At 2257 K and 1.00 atm total pressure, water is 1.77 per cent dissociated at equ

Question

At 2257 K and 1.00 atm total pressure, water is 1.77 per cent dissociated at equilibrium by way of the reaction 2 H2O(g) 2H2(g) + O2(g). Calculate the following. K Delta r G0 kJ mol-1 Delta r G at this temperature kJ mol-1 In the gas-phase reaction 2A + B 3C + 2D, it was found that, when 1.00 mol A , 2.00 mol B, and 1.00 mol D were mixed and allowed to come to equilibrium at 25 degree C, the resulting mixture contained 0.90 mol C at a total pressure of 1.00 bar. Calculate the following. the mole fractions of each species at equilibrium A B D D Kx K Delta rG* J mol-1 The standard Gibbs energy of formation of NH3(g) is -16.5 kJ mol-1 at 298 K. What is the reaction Gibbs energy when the partial pressures of the N2, H2, and NH3 (treated as perfect gases) are 3.0 bar, 1.0 bar, and 4.0 bar, respectively? What is the spontaneous direction of the reaction in this case? toward products toward reactants

Explanation / Answer

1)dG' = -2.3 RT logK where T = 25C = 298 K , dG' = -2.3x 8.314 x298 log(2.85 x10^-6) =-31600Joules = -31.6 KJ, dG = -2.3 x 8.314 x 2257 log(2.85 x10^ -6) = 239322 Joules = 239.3 KJ , 2)(b) Kx = [(0.2^3)(0.35^2)/(0.087^2)(0.37)] x (0.986)^2 = 0.34 ( 1bar = 0.986 atm) , (c) K= Kx(P/RT)^dn , where dn = moles of products- moles of reactansts = 3+2-2-1 = 2, K = 0.34(0.986/0.0821x298)^2 = 0.00055, d) dG' = -2.3 x8.314 x298 x log(0.00055) = 18575 Joules = 18.575 KJ, 4) Kp = (4x0.986)^2/(3 x0.986)(0.986)^3 = 5.485 , Kc = KP(RT)^dn , dn = 2-3-1 = -2 , KC = 5.485/(0.0821 x298)^2 = 0.0091, gD = -2.3 x 8.314 x298 log(0.0091) = 11666 Joules = 11.666 KJ

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