Chem 25 Chapter 10

What is the entropy change when 367 J of energy is reversibly transferred to a sample of water at 25°C?

1.23 J/K Explanation: The entropy change can be calculated using the formula below. ΔS=heat/T Convert the given T into Kelvin and plug the numbers into the formula. ΔS=367J/298.15K ΔS=1.23 J/K

Determine ∆S for the phase change of 1.23 moles of water from solid to liquid at 0°C. (∆H = 6.01 kJ/mol)

27.1 J/K Explanation: ΔS=heat/T ΔS=1.23 mol(6010J/molK)/273.15K ΔS=27.1 J/K

What is the value of n in the Nernst equation for the reaction Al(s) + 3 Ag⁺(aq) ⟶ Ag(s) + Al³⁺(aq).

3 Explanation: Three electrons are transferred from Al to 3 Ag⁺ to form Al³⁺ and 3 Ag, so the value of n in the Nernst equation is 3.

What is the entropy when 1.06 moles of CCl₂F₂ vaporize at 25°C? [∆H(vap) = 17.2 kJ/mol at 25°C]

61.2 J/K Explanation: ΔS=heat/T ΔS=1.06 mol(17200J/molK)/298.15K ΔS=61.2 J/K

For a reaction for which ∆H = +29.3 kJ/mol and ∆S = +106 J/mol･K, which of the following statements is true? A) The reaction is spontaneous above 276 K. B) The reaction is spontaneous below 276 K. C) The reaction will never reach equilibrium. D) The reaction will never be spontaneous.

A) The reaction is spontaneous above 276 K. Explanation: Spontaneity is determined by the free energy. When ΔG is negative, it is spontaneous. ΔG=ΔH−TΔS Plug in the given values. ΔG=(29300)−(276)(106) ΔG=44 At values less than 276, the reaction is not spontaneous. Now, try a higher temperature. ΔG=(29300)−(277)(106) ΔG=−62 At values greater than 276, the reaction is spontaneous.

If a chemical reaction has a negative ∆H and a positive ∆S, then A) it will be spontaneous at all temperatures. B) it will be non-spontaneous at all temperatures. C) it will be spontaneous at high temperatures and non-spontaneous at low temperatures. D) it will be spontaneous at low temperatures and non-spontaneous at high temperatures. E) the absolute entropy of the reactants will be less than 0.

A) it will be spontaneous at all temperatures. Explanation: A negative ∆G means a process is spontaneous, and a positive ∆G is always non-spontaneous. ∆G = ∆H - T∆S and T is always positive since it is in kelvin. A chemical reaction having a negative ∆H and a positive ∆S will always have a negative ∆G and will always be spontaneous.

A reaction which is endothermic and has an overall increase in entropy is A) spontaneous only at high T B) spontaneous only at low T C) always spontaneous as written. D) always spontaneous in the reverse direction.

A) spontaneous only at high T Explanation: Spontaneity is determined by the free energy. When ΔG is negative, it is spontaneous. ΔG=ΔH−TΔS The problem indicates that the ΔS is positive. If the reaction is endothermic, this means that the ΔH is positive. The only way for ΔG to be negative is if the temperatures are very high.

For which process is ΔS positive? Group of answer choices A. A spontaneous endothermic process at a constant temperature and pressure B. An ideal gas being compressed reversibly at a constant temperature and pressure C. Water freezing at -20°C D. A precipitation reaction

A. A spontaneous endothermic process at a constant temperature and pressure Explanation: The thermodynamic quantity of the system, the decrease in whose value during a process is equal to useful work done by the system, i.e. ΔG=ΔH−TΔSΔG=ΔH-TΔS, where H = enthalpy, S= entropy and T= absolute temperature. If an endothermic reaction occurs spontaneously i.e. ΔH=+veΔH=+ve then for reaction to be spontaneous ΔGΔG should be negative that is only possible when ΔS>0ΔS>0 at constant temperature and pressure.

Which of the following substances has the higher positional probability per mole? Group of answer choices: A. Gaseous water at 100°C. B. Because the temperatures are the same, the positional probabilities are the same C. Solid water at 100°C. D. Liquid water at 100°C.

A. Gaseous water at 100°C. Explanation: Changes of state: positional probability increases: - From solid to liquid to gas Much smaller volume in the solid state than in the gas - Solid state: molecules close together - Gaseous state: molecules are fart apart, many more positions available Positional probability increases with mixing as well as decrease in pressure and increase in volume.

For any spontaneous process, Group of answer choicesOnly A. Only ΔSuniv must be positive. B. ΔSsys and ΔSsurr must be positive. C. ΔSuniv and ΔSsys must be positive. D. Only ΔSsys must be positive. E. ΔSuniv, ΔSsys, and ΔSsurr must all be positive.

A. Only ΔSuniv must be positive. *See Page 5 of Lecture 10 Slides

At constant pressure, the reaction 2NO2(g) = N2O4(g) is exothermic. The reaction is Group of answer choices A. spontaneous at low temperatures, but not at high temperatures. B. always spontaneous. C. spontaneous at high temperatures, but not at low temperatures.n D. ever spontaneous.

A. Spontaneous at low temperatures, but not at high temperatures. Explanation: At T= const, exothermic process causes heat to flow into the surroundings, increasing random motions. An exothermic reaction is when a reaction has a net release of heat, system loses heat (ΔH is negative). Spontaneous means that the reaction happens without any added help (ie. any extra energy).

Which of the following has the smallest standard molar entropy, S° (298.15 K)? A) CaSO₄(s) B) C(diamond) C) Xe(g) D) C₆₀(s) E) Br₂(l)

B) C(diamond) Explanation: Perfect crystalline structures have lower entropy than all other states of matter because they are highly ordered. As molecules get larger, they have more entropy. Liquids by comparison have less order and have higher entropies than solids.

Which of the following reactions would have the most positive ∆S° value? A) SO₂(g) + Na₂O(s) ⟶ Na₂SO₃(s) B) CO₂(s) ⟶ CO₂(g) C) Fe³⁺(aq) + SCN⁻(aq) ⟶ FeSCN²⁺(aq) D) N₂(g) + 3 H₂(g) ⟶ 2 NH₃(g) E) 2 NO(g) + Cl₂(g) ⟶ 2 NOCl(g)

B) CO₂(s) ⟶ CO₂(g) Explanation: The most positive change in entropy would indicate that there is more disorder at the end of the reaction. Option B goes from one mole of a highly ordered solid to one mole of a very disordered gas.

Consider the following half-reactions: I₂(s) + 2e⁻ → 2 I⁻(aq) E°(red) = 0.54 V Cd²⁺(aq) + 2 e⁻ → Cd(s) E°(red) = -0.40 V. Will the electrochemical cell represented by I⁻ | I₂ || Cd²⁺ | Cd be galvanic or electrolytic under standard conditions? A) Galvanic B) Electrolytic C) Not enough information

B) Electrolytic

Which of the following has the largest standard molar entropy, S° (298.15 K)? A) He(g) B) H₂(g) C) NaCl(aq) D) KBr(s) E) Hg(l)

B) H₂(g) Explanation: Perfect crystalline structures have an entropy of 0 because they are highly ordered. Gases have high entropy because they are very disordered and are random. Larger or heavier molecules have higher entropies. H₂ has a larger standard molar entropy than He when both are in gaseous form because the molecular is larger. Disorder increases because the solid is now more disordered as an aqueous solution and one molecule became two ions, so the solution is likely more disordered than a liquid.

Which of the following is true for a process where ∆S(universe) < 0 at 298 K? A) It is exothermic B) It is nonspontaneous. C) It is product-favored. D) It is spontaneous. E) It is endothermic.

B) It is nonspontaneous. Explanation: Any spontaneous process must increase the entropy of the universe, according to the second law of thermodynamics.

If ∆H > 0 and ∆S < 0, a reaction will be: A) Spontaneous B) Nonspontaneous C) At equilibrium D) Not enough info

B) Nonspontaneous Explanation: The free energy can be calculated using this formula. ΔG=ΔH−TΔS If ∆H > 0 and ∆S < 0, a reaction will be nonspontaneous because the ∆G will be positive at all temperatures.

If the Gibbs free energy for an equilibrium is a large, negative number, the equilibrium constant is expected to be _____ A) a large, negative value B) a large, positive value C) a small, negative value D) a small, positive value E) Zero

B) a large, positive value Explanation: Since ∆G° = - RTlnK, when ∆G° is a large negative number, K must be a large positive number, indicating the reaction is spontaneous and the equilibrium lies far to the right.

If a chemical reaction has a positive ∆H and a negative ∆S, then A) it will be spontaneous at all temperatures. B) it will be non-spontaneous at all temperatures. C) it will be spontaneous at high temperatures and non-spontaneous at low temperatures. D) it will be spontaneous at low temperatures and non-spontaneous at high temperatures. E) the absolute entropy of the products will be less than 0.

B) it will be non-spontaneous at all temperatures. Explanation: A negative ∆G means a process is spontaneous, and a positive ∆G is always non-spontaneous. ∆G = ∆H - T∆S and T is always positive since it is in kelvin. A chemical reaction having a positive ∆H and a negative ∆S will always have a positive ∆G and will always be non-spontaneous.

Cl₂ is a stable diatomic molecule. It can be decomposed to form two Cl atoms as shown below. Cl₂ → 2Cl(g) Predict the change in entropy (∆S) of this reaction. A) ∆S is (-) B) ∆S is (+) C) ∆S = 0 D) ∆S cannot be predicted for this reaction.

B) ∆S is (+) Explanation: ∆S is (+). Entropy is positive when the degrees of freedom of the system increases. In this system, every mole of reactant yields two moles of product so the entropy and degrees of freedom are increasing.

Which of the following is true for a system at equilibrium? A) ∆S°(sys) = ∆S°(surr) B) ∆S°(sys) = -∆S°(surr) C) ∆S°(sys) = ∆S°(surr) = 0 D) ∆S°(sys) = ∆S°(univ) E) ∆S°(surr) = ∆S°(univ)

B) ∆S°(sys) = -∆S°(surr) Explanation: At equilibrium the change in entropy of the system is equal to the negative change in entropy of the surroundings.

For the vaporization of a liquid at a given pressure, Group of answer choices A. ΔG is negative at all temperatures. B. ΔG is positive at low temperatures, but negative at high temperatures. C. ΔG is positive at all temperatures. D. ΔG is negative at low temperatures, but positive at high temperatures.

B. ΔG is positive at low temperatures, but negative at high temperatures. *See Page 6 Lecture Slide 10

Evaluate the validity of the following statement: Spontaneous processes are ones that occur quickly and have a low activation energy. A) False. Spontaneous processes always require an input of energy to overcome the activation energy, but always react quickly. B) False. Spontaneous processes can occur slowly, but always have a low activation energy. C) False. Spontaneous reactions can react slowly and can have a high activation energy. D) False. Spontaneous processes always react slowly and always have a high activation energy. E) True.

C) False. Spontaneous reactions can react slowly and can have a high activation energy. Explanation: Although a spontaneous reaction has negative free energy and is thermodynamically favorable, the kinetics are not necessarily fast. Rusting and the conversion of diamond to graphite are both spontaneous processes that occur slowly and have a high activation energy.

Which of the following is true about all spontaneous processes? A) It releases energy as heat. B) It will occur quickly. C) It will continue on its own once begun. D) It is never endothermic.

C) It will continue on its own once begun.

Which of the following indicates a spontaneous reaction under standard conditions? A) K = 8.6 x 10⁻² B) K = 7.9 x 10⁻⁸ C) K = 2.2 x 10² D) K = 9.2 x 10⁻⁴⁵ E) K = 0.14

C) K = 2.2 x 10² Explanation: A spontaneous reaction has an equilibrium constant that is greater than 1. K = 2.2 x 10², is the only choice where K > 1.

Which of the following is true for a perfect crystal at absolute zero? A) ∆S < 0 B) ∆S > 0 C) S = 0 D) S > 0 E) S < 0

C) S = 0 Explanation: The third law of thermodynamics states that the entropy for a perfect crystal is 0 at absolute zero temperature in Kelvin.

For a reaction for which ∆H = -64.2 kJ/mol and ∆S = 285 J/mol･K, which of the following statements is true? A) The reaction is spontaneous only above 225 K. B) The reaction is spontaneous below 225 K. C) The reaction will never reach equilibrium. D) The reaction is always at equilibrium.

C) The reaction will never reach equilibrium. Explanation: Any reaction with a negative ΔH and positive ΔS will be spontaneous at all temperatures. The reverse is also true: any reaction with a positive ΔH and a negative ΔS will be non-spontaneous at all temperatures. Spontaneity is determined by the free energy. When ΔG is negative, it is spontaneous. ΔG=ΔH−TΔS Plug in the given values. ΔG=(−64200)−(225)(285) ΔG=−128325 At values greater than 276, the reaction is spontaneous. Now try at absolute 0. ΔG=(−64200)−(0)(285) ΔG=−64200 This reaction is going to be spontaneous at all temperatures and therefore will never have a ΔG of 0 to reach equilibrium.

A reaction which is exothermic and has an overall increase in entropy is A) spontaneous only at high T B) spontaneous only at low T C) always spontaneous D) spontaneous in the reverse direction.

C) always spontaneous Explanation: Spontaneity is determined by the free energy. When ΔG is negative, it is spontaneous. ΔG=ΔH−TΔS The problem indicates that the ΔS is positive. If the reaction is exothermic, this means that the ΔH is negative. ΔG is going to be negative no matter the temperature.

If a chemical reaction has a positive ∆H and a positive ∆S, then A) it will be spontaneous at all temperatures. B) it will be non-spontaneous at all temperatures. C) it will be spontaneous at high temperatures and non-spontaneous at low temperatures. D) it will be spontaneous at low temperatures and non-spontaneous at high temperatures. E) the absolute entropy of the products will be less than 0.

C) it will be spontaneous at high temperatures and non-spontaneous at low temperatures. Explanation: A negative ∆G means a process is spontaneous, and a positive ∆G is always non-spontaneous. ∆G = ∆H - T∆S and T is always positive since it is in kelvin. A reaction where ∆H and ∆S are both positive will be non-spontaneous at low temperatures when ∆H dominates and will be spontaneous at high temperatures when T∆S dominates.

Which of the following is always true for an equilibrium? A) ∆S = 0 B) ∆H = 0 C) ∆G = 0 D) T = 298 K E) K = 0

C) ∆G = 0 Explanation: A criterion for equilibrium is that ∆G = 0.

Which of the following is true for a reaction when K is much less than one? A) ∆G° = K B) ∆G° = large negative number C) ∆G° = large positive number D) ∆G° = small negative number E) ∆G° = small positive number

C) ∆G° = large positive number Explanation: When K is much less than one, the equilibrium lies far to the left and the reaction is very unfavorable. Since ∆G° = - RTlnK, ∆G° will be large and positive.

ΔSsurr is _______ for exothermic reactions and ______ for endothermic reactions. Group of answer choices A. More information is needed to answer this question. B. unfavorable; favorable C. favorable; unfavorable D. favorable; favorable E. unfavorable; unfavorable

C. favorable; unfavorable Explanation: Since an exothermic process is a process in which the flow of energy into surroundings increases random motions, it also increases the entropy of surroundings In an exothermic process, ΔSsurr is positive In an endothermic process, ΔSsurr is negative - Ex. Vaporization of water ΔSsurr<0 (You have to put in heat to boil)

When a stable diatomic molecule spontaneously forms from its atoms at constant pressure and temperature, what is true about ΔS Group of answer choices A. ΔS is positive. B. There is not enough information to answer this question. C. ΔS is negative. D. ΔS is zero.

C. ΔS is negative. For a molecule to spontaneously form, ΔG would have to be negative, ΔH would have to be negative, and ΔS would also have to be negative.

Which of the followings statements is (are) true for a galvanic (voltaic) cell when E° = +1.00 V? 1. The reaction is spontaneous 2. At equilibrium, K = 1. 3. ∆G° is negative. A) 1 only B) 2 only C) 3 only D) 1 and 3 only E) 2 and 3 only

D) 1 and 3 only Explanation: Since ∆G°=-nFE°, ∆G°=-nF when E° = +1.00 V, so "1. The reaction is spontaneous", and 3. "∆G° is negative", are true statements. K does not equal 1 at equilibrium because K=e^nFE°/RT=e^n×96500×1.0/8.314× 298 =e^38.9n

How many of the following processes are nonspontaneous? climbing stairs dissolving salt in water converting table salt to its elements 2 H₂O(l) → 2 H₂(g) + O₂(g) A) 0 B) 1 C) 2 D) 3 E) 4

D) 3 Explanation: Climbing stairs, converting table salt to its elements, and the decomposition of water into its elements are nonspontaneous processes. Dissolving salt in water is a spontaneous process. Nonspontaneous processes can happen, they just require the input of energy.

Which of the following reactions would have the most negative ∆S° value ? A) N₂(g) + 3 Cl₂(g) ⟶ 2 NCl₃(g) B) CaSO₄ ･ 7 H₂O(s) ⟶ CaSO₄(s) + 7 H₂O(g) C) K(s) + O₂(g) ⟶ KO₂(s) D) NH₃(g) + HBr(g) ⟶ NH₄Br(s) E) 2H₂(g) + O₂(g) ⟶ 2 H₂O(g)

D) NH₃(g) + HBr(g) ⟶ NH₄Br(s) Explanation: The most negative change in entropy would indicate that there is less disorder at the end of the reaction. Option D goes from two moles of very disordered gases to one mole of highly ordered solid.

What will the sign on ∆S be for the following reaction and why? 2 Mg (s) + O₂ (g) → 2 MgO (s) A) Positive, because there is a solid as a product. B) Positive, because there are more moles of reactant than product. C) Positive, because it is a synthesis reaction. D) Negative, because there are more moles of gas on the reactant side than the product side. E) Negative, because there are more moles of reactant than product.

D) Negative, because there are more moles of gas on the reactant side than the product side. Explanation: The sign of ∆S can be determined by comparing the number of moles of gas on the reactant and product side. Since the number of moles of gas is decreasing, entropy is decreasing, and thus is negative.

If ∆H < 0 and ∆S < 0, a reaction will be: A) Spontaneous B) Nonspontaneous C) At equilibrium D) Not enough information

D) Not enough information Explanation: The free energy can be calculated using this formula. ΔG=ΔH−TΔS If ∆H < 0 and ∆S < 0, the spontaneity of the reaction depends on the temperature. If the temperature is low, ∆G will be negative and the reaction will be spontaneous. If the temperature is high, ∆G will be positive and the reaction will be nonspontaneous.

Which is true for the endothermic reaction 2 SO₃(g) → 2 SO₂(g) + O₂(g)? A) Spontaneous at all T B) Spontaneous at low T C) Non-spontaneous at any T D) Spontaneous at high T

D) Spontaneous at high T Explanation: The free energy can be calculated using this equation. ΔG=ΔH−TΔS If both ∆H and ∆S are positive, then T must be large in order for ∆G to be negative.

Which law states that the entropy of a perfect crystalline substance is zero at the absolute zero of temperature? A) Zeroth law of thermodynamics B) First law of thermodynamics C) Second law of thermodynamics D) Third law of thermodynamics E) Law of conservation of energy

D) Third law of thermodynamics Explanation: The third law of thermodynamics states that the entropy of a perfect crystalline substance is zero at the absolute zero of temperature.

The second law of thermodynamics states that _____ A) no engine can be 100% efficient. B) transferring energy from a cold to a hot bath requires work. C) the entropy of the universe increases with a spontaneous process. D) all of these are true.

D) all of these are true.

If a chemical reaction has a negative ∆H and a negative ∆S, then A) it will be spontaneous at all temperatures. B) it will be non-spontaneous at all temperatures. C) it will be spontaneous at high temperatures and non-spontaneous at low temperatures. D) it will be spontaneous at low temperatures and non-spontaneous at high temperatures. E) the absolute entropy of the products will be less than 0.

D) it will be spontaneous at low temperatures and non-spontaneous at high temperatures. Explanation: A negative ∆G means a process is spontaneous, and a positive ∆G is always non-spontaneous. ∆G = ∆H - T∆S and T is always positive since it is in kelvin. A reaction where ∆H and ∆S are both negative will be spontaneous at low temperatures when ∆H dominates and will be non-spontaneous at high temperatures when T∆S dominates.

For which process is ΔS negative? Group of answer choices A. Mixing 5 mL ethanol with 25 mL water B. Grinding a large crystal of KCl to powder C. Raising the temperature of 100 g Cu from 275 K to 295 K D. Compressing 1 mol Ne at constant temperature from an internal pressure of 0.5 atm to 1.5 atm E. Evaporation of 1 mol of CCl4(l)

D. Compressing 1 mol Ne at constant temperature from an internal pressure of 0.5 atm to 1.5 atm Explanation: The entropy of a system, usually denoted by S, is a thermodynamic quantity used in describing the level of disorderliness or randomness of a system. The higher the order of a system, the less entropy it has. In contrast, the more disordered the system is, the higher is its entropy. An entropy change, ΔS, that is negative implies that the entropy of the system decreased, i.e. it becomes more ordered. Let us look at each process: a. An evaporation process is the conversion of a liquid substance to its gaseous state. With a gas having a higher level of disorder than a liquid, the entropy of the system increased resulting in a positive ΔS. b. When two substances are mixed, the level of disorderliness of the mixed substances is increased. Thus, the entropy increased and ΔS is positive. c. Compressing a gas under constant temperature suggests that its volume is decreased. When the volume is decreased, there will be less space for the gas particles to move resulting in a more ordered gas particles. Thus, the entropy decreased and its ΔS is negative. d. An increase in temperature results in the particles of a substance gaining velocity which, in turn, increases the level of randomness of the particles. Thus, ΔS is positive for the given process. e. Pulverizing crystal results in smaller but more particles resulting in a higher disorder. Thus, ΔS is positive for this process.

A liquid is vaporized at its boiling point at constant pressure. What is true about ΔG Group of answer choices A. ΔG > 0 B. ΔG < 0 C. ΔG = ΔS D. ΔG = 0

D. ΔG = 0 Explanation: Vaporization at the boiling point is known simply as boiling. The temperature of a boiling liquid remains constant until all of the liquid has been converted to a gas. The energy required for vaporization offsets the increase in disorder of the system. Thus ΔG = 0, and the liquid and vapor are in equilibrium, as is true of any liquid at its boiling point under standard conditions.

Which of the following are spontaneous processes? 1. Ice melting at 1atm and 273 K (assume only ice is initially present). 2. Heat flowing from a hot object to a cold object. 3. An iron bar rusting. A) 1 only B) 2 only C) 3 only D) 1 and 3 only E) 1, 2, and 3

E) 1, 2, and 3 Explanation: All three of these processes are spontaneous and occur naturally.

Which of the following processes is spontaneous? A) Heat coming out of an open oven. B) A puddle evaporating in the sun. C) A car rusting. D) Water turning into ice in a freezer. E) All of the above

E) All of the above Explanation: All of these processes are spontaneous and occur naturally.

A galvanic cell Zn | Zn²⁺ || Ni²⁺ | Ni runs spontaneously. If a current is imposed to turn this into an electrolytic cell, which of the following will occur? A) Zn(s) still gets oxidized at the same rate B) Zn(s) gets oxidized at a faster rate C) Ni²⁺ gets oxidized D) Ni²⁺ gets reduced E) Zn²⁺ gets reduced

E) Zn²⁺ gets reduced Explanation: An electrolytic cell runs in the opposite direction of a galvanic cell. Applying a current to the galvanic cell Zn | Zn²⁺ || Ni²⁺ | Ni would convert the Zn anode to a cathode and cause Zn²⁺ to be reduced.

In which case must a reaction be spontaneous at all temperatures? Group of answer choices A. ΔH is positive, and ΔS is negative. B. ΔH = 0, and ΔS = 0. C. ΔH is positive, and ΔS is positive. D. ΔH is negative, and ΔS is negative. E. ΔH is negative, and ΔS is positive.

E. ΔH is negative, and ΔS is positive. *See Chem 101/Page 6 Lecture 10 Slides