×(1000 J/1kJ) 446 J/K

Consider the universe =system + surrounding Since most spontaneous processes are _________________ S universe = S system + S surr. >0 _______________ processes therefore will not increase entropy S universe = S system + S surr. = 0

irreversible reversible

Consider a process in which an ideal gas changes from state 1 to state 2 in such a way that its temperature changes from 300 K to 200 K. Does the change in the internal energy, ΔE, depend on the particular pathway taken to carry out this change of state? Do the heat transferred between a system and its surroundings, q, and the work done by or on the system, w, depend on the particular pathway taken to carry out this change of state? Does the temperature change depend on whether the process is reversible or irreversible?

no yes no

Increase in entropy will be positive or negative?

positive

___________ processes are those that reverse direction when an infinitesimal change is made in some property.

reversible

Consider the following reaction: 2NO2(g) ⟶ N2O4(g) Using the following data, calculate ΔG∘ at 298 K. ΔG∘(NO2(g)) = 51.84 kJ/mol ΔG∘(N2O4(g))= 98.28 kJ/mol Calculate ΔG at 298 K if the partial pressures of NO2 and N2O4 are 0.39 atm and 1.63 atm , respectively.

ΔG∘ =-5.40kJ The standard free energy change for the reaction is the sum of product standard free energies minus the sum of reactant standard free energies, which is summarized in the calculation below. ΔG=∑nΔG∘products−∑mΔG∘reactants ΔG∘(N2O4(g))−2ΔG∘(NO2(g)) 98.28−2(51.84) ΔG =0.48kJ With the partial pressures provided, the reaction quotient (Q) can be calculated and used to determine the free-energy change for these conditions: ΔG=ΔG∘+RTlnQ. Based on that relationship, you can predict that ΔG will be more positive than the standard free energy for the reaction (i.e., less spontaneous) because the partial pressure of the product (1.63 atm ) is about four times greater than that of the reactant (0.39 atm), resulting in a value of Q that is greater than 10. The calculation of the free energy for these conditions is summarized below. ΔG=ΔG∘+RTlnQ ΔG∘+(0.008314 kJ/K)(298 K)ln(1.63/0.392)

What is the standard Gibbs free energy for this reaction? Assume the commonly used standard reference temperature of 298 K. The chemical reaction that causes magnesium to corrode in air is given by 2Mg+O2→2MgO in which at 298 K ΔH∘rxn= −1204 kJ ΔS∘rxn= −217.1 J/K What is the Gibbs free energy for this reaction at 5958 K? Assume that ΔH and ΔS do not change with temperature. At what temperature Teq do the forward and reverse corrosion reactions occur in equilibrium?

ΔG∘rxn=-1139 kJ G=H−TS It makes sense that the rusting reaction has a negative Gibbs free energy value because rusting occurs spontaneously. However, notice that the entropy change for this reaction is not favored (7 moles going to 2 moles and a negative ΔS value). The reason that this reaction is nonetheless spontaneous at room temperature is the large negative value for enthalpy. ΔGrxn=89 kJ This positive value for ΔG means that, at extremely high temperatures, the corrosion reaction will actually go in reverse, converting magnesium oxide into magnesium and oxygen. Teq=5546 K This value means that 5546 K is the temperature at which this reaction changes from product-favored (spontaneous) to reactant-favored (nonspontaneous). As in this example, when ΔH and ΔS are both negative, the reaction is spontaneous at all temperatures below Teq. For a reaction in which ΔH and ΔS are both positive, the reaction is spontaneous at all temperatures above Teq.

Calculate the standard free-energy change, ΔGo, for a reaction for which ΔHo = 15.6 kJ and ΔSo = 125 J/K. Assume 25°C. -4.65 × 103 kJ −3.72 × 104 kJ −21.7 kJ +12.5 kJ

−21.7 kJ These state functions are related by the equation: .ΔGo = ΔHo - T ΔSo.