The standard free energy change for the following reaction is Delta G degree = -

Question

The standard free energy change for the following reaction is Delta G degree = -225.8 kJ at 25 degree C: 2 KClO3(s) doubleheadarrow 2 KCl(s) + 3 O2(g) Calculate the standard molar entropy S degree for O2 at 25 degree C given the following data - S degree (KClO3) = 143 J.K^-1 mol^-1 S degree (KCl) = 82.6 J.K^-1.mol6-1 Delta H degree (KClO3) = -397.7 kJ.mol^-1 Delta H degree f (KCl) = -436.7 kJ.mol^-1 The Second Law of Thermodynamics states that "the total entropy of a system and its surroundings always increases in a spontaneous process". Show that the reaction above supports the Second Law. Calculate the pressure of O2 at equilibrium, giving your answer in the correct units.

Explanation / Answer

First, finding the enthalpy of the reaction = -436.7-(-397.7) = -39 KJ/mol

Change in gibbs free energy = chnge in enthaly - T (change in entropy of reaction)

-225.8 = 39 - 298(change in entropy of reaction)

Changein entropy of reaction = 264.8/298 =0.889 kJ/K mol = 889 J/ K mol

A) But chnge in entropy = entropy of products -entropy of reactants

889 = (2)(82.6)+ (3)(x) - (2)(143)

(889-165.2+286)/3 = x

Entropy of oxygen , x = 336.6 J/ K mol

b) Entropy of a system is nothing but stability of the system. The above reaction is a spontaneous reaction and spontaneous reaction reaches a stabilized state by releasing the energy it has in the form of free energy. Hence entropy (stability) of the system andsurroundings increases.

C) at equilibrium , PV = nRT

Here, n = 3 , V = 22.4 L, R = 0.0821 Latm/mol K, T = 298 K for O2

Substituting above values, P = 73.3974/22.4 = 3.276 atm

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