ame: 2 Quiz 4 ThR-8.314 J/mole K 25 points DUE 21 NOV, TUESDAY 12:00 Consider th

Question

ame: 2 Quiz 4 ThR-8.314 J/mole K 25 points DUE 21 NOV, TUESDAY 12:00 Consider the following synthesis of HBr at 25 °C. AG (kJ/mole) 0 Compound | (J/mole K) | (kJ/mole) 31 36 131 245 199 H2 (g) Br2(g)- + 2 HBr@ [BT2(g) HBro) Sedron -53 enthalpy for the reacti mele wele )-31 =-103k] 2). Can you predict the AS° for the system? Why or Why not? Yes or NO 3). Is the reaction spontaneous at standard conditions? Based upon what? ). Determine the Gibbs free energy at 2130. 5). Determine the equilibrium constant K, for the reaction at 213 C 1% Does the reaction lie towards reactants or products? .

Explanation / Answer

H2(g) + Br2(g) ---> 2HBr(g)

1) DH0rxn = (2*DH0f,HBr)-(DH0f,H2 + DH0f,Br2)

          = (2*-36)-(0+31)

          = -103 kj

2) no,it is not possible,because no of mol of products = reactants.

   DS0 = (199)-(245+131)

       = -177 kj

3) at STP , T = 273.15 k

    DG0 = DH0-TDS0

        = (-103*10^3)-(273.15*-177)

        = -54.65 kj

as DG0 = - ve, the process is spontaneous.

4) DG0 = DH0-TDS0

        = (-103*10^3)-((213+273.15)*-177)

        = -16.95 kj

5) equilibrium constant Kp =

DG0 = - RTlnKp

-16950 = -8.314*486.15lnKp

Kp = 66.3

reaction lie towards reactants.

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