Question

For the following reactions assume that dH and dS are independent of temperature and determine teh temperature at which the reaction becomes spontaneous. a.) C(gr)--> C(diamond) . dHf* kJ mol-1 respectively is: 0 ; 1.895 & dS* J K-1 mol-1 respectively is: 5.740 ; 2.377 . [Please explain the answer is none] . b.) 4Fe(s) +3 O2(g) --> 2Fe2O3 (s). dH* kJ mol-1 respectively for this reaction is: 0; 0 ; -824.2. dS* J K-1 mol-1 respectively is: 27.28 ; 205.03 ; 87.40 . [Please explain the answer is < 3000 K] . c.) C6H12O6(s) +6O2(g)--> 6CO2(g) +6 H20(l). dH* kJ mol-1 respectively is : -1268 ; 0 ; -393.5 ; -285.8. dS*J K-1 mol-1 respectively is: 212 ; 205.0 ; 213.6 ; 69.6 . [Please explain the answer is all] . d.) N2O4 (g) --> 2NO2 (g).dH* kJ mol-1 repectively is : 9.16 ; 33.18. dS* J K-1 mol-1 respectively is: 304.18 ; 239.95. [Please explain the answer is >326 K ]

Answer 1

a.) C(gr)--> C(diamond) . dHf* kJ mol-1 respectively is: 0 ; 1.895 & dS* J K-1 mol-1 respectively is: 5.740 ; 2.377 . [Please explain the answer is none] .

solution :- lets calculate the delta H and Delta S reaction

delta H rxn = sum of delta Hf product – sum of delta Hf reactant

                    = [1.895*1]-[0*1]

                    = 1.895 kJ

Delta S rxn = sum of delta S product – sum of delta S reactant

                     = [2.377*1] – [ 5.740*1]

                     = -3.363 J per mol

Delta G is need to less than zero for the spontaneous reaction

So lets assume delta G is zero

So the delta G= Delta H – Tdelta S

T= Delta H/ delta S

= 1.895 kJ / (-3.363 J * 1 kJ / 1000 J)

= -563 K

b.) 4Fe(s) +3 O2(g) --> 2Fe2O3 (s). dH* kJ mol-1 respectively for this reaction is: 0; 0 ; -824.2. dS* J K-1 mol-1 respectively is: 27.28 ; 205.03 ; 87.40 . [Please explain the answer is < 3000 K] .

Solution :- delta H rxn = sum of delta Hf product – sum of delta Hf reactant

Delta H rxn = [-824.2 * 1] –[(0*4)+3*0)]

                     = -824.2 kJ

Delta S rxn = sum of delta S product – sum of delta S reactant

                    =[87.40*2]-[(27.28*4)+(205.03*3)]

                    =-636.81 J

T = delta H /delta S

   = -824.2 kJ / (-636.81 J * 1 kJ / 1000 J)

= 1294 K

c.) C6H12O6(s) +6O2(g)--> 6CO2(g) +6 H20(l). dH* kJ mol-1 respectively is : -1268 ; 0 ; -393.5 ; -285.8. dS*J K-1 mol-1 respectively is: 212 ; 205.0 ; 213.6 ; 69.6 . [Please explain the answer is all] .

Solution :-

delta H rxn = sum of delta Hf product – sum of delta Hf reactant

                    = [(-393.5*6)+(-285.8*6)]-[-1268*1]

                    = -2807.8 kJ

Delta S rxn = sum of delta S product – sum of delta S reactant

                     =[(213.6*6)+(69.6*6)]-[212*1)+(205*6)]

                     = 250 J

T= delta H/ delta S

= -2807.8 kJ / (250 J per K * 1 kJ / 1000 J)

=-11231 K

d.) N2O4 (g) --> 2NO2 (g).dH* kJ mol-1 repectively is : 9.16 ; 33.18. dS* J K-1 mol-1 respectively is: 304.18 ; 239.95. [Please explain the answer is >326 K ]

Solution :-

delta H rxn = sum of delta Hf product – sum of delta Hf reactant

                     = [33.18*2]-[9.16*1]

                     = 57.2 kJ

Delta S rxn = sum of delta S product – sum of delta S reactant

                     =[239.95*1]-[304.18*1]

                    = -64.23 J/ K

T= delta H/ delta S

= 57.2 kJ / (-64.23 J per K * 1 kJ / 1000 J)

= -890 K