For the reaction CuS(s) + H2(g) H2S(g) + Cu(s) G°f (CuS) = 53.6 kJ/mol G°f (H2S)

ID: 972738 • Letter: F

Question

For the reaction CuS(s) + H2(g) H2S(g) + Cu(s) G°f (CuS) = 53.6 kJ/mol G°f (H2S) = 33.6 kJ/mol H°f (CuS) = 53.1 kJ/mol H°f (H2S) = 20.6 kJ/mol

a. Calculate G° and H° at 298 K and 1 atm pressure.

b. Will this reaction proceed spontaneously at 298 K and 1 atm pressure?

c. Calculate the equilibrium constant for this reaction at 298 K.

d. Calculate S° at 298 K and 1 atm pressure.

e. Calculate G at 798 K and 1 atm pressure (assume S° and H° do not change with temperature).

f. Calculate the equilibrium constant at 798 K and 1 atm pressure.

delG0= delG of H2S- delG of CuS= -33.6+53.6= 20 Kj

delH0= -20.6+53.1=32.5 Kj

since delG0 is +ve, the reaction is not spontaneous

delG0= -RTlnK

lnK= -delG0/RT= -20*1000/(8.314*298.15)= -8.068

K= 0.000313

delG = delH- TdelS

T delS= delH- delG= 32.5-20 =12.5 Kj

delS= 12.5*1000/298=41.94 J/K

ln(K2/K1) = (delH/R)*(1/T1-1/T2)

ln(K2/0.000313)= (32.5*1000/8.314)*(1/298-1/798) =8.22

K2/0.000313= 3714.5

K2= 3714.5*0.000313=1.1626

delG= delGo+RTlnK= 20*1000+8.314*798*ln(1.1626)=20999.5 Kj

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.