Question

2XY2(g) + Y2(g) -> 2XY3(g) (delta)Hf(kj/mol) S(J/K*mol) XY2(g) -296.3 262.3 Y2(g) 0 205.4 XY3(g) -403.4 275.2...

2XY2(g) + Y2(g) -> 2XY3(g)

(delta)Hf(kj/mol) S(J/K*mol)

XY2(g) -296.3 262.3

Y2(g) 0 205.4

XY3(g) -403.4 275.2

Determine the value of the equilibrium constant at 25*C.

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Answer #1

DH0rxn = 2*DH0f,XY3 - (1*DH0f,Y2 + 2*DH0f,XY2)

        = (2*-403.4)-(1*0+2*-296.3)

        = -214.2 kj

DS0rxn = 2*S0f,XY3 - (1*S0f,Y2 + 2*S0f,XY2)

         = (2*275.2)-(1*205.4 + 2*262.3)

         = -179.6 j/mol.k

dg0 = DH0rxn - TDS0rxn

   T = 298 k

dG0 = (-214.2*10^3)-(298*-179.6)

      = -160.68 kj

dG0 = - RTlnK

-160680 = -8.314*298lnk

equilibrium constant (K) = 1.46*10^28

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