Question

Use the Table of Thermodynamic Data to decide which of the following reactions have K>1 at 298 K:

HCl(g)+NH3(g) <=> NH4Cl(s)

2 Al2O3(s) + 3 Si(s)<=>  3 SiO2(s) + 4 Al(s)

Fe(s) + H2S(g) <=> FeS(s) + H2(g).

Which of the products of these reactions are favored by a rise in temperature?

Answer 1

we know that

dGo = -RTlnK

so

to have K > 1 , dGo < 0

1)

HCl + NH3 --> NH4Cl

dGo rxn = dGo products - dGo reactants

so

dGo rxn = dGo NH4Cl - dGo HCl - dGo NH3

dGo rxn = -203.89 + 95.3 + 16.4

dGo rxn = -92.19

as dGo < O , K > 1

2)

2 Al203 + 3 Si ---> 3 SiO2 + 4 Al

dGo rxn = 3 x dGo SiO2 + ( 4 x dGo Al) - ( 2 x dGo Al203) - ( 3 x dGo Si)

dGo rxn = ( 3 x -856.4) + ( 4 x 0) - ( 2 x -1582.3) - ( 3 x 0)

dGo rxn = 595.4

as dGo > 1 , so K < 1

3)

Fe + H2S ---> FeS + H2

dGo rxn = dGo FeS + dGO H2 - dGo Fe - dGo H2S

dGo rxn = -100.4 + 0 - 0 + 33.4

dGo rxn = -67

as dGo < O we get K > 1

Now

we know that

for endothermic reactions ( dH > O )

the products are favored by a rise in temperature

1)

HCl + NH3 ---> NH4Cl

dH rxn = dHfo NH4Cl - dHfo HCl - dHfo NH3

dH rxn = -314.43 + 92.3 + 46.11

dH rxn = -176.02

as dH < 0 , the products are not favored when rise in temperature

2)

2Al203 + 3 Si ---> 3 SiO2 + 4 Al

dH rxn = 3 x dHfo SiO2 + ( 4 x dHfo Al) - ( 2 x dHfo Al203) - ( 3 x dHfo Si)

dH rxn = ( 3 x -910.94) + ( 4 x 0) - ( 2 x -1675.7) - ( 3 x 0)

dH rxn = 618.58

as dH > 0 , the products are favored when rise in temperature

3)

Fe + H2S ---> FeS + H2

dH rxn = dHfo FeS + dHfo H2 - dHfo Fe - dHfo H2S

dH rxn = -100 + 0 - 0 + 20.63

dH rxn = -79.37

as dH < 0 , the products are not favored when rise in temperature