Unit 5 Thermodynamics- AP Chem- Doc Bennett

Determine the sign of entropy solid to gas

+

Determine the sign of ΔG, whether it is spontaneous, and at what temperature conditions is it spontaneous ΔH: - ΔS: -

+ or -; depends on temperature; low temperatures

Determine the sign of ΔG, whether it is spontaneous, and at what temperature conditions is it spontaneous ΔH: + ΔS: +

+ or -; depends on temperature; high temperature

Determine the sign of ΔG, whether it is spontaneous, and at what temperature conditions is it spontaneous ΔH: + ΔS: -

+; no; never

Determine the sign of entropy liquid to solid

-

2PCl3(g)+O2(g)→2POCl3(g)2PCl3(g)+O2(g)→2POCl3(g) CHART Substance |Approximate PCl3(g)=310 O2(g)=210 POCl3(g)=330 The oxidation of PCl3(g)) is represented by the equation above, and the table provides the approximate values of the absolute molar entropies, S°, for these substances. Based on the information given, what is the approximate ΔS° for the reaction?

-170 J/(mol rxn K) (HOW: ΔS°rxn=Σ(moles of substance×S°)products−Σ(moles of substance×S°)reactants; therefore, ΔS°rxn=(2×330)−[(2×310)+210]=−170 J/(molrxn⋅K))

Determine the sign of ΔG, whether it is spontaneous, and at what temperature conditions is it spontaneous ΔH: - ΔS: +

-; yes; at all temperatures

Which has greater entropy? A) solid to gas B) Gas to solid C) 7 moles to 6 moles D) 6 moles to 3 moles

A

4Fe(s)+3O2(g)⇄2Fe2O3(s)ΔH=−1,650kJ/molrxn The oxidation of Fe(s) is represented by the chemical equation above. Which of the following correctly explains whether or not the reaction is thermodynamically favorable? A) There are more particles (including particles in the gas state) in the reactants than in the product, thus ΔSrxn<0. Because ΔH is large and negative, the reaction will be thermodynamically favorable at low temperatures. B) There are more particles (including particles in the gas state) in the reactants than in the product, thus ΔSrxn<0. Because ΔH is large and negative, the reaction will be not be thermodynamically favorable at any temperature. C)There are more particles (including particles in the gas state) in the reactants than in the product, thus ΔSrxn>0. Because ΔH is large and negative, the reaction will be thermodynamically favorable at all temperatures. D)There are more particles (including particles in the gas state) in the reactants than in the product, thus ΔSrxn>0. Because ΔH is large and negative, the reaction will be not be thermodynamically favorable at any temperature.

A (WHY: Because ΔG=ΔH−TΔS<0 is the criterion for a process to be thermodynamically favorable, for ΔG to be negative, the term −TΔS should have a much smaller magnitude than ΔH. This will be true at low temperatures.)

K (s) + 1/2 Cl2(g) → KCl(s) ΔH° = −437 kJ/molrxn The elements K and Cl react directly to form the compound KCl according to the equation above. It is observed that the reaction producing KCl from its elements goes essentially to completion. Which of the following is a true statement about the thermodynamic favorability of the reaction? A) The reaction is favorable and driven by an enthalpy change only. B) he reaction is unfavorable and driven by an entropy change only. C) The reaction is favorable and driven by both enthalpy and entropy changes. D)The reaction is unfavorable due to both enthalpy and entropy changes.

A (WHY: This option is correct. It predicts correctly that the sign of the free energy change would be negative (thermodynamically favorable since it goes to completion) and driven by an enthalpy change, since the overall reaction listed in the stimulus is exothermic. The reaction is not driven by an entropy change because the entropy of the system decreases as reactants convert to products.)

Which of the following reactions is not thermodynamically favored at low temperatures but becomes favored as the temperature increases? A Reaction: 2 CO(g) + O2(g) → 2 CO2(g) ∆H°: -566 ∆S°: -173 B Reaction: 2 H2O(g) → 2 H2(g) + O2(g) CO2(g) ∆H°: 484 ∆S°: 90.0 C Reaction: 2 N2O(g) → 2 N2(g) + O2(g) ∆H°: -164 ∆S°: 149 D Reaction: PbCl2(s) → Pb2+(aq) + 2 Cl ̄(aq) ∆H°: 23.4 ∆S°: -12.5

B (WHY: ΔG = ΔH - T ΔS. At low temperature, ΔG is dominated by ΔH, and since ΔH > 0, the reaction is not favored. At high temperatures, ΔG is dominated by -T ΔS, and since ΔS > 0, the reaction is favored.)

When water is added to a mixture of Na2O2(s) and S (s) , a redox reaction occurs, as represented by the equation below. 2 Na2O2(s) + S(s) + 2 H2O(l) = 4 NaOH(aq) + SO2(aq) ΔH°298 = -610 kJ/mol; ΔS°298= -7.3J/Kmol Which of the following statements about the thermodynamic favorability of the reaction at 298 K is correct? A)It is thermodynamically unfavorable. B) B It is thermodynamically favorable and is driven by ΔS° only. C) It is thermodynamically favorable and is driven by ΔH° only. D) It is thermodynamically favorable and is driven by both ΔH° and ΔS°.

C

Which of the following equations represents a reaction for which the standard entropy change is positive (ΔS° > 0) ? A) 3 O2(g) → 2 O3(g) B) 2 H2(g) + O2(g) → 2 H2O(l) C) CaCO3(s) → CaO(s) + CO2(g) D) I2(g) + 2 K(s) → 2 KI(s)

C

H(g)+Cl(g)→HCl(g) The formation of HCl(g) from its atoms is represented by the equation above. Which of the following best explains why the reaction is thermodynamically favored? A) ΔG>0⁢ because energy is released as the bond between the H and Cl atoms forms, and entropy increases because the number of gaseous product particles is less than the number of gaseous reactant particles. B) ΔG>0because energy is absorbed as the bond between the H and Cl atoms forms, and entropy decreases because the number of gaseous product particles is less than the number of gaseous reactant particles. C) ΔG<0⁢ because although energy is absorbed as the bond between the H and Cl atoms forms, entropy increases because the number of gaseous product particles is less than the number of gaseous reactant particles. D)ΔG<0 because although entropy decreases because the number of gaseous product particles is less than the number of gaseous reactant particles, energy is released as the bond between the H and Cl atoms forms.

D (WHY: Although entropy decreases, not increases, as the reaction proceeds (ΔS<0), the amount of energy that is released (ΔH<0) as the bonds between the H and Cl atoms form is large enough to make the reaction thermodynamically favored (ΔG<0).)

CH4(g)+2O2(g)⇄CO2(g)+2H2O(g) ΔH°rxn=−803kJ/molrxn ΔS°rxn=−5J/(molrxn⋅K) The chemical equation above represents the exothermic reaction of CH4(g) with O2(g). Which of the following best helps to explain why the reaction is thermodynamically favored (ΔG<0) at 2000K and 1atm? A) The total number of gaseous product molecules is less than the total number of gaseous reactant molecules, thus ΔS<0. B) The total number of gaseous product molecules is greater than the total number of gaseous reactant molecules, thus ΔS>0. C) The amount of energy released when the product bonds form is much less than the amount of energy needed to break the reactant bonds. D) The amount of energy released when the product bonds form is much greater than the amount of energy needed to break the reactant bonds.

D (WHY: ΔG for a process that is thermodynamically favored, and ΔG=ΔH−TΔS. The value of ΔS for the reaction is small because the number of moles of gas is the same on both sides of the equation. ΔH for the process is negative and large because of the formation of much stronger bonds in the products than existed in the reactants. When the relatively large negative term ΔH is added to the relatively small term −TΔS, ΔG<0 and the process is thermodynamically favored.)

first law of thermodynamics

Energy cannot be created or destroyed

A cube of ice is added to some hot water in a rigid, insulated container, which is then sealed. There is no heat exchange with the surroundings. What has happened to the total energy and the total entropy when the system reaches equilibrium?

Energy remains constant and entropy increase

Quantity that would be zero for a pure, perfect crystal at 0 K

Entropy

When K=1, 1n K is zero and ΔG°rxn is zero. What are the qualities of the reaction under standard conditions? (pg 43 in notes)

Equilibrium

For which of the following processes would ΔS have a negative value? I. 2 Fe2O3(s) → 4 Fe(s) + 3 O2(g) II. Mg2+ + 2 OH- → Mg(OH)2(s) III. H2(g) + C2H4(g) → C2H6(g)

II and III only

The reaction represented above goes essentially to completion. The reaction takes place in a rigid, insulated vessel that is initially at 600 K What can be inferred about ∆S° for the reaction at 600 K?

It must be positive, since ∆G° is negative and ∆H° is positive.

When K>1, 1n K is positive and ΔG°rxn is negative. What are the qualities of the reaction under standard conditions? (pg 43 in notes)

Spontaneous in the forward direction

When K<1, 1n K is negative and ΔG°rxn is positive. What are the qualities of the reaction under standard conditions? (pg 43 in notes)

Spontaneous in the reverse direction

What are the two ways in which energy is "lost" from a system?

When converted to heat (q) and when used to do work (w)

According to the first law of thermodynamics, the total energy of the universe can or cannot change?

cannot

what is spontaneity?

chemical potential energy of the system

What are the 2 factors that determine whether a reaction is spontaneous?

enthalpy and entropy change

For this type of reaction "lost" heat from the systems goes into the surrounds

exothermic

3rd law of thermodynamics

for a perfect crystal at absolute zero, the absolute entropy= 0 J/molK

entropy is higher for larger or smaller molar mass?

larger

entropy is higher for more or less constrained structures?

less

In every energy transition, energy is lost or gained?

lost (this is bc of heat tax)

entropy is higher for higher or lower pressures?

lower

entropy his higher for more or less moles?

more

entropy is higher for more or less atoms?

more

entropy is higher to more and less complex compounds?

more

2nd law of thermodynamics

the total entropy change of the universe must be positive for a process to be spontaneous

2 H2O2(aq) → 2 H2O(l) + O2(g) ΔH° = −196 kJ/molrxn The decomposition of H2O2(aq) is represented by the equation above. A student monitored the decomposition of a 1.0 L sample of H2O2(aq) at a constant temperature of 300. K and recorded the concentration of H2O2 as a function of time. The results are given in the table below. Time (s) | [H2O2}] 0 | 2.7 200 | 2.1 400 | 1.7 600 | 1.4 The reaction is thermodynamically favorable. The signs of ΔG°

ΔG°: negative ΔS°: positive

When solid ammonium chloride, NH4Cl(s), is added to water at 25oC, it dissolves and the temperature of the solution decreases. What are the values of ΔH and ΔS for the dissolving process?

ΔH is positive ΔS is positive

What are the ideal conditions for spontaneity?

ΔH: - ΔS: +

2 H2(g) + O2(g) → 2 H2O(g) For the reaction represented above at 25°C, what are the signs of ΔH°, ΔS°, and ΔG°?

ΔH°: - ΔS°: - ΔG°: -

2AgNO3(aq)+CaCl2(aq)→2AgCl(s)+Ca(NO3)2(aq) CHART substance | Approximate S AgNO3(aq)=220 CaCl2(aq)=60 AgCl(s)=96 Ca(NO3)2(aq)=240 The reaction between AgNO3 and CaCl2 is represented by the equation above, and the table provides the approximate S° values for the reactants and products. What is the approximate ΔS° for the reaction?

−68J/(molrxn⋅K) (How: ΔS°=Σ(moles of substance×S°)products−Σ(moles of substance×S°)reactants ΔS°=[(2×96)+240]−[(2×220)+60]=−68J/(molrxn⋅K))