84. Calculate AG\" of the reaction N2 (g) + 31 12 g) 2NH, (g) at 25 from data in

Question

84. Calculate AG" of the reaction N2 (g) + 31 12 g) 2NH, (g) at 25 from data in the following table (data are given at 25 oC) (a) Substance (c) H2 (d) NII, 0.0 0.0 -16.7 191.6 143.8 192.5 85. Calculate 1je for the reaction: NO (g) +O(g) NO2(g) Given the following information: NO (g) +O3 (g) NOz (g) + O2 (g 0, (g) 3/2O2(g) 02 (g) 20(g) 1Mz . 198.9 KJ dHe=-142.3 KJ AHe= 495.0 KJ 86. Use thermodynamic data in the following table that was determined at 25 °C Substance NO (g NO2 (2) NOCI (g) ,G" (KJ/mol) 86.7 51.8 66.3 103.6 So (J/mol K) 211 240 264 220 to calculate 1 10 at 25°C for the following reaction: Nd)(g) + NO2 () NO(g) 87. Given the reactions (1) and (2) below, determine the A,U" for the reaction (3). are derived at 298K.Assume all gases are perfect The following data in (1) and (2) for Ad (2) 21 lago +Osuo 21100 Arlr 184.62 kJ.mol Arlr 483.64 kJ.mol (3) 41 1Cou+0200 20kg) + 2H0

Explanation / Answer

84)

delta G0 = delta H0 - T delta S0

delta H0 = sumof the heat of formation of products - reactants

           = [2*(-16.7 kJ.mol)] - [0+3(0)] = - 33.4 kJ/mol K

delta S0 = sumof the entropy of products - reactants = [2*(192.5 J/mol K)] - [191.6 J/mol K+3(143.8J/mol K )]

             = 385 J/mol K - 623J/mol K = - 238 J/mol K

delta G0 = delta H0 - T delta S0

           = - 33.4 kJ/mol K - [298 K * (- 238 J/mol K)] = - 33.4 kJ/mol K + 70.92 kJ/mol K

delta G0 = 37.72 kJ/mol. K

** Note : Answering only first question as per the policy

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