Question

AG,O (25°C kJ/mol) ДнР (25°C kJ/mol) So Substance (25°C J/mol-K) NaNO, (s) -467.9 -367.0 116.5 NaNO2 (s) -358.7 -284.6 103.8 Cl2O (g) CI2OS (g) PbO (s) 80.3 97.9 266.2 -212.5 -198.3 309.8 -217.32 -187.89 68.7 CI2CRO2 (I) CI2CRO2 (g) -579.5 -510.8 221.8 -538.1 -501.6 329.8

Answer 1

At 298 K:
PbCl2(s) + 3NaNO2(s) ----> PbO(s) + NaNO3(s) + 2NO(g) + NaCl(s)

DH0rxn = (1*DH0f,PbO(s) + 1*DH0f,NaNO3(s) + 2*DH0f,NO(g) + 1*DH0f,NaCl(s)) - (1*DH0f,PbCl2(s) + 3*DH0f,NaNO2(s))

       = (1*-217.32 + 1*-467.9 + 2*90.25 + 1*-411.153)-(1*-359.41+3*-358.7)
     
       = 519.6 kj

DS0rxn = (1*S0f,PbO(s) + 1*S0f,NaNO3(s) + 2*S0f,NO(g) + 1*S0f,NaCl(s)) - (1*S0f,PbCl2(s) + 3*S0f,NaNO2(s))

       = (1*68.7 + 1*116.5 + 2*86.55 + 1*-384.14)-(1*-314.1+ 3*103.8)

       = -23.14 j/k

DG0 = DH0-TDS0

T = 298 k

DG0 = (519.6)-(298*-23.14*10^-3)

    = 526.5 Kj

DG0 = - RTlnK

526.5*10^3 = -8.314*298lnK

K = equilibrium constant = 5.12*10^-93

At 315 K

PbCl2(s) + 3NaNO2(s) ----> PbO(s) + NaNO3(s) + 2NO(g) + NaCl(s)

DH0rxn = 519.6 kj

DS0rxn = -23.14 j/k

(DH0,DS0 are taken as Temperature independent)

DG0 = DH0-TDS0

T = 315 k

DG0 = (519.6)-(315*-23.14*10^-3)

    = 526.9 Kj

DG0 = - RTlnK

526.9*10^3 = -8.314*315lnK

K = equilibrium constant = 4.21*10^-88