Question

Please asnwer the following question completely and correctly. Please write neatly and show all work.

1. Calculate AG (375K) for the reaction: 2CO(g) + O2(g) →202 (g) from the value of AG (298.15K) and AH (298.15K). Assume A His constant in such temperature range.

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Answer 1

Equation : 200(8) + O2(g) → →200 (8) [2xAG (COCO) Free energy change of the reaction at 298.15K, AG= -2xAG(com) – 1x AH (0.6)] 1 -{{2x-1313 (1 xo ] = -514.2 kJ/mol = -514.2x10' Jimol - - - - - - - - - - - - - -------------- Equation : 200(8) + 0,(8) —> 200,(8) Standard free energy change of the reaction, AG' = - RTin Ka _AG -514.2x10 Jim ol =2074 In kq-RT -- 8.314 J K-mol-1x 298.15 K) Kgge = 2014 = 1.2x10 2xAH (COD)) Enthalpy change of the reaction at 298.15K, AH) = AH(coco) – 1x AH (0) mol |(2 mol x-3935 det) 1 -{{2molx-110,5 hod) + (1 mol xo hold)] = -566 kJ/mol Temperature, T = 375K According to VantHoff equation, K at 375 K ( K ) -566x10Jim ol ( 1 1 --468 375 =-46.8 1.2x10908.314 JK-mol-1 298.15 K375 = €4684.8x10-21 1.2 x 100 K37 = 1.2x10" (4.8x10-21) = 5.8x1069 Calculation of AG° at 375 K AC°=R7ın K =-(83142. mal](375 K)in (5.8x10® ) = -5.01x10% I, 1kJ =-5.01x10 mol 1000 J = -501 kJ/mol