Question

1. The standard enthalpy of combustion of glucose (C6 H12O6 (s)) to gaseous CO2 and H2 O is -2538.5 kJ/mol at 298 K. The corresponding standard free energy is -2826.0 kJ/mol. This observation appears to indicate that the amount of useful work that can be extracted from the combustion of glucose at constant pressure and temperature is actually larger than the heat released when the combustion performs no useful work. (a) Explain, on the basis of the second principles of thermodynamics ( dSuniverse≥0 ), how this is possible. (b) Compute the free energy of reaction at 298 K if the partial pressure of oxygen gas is 0.30 bar, the partial pressure of CO2 is 0.0010 bar and the partial pressure of water vapor is 0.100 bar. (c) Estimate the standard free energy of reaction at T=273 K

2.The constant pressure molar heat capacity of nitrogen is given by the expression Cp ,m=(27.0+5.90×10−3 T−0.34×10−6 T 2 ) J/( K mol ) Calculate the value of S for heating 1 mol of nitrogen from 25.0 ºC to 125 ºC at constant pressure

3. A household refrigerator (see diagram) is essentially a device that uses electrical work to transfer heat from a cold system (the contents of the refrigerator) to a warmer system (the kitchen). (a) Explain why, on the basis of the second law of thermodynamics ( dSuniverse≥0 ) some minimum amount of electrical work is necessary for the refrigeration process to occur, and (b) compute the minimum amount of work necessary to transfer 1 kJ of heat from the refrigerator at 5 ºC to the kitchen at 25 ºC. (a) 1 dq1 dq2 compressor dw fridge kitchen Thot Tcold electric grid ref

4.10 kJ of heat is absorbed by a gas dissociation reaction occurring reversibly and isothermally at 300 K in an elastic balloon. The gas mixture expands producing 5 kJ of work against the tension of the balloon and the pressure of the atmosphere. Calculate the change in entropy of the gas mixture in the balloon and the change in entropy of the universe.

5. One mole of an ideal gas is isothermally expanded from 5.0 L to 10.0 L at 300 K. Compute the entropy changes for the system, surroundings, and the universe if the process is carried out (a) reversibly by adjusting the pressure of the surroundings to match the internal pressure of the gas, and (b) irreversibly, freely expanding in a vacuum.

6. The molar heat of vaporization of ethanol is 39.3 kJ/mol, and the boiling point of ethanol is 78.3 ºC. Calculate the entropy of vaporization of 0.50 moles of ethanol.

Answer 1

1.(a) G=H-TS Thus -2826.0=-2538.5-298*S Solving this equation we get S=964.7 JK-1mol-1 This high positive value of entropy change is indicating that it will contribute much to make the free energy change highly negative(as enthalpy change is also negative) which is the requirement for spontaneity.

Heat change at constant pressureH=-2538.5 kJmol-1(as given), Thus heat change at constant volume U=H-n_gRT. Combustion reaction of glucose is C₆H₁₂O₆(s)+6O₂(g)=6CO₂(g)+6H₂O(g) So n_g=6

Putting all the values we get U=(2538.5-6*8.314*10^-3*298)Jmol^-1=2523.6(only value considered without sign) So it's less than enthalpy change.

(b) G=-RTlnKP=-2.303*8.314*298log{[p_CO2⁶*p_H2O⁶]/p_O2⁶}=-2.303*8.314*298*6log{(0.0010*0.1)/0.3}=+51.688 kJ mol-1

(c) At equilibrium G=0 So, the standard free energy change of the reaction at 273K =0

2.Entropy change at constant pressureS=nC_PdT from lower limit T₁ to upper limit T₂, Putting all the values S=27*(398-298)+5.9*10^-3(398+298)(398-298)/2-1/3*0.34*10^-6[398^3-298^3]=(2700+50*5.9*10^-3*696-0.34*12.193)JK-1mol-1=2901.24JK-1mol-1

4. From 1st law of thermodynamics U=q+w and we know entropy change in reversible isothermal processS_system=q_rev/T=10kJ/300K=33.33 JK^-1

As the process is isothermal and reversible so total entropy change ie S_total=S_system+S_surr=0

5.(a) Entropy change in isothermal reversible process S_iso-rev=RlnV₂/V₁=8.314*ln(10/5)=JK-1=5.762 JK^-1 It's the required S_system

As for isothermal reversible process S_system=-S_surr So S_surr=-5.762JK^-1 and S_universe=0

(b) For free expansion there is no work done. Thus S_system=S_total(S_surr=0) So S_system=S_universe=5.762 JK-1

6. Entropy of vaporisation S_vap=q_rev/T=39.3*0.5/(78.3+273) kJK^-1mol^-1=55.93 JK^-1mol^-1