Constants 1 Periodic Table Part A Estimate the value of the equilbrium constant

Question

Constants 1 Periodic Table Part A Estimate the value of the equilbrium constant at 645 K for each of the following reactions ?G? for BrCl(g)is-1 0 kJ/mol, The standard molar entropy, S" for BrCI(g) is 240 0 J/mol-K. 2NO2(g) N,04(g). ??¡for N,01(g)is 9.16 kJ/mol, Express your answer using three significant figures You may want to reference (Pages 873-878) Section 18.10 while completing this problem Submit Reguest Answer Part B Bra(8) + Cla () 2BrCKg) AHi for BrCl(g) is 146 kJ/mol. Express your answer using three significant figures Submit Request Answer

Explanation / Answer

part A

DG0 = DH0-TDS0

DH0 = (1*DH0,N2O4)-(2*DH0,NO2)

      = (1*9.16)-(2*33.2)

      = -57.24 Kj

DS0 = (1*S0,N2O4)-(2*S0,NO2)

     = (304.3)-(2*240.1)

     = -175.9 J/K

dG0 = DH0-TDS0

T = 645 k

DG0 = (-57.24)-(645*-175.9*10^-3)

    = 56.2 kj

DG0 = - RTlnK

56.2*10^3 = -8.314*645lnk

K = equilibrium constant = 2.809*10^-5

part B

DG0 = DH0-TDS0

?Hf = [1?Hf(Br2 (g)) + 1?Hf(Cl2 (g))] - [2?Hf(BrCl (g))]

   = [1(30.91) + 1(0)] - [2(14.64)]

      = 1.63 kJ

?S0 = [1?Sf(Br2 (g)) + 1?Sf(Cl2 (g))] - [2?Sf(BrCl (g))]

     = [1(245.35) + 1(222.97)] - [2(239.99)]

     = -11.66 J/K

dG0 = DH0-TDS0

T = 645 k

DG0 = (1.63)-(645*-11.66*10^-3)

    = 9.15 kj

DG0 = - RTlnK

9.15*10^3 = -8.314*645lnk

K = equilibrium constant = 0.18

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