Chapter 10 – Chem 1B – CLAS
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drop-in: T/R 2-3, F 11-1 (SRB 3274)
1. Which has the greatest entropy?
a. 1 mol of He at0.5 atm and 25°Cor 1 mol of He at1 atm and 25°C ⇒ as P ↓ V ↑ S ↑
b. 1 mol of Ne at STP or1 mol of CH4 at STP⇒ as #bonds/molecule ↑ S ↑
c. 1 mol of Cl2 at 1 atm and 25°Cor 1 mol of Br2 at 1 atm and 25°C⇒Cl2 at 1 atm and 25°C is a gas which has more S thanBr2 at 1 atm and 25°C which is a liquid
2. Predict if ∆Ssys and ∆Ssurr is positive or negative for the following under standard conditions.
a. melting ice
∆Ssys > 0, ∆Ssurr 0
b. photosynthesis⇒ 6 CO2 (g) + 6 H2O (l) → C6H12O6 (s) + 6 O2 (g)∆Ssys 0, ∆Ssurr 0
3. One mole of an ideal gas at 25°C undergoes a reversible expansion from 125.0 L to 250.0 L. Which statement is correct?
a. ∆Sgas = 0
b. ∆Ssurr= 0
√c. ∆Suniv = 0⇒ reversible is another way of saying that the system is at equilibrium
4. A 100-mL sample of water is placed in a coffee cup calorimeter. When 1.0 g of an ionic solid is added, the temperature decreases from 21.5°C to 20.8°C as the solid dissolves. Which of the following is true for the dissolving of the solid?
a. ∆H < 0 false⇒ it was observed that the T of the surroundings cooled
√ b. ∆Suniv > 0 true ⇒ because the solid dissolved spontaneously
c. ∆Ssys0 false⇒ aqueous ions have more S than in a solid
d. ∆Ssurr0 false⇒ since the dissolving was endothermic
e. none of these
5. One mole of an ideal gas is compressed reversibly at 607.4 K from 5.60 atm to 8.90 atm. Calculate ∆S for the gas.
a. 2.34 J/K
b. – 2.34 J/K
c. – 3.85 J/K
d. 3.85 J/K
e. 0 J/K
If there’s a change in volume ⇒ ΔS = nRln or ΔS = nRln ⇒
ΔS = (1mol)(8.314J/molK)ln = – 3.85 J/K
6. Calculate the change in entropy (in J/K) for a process in which 54 g of ice at – 5 °C is mixed with 112 g of liquid water at 100°C in a perfectly insulated container. The specific heat capacities of ice and liquid water is 2.03 J/g°C and 4.18 J/g°C respectively. The heat of fusion for water is 6.01 kJ/mol.
∆Stotal = ∆Sicecube + ∆Shot water
Multiple steps are taking place ⇒ first the ice cube will go from -5 °C → 0 °C, second the ice cube will melt at 0 °C, then the melted water will go from 0 °C → unknown Tf, all the while the warm water will be cooling from 100 °C → unknown Tf
When temperature is changing ⇒ ∆S = mCln and q = mCΔT
When phase is changing ⇒ ∆S = and q = nΔHfus
∆Stotal= mCln+ mCln + mCln
Tf is unknown⇒ qtotal = mCΔT + nΔHfus + mC∆T + mC∆T= 0
(54 g)(2.03J/g°C)(0°C--5°C) + ((54g)(4.18 J/g°C)(T2 - -5°C) + (112 g)(4.18 J/g°C)(T2 - 0°C) ⇒ T2 = 40.7°C
∆Stotal = (54 g)(2.03J/gK)ln + (112 g)(4.18J/gK)ln = 18.4J/K
7. Consider the process: A (l) at 75°C → A (g) at 155°C
which is carried out at constant pressure. The total ΔS for this process is 75.0 Jmol-1K-1. For A (l) and A(g) the Cp values are 75.0 Jmol-1K-1 and 29.0 Jmol-1K-1 respectively. Calculate ΔHvap at 125°C (its boiling point).
3 steps⇒ goes from 75°C → 125 °C, vaporizes at 125 °C, then goes from 125 °C→155 °C
When temperature is changing ⇒ ∆S = nCln
When phase is changing ⇒ ∆S =
∆Stotal= nCln+ nCln
since∆Stotalis in units of per mole then there is 1 mol of A
∆Stotal= (1mol)(75 J/molK)ln + + (1mol)(29 J/molK)ln = 75J/molK
⇒ΔHvap = 25 kJ/mol
8. Indicate true or false for each of the following statements.
a. falseSpontaneous reactions must have a positive ΔSº for the reaction.
b. falseWhen the change in free energy is less than zero for a chemical reaction, the reaction must be exothermic.
c. trueFor a spontaneous reaction, if ΔSº < 0 then the reaction must be exothermic.
9. The following graph of G° versus temperature (T) corresponds to which of the following situations?
a. H° < 0 andS° > 0
b. H° > 0 and S° < 0
c. H° > 0 and S° > 0
d. H° < 0 and S° < 0
10. Which of the following is true for the dissociation of fluorine?
F2 (g) → 2 F (g)
a. spontaneous at all temperatures
b. spontaneous at high temperatures
c. spontaneous at low temperatures
d. never spontaneous
11. At 1 atm, the freezing point of mercury is –39 °C. Which of the following is true regarding the freezing of mercury at –30 °C and 1 atm?
a. Ssurr > 0, Suniv > 0
b. Ssurr > 0, Suniv = 0
c. Ssurr < 0, Suniv > 0
d. Ssurr < 0, Suniv < 0
e. Ssurr > 0, Suniv < 0
12. Given the following data, calculate the normal boiling point for formic acid (HCOOH).
∆Hfo (kJ/mol) S° (J/K·mol)
HCOOH (l) – 410. 130.
HCOOH (g) – 363 251
a. 2.57 K
b. 1730°C
c. 388°C
d. 82°C
e. 115°C
Boiling points (and freezing points) aretemperatures at equilibrium (∆G=0)
T = =
T =
T = 388K or 115 °C
13. Consider the following reaction at 25 oC: CO (g) + H2O (g) → H2 (g) + CO2 (g)
For this reaction ΔHo= –5.36 kJ and ΔSo= –109.8 J /K. At what temperatures will the reaction be spontaneous?
a. T > 48.8 K
b. T < 48.8 K
c. T > 20.5 K
d. T < 20.5 K
e. Spontaneous at all temperatures.
Since both ΔHoandΔS° are negativethe reaction will be spontaneous at T
T <
T < 48.8 K
14. Consider the following reaction.
2 POCl3 (g) → 2 PCl3 (g) + O2 (g)
The standard free energies of formation at 25 °C. ΔGf° POCl3 (g) = –502 kJ/mol, ΔGf° PCl3 (g) = –270 kJ/mol
Indicate true or false.
a. TrueThe entropy change for the reaction is positive.
b. TrueThe enthalpy change for the reaction is positive.
c. TrueThe reaction is non-spontaneous at standard conditions and 25 °C but will eventually become spontaneous if the temperature is increased.
d. TrueThe equilibrium constant for the reaction at 298 K is less than 1.
e. FalseIncreasing the pressure of POCl3 will cause an increase in ΔG.
15. Consider the following reaction at 800 K: 2 NF3 (g) → N2 (g) + 3 F2 (g)
At equilibrium, the partial pressures are PN2= 0.040 atm, PF2= 0.063 atm and PNF3= 0.66 atm. Which of the following is true about the value of ∆G°?
a. is a positive number
b. is a negative number
c. is equal to zero
d. is independent of the temperature
e. can not be predicted from this data
K = = 2.3 x 10 –5⇒ if K < 1 then ΔG° >0
16. Use the following reaction to answer the following: 2 SO2 (g) + O2 (g) → 2 SO3 (g)
(∆Gf° SO2 (g) = – 300 kJ/mol, ∆Gf° SO3 (g) = – 321 kJ/mol)
a. Calculate ∆G°
∆G°rxn = Σ ∆G°f(products) –Σ ∆G°f(reactants)
∆G°rxn = (2 mol SO3)(– 321 kJ/mol) - (2 molSO2)(– 300 kJ/mol) = - 42kJ
b. What is the equilibrium constant at 25 °C?
ΔG° = –RTlnKeq ⇒ -42,000 J = -(8.314 J/molK)(298K)lnKeq⇒Keq= 2.3x107
c. At 25 °C the initial pressures for SO2, O2 and SO3 are 0.001 atm, 0.002 atm and 40 atm respectively. Will SO3 be consumed or will SO3 be formed?
ΔG = ΔG° + RTlnQ
ΔG = -42,000 J + (8.3145 J/molk)(298 K)ln
ΔG = 25.9 kJ
sinceΔG > 0 the reaction will go backward and therefore SO3 will be consumed
17. Consider the following reaction and thermodynamic data:
2 NO (g) + O2 (g) → 2 NO2 (g) Ho = –190 kJ Go = –71 kJ at 600 K
Calculate the equilibrium constant (K) for this reaction at 370 K.
a.2.9x1016
b. 7.1x1013
c. 1.9x10–7
d. 6.1x105
e. 3.7x103
= ⇒ΔG°= –RTlnKeq ⇒ -71,000J = - (8.314J/molK)(600K)lnKeq
Keq = 1.52x106
=
K2 = 2.9x1016