Question

Pi T2 = P₂ T 12) Calculate the standard free energy change at 25°C for the reaction 2 NO(g) + O2(g) → 2 NO2(g). 12 46°E AH-TAS Species | AHºf, kJ/mol NO () 91.3 02 (8) 0.0 NO2 (8) 5, J/(K.mol) 210.7 205.0 240.0 33.2 A) -4.7 kJ (B) -72.6 kJ C) -157.8 kJ D) -532.6 kJ 13) The Boltzmann formula is S=k In W. A perfect crystal has a molar entropy of 0 at absolute zero because A) W=NA B) W=1. C) k = 1. D) W=0.

Answer 1

The balanced equation is 2 NO (g) + O2 (g) -------------> 2 NO2 (g) AH,° = AH (products) – AH° (reactants) AH,° = [2 mol x AHº (NO2 (g)) – [2 mol AH° (NO (g)) + 1 mol AH° (O2 (g))] AH, = [2 mol (33.2 kJ/mol )]- [2 mol x91.3 kJ/mol ) + 1 mol x 0)] AH,° = - 116.2 kJ AH°rxn =- 116.2 kJ AS,' = 5° (products) - Sº (reactants) AS,' = = [2 mol % ASO (NO2(g)) - [2 mol X AS° (NO (g)) + 1 mol ASO (02 (g))] AS,° = [2 mol x 240 J/mol.K]-[2 mol x 210.7J/mol.K + 1 mol x 205 J/mol.K] AS,' = - 146.4 J/K = -0.1464 kJ/K AS,' = -0.1464 kJ/K We know that AG,° = AH," - TAS, AG, = -116.2 kJ - (298 KX-0.1464 kJ/K) = - 72.5728 kJ AG,° = - 72.6 kJ Option B is the right answer.

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13.

W is the number of microstates in which the energy of the molecules in a system can be arranged on energy levels.

At absolute zero, W = 1

So, S = 0