10 0001 11.55 PM 12.2/100 10/16/2017 12:23 PM Periodic Table Print lCalculator G

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10 0001 11.55 PM 12.2/100 10/16/2017 12:23 PM Periodic Table Print lCalculator Gradebook Question 5 of 23 Chemical Principles 7th Edition MHE/Freeman presanted by Sapling Leaming Map Calculate the equilibrium constant. K, at 25.0 'C for each of the following Compound AGT u mot ) reactions. Use the thermodynamic information provided in the table. a. The hydrogenation of acetylene to ethane. CHtCl(g) 48.50 CH4g)-50.72 CHalg) 209.2 CHelg) 32 82 Clhig) 0.00 H2lg) 0.00 H0)-237.13 HCI(g)-95.30 HNO (ag) -111.25 NO(g) 86.56 Number b. The final step in the industrial production of nitric acid. 0 3NO2(g) + H2O(1) 2HNO, aq)+No) NOg) 1.31 O Hel 0 Number 0 o Web K- 0 O Techn eEut 0 Give up & View SoAutonCheck Answer 0Nen OPrevious 0 Hint about us careers privacy poicy terms ofuse contacts hea

Explanation / Answer

Solution:- (a) C2H2(g) + 2H2(g) <-----> C2H6(g)

deltta G0 for the reaction = delta Gf0(products) - delta Gf0(reactants)

deltta G0 for the reaction = -32.82 - [209.2 + 2(0.00)]

deltta G0 for the reaction = -32.82 - 209.2 = -242.0 kJ/mol

delta G0 = - RT ln k

-242.0 = - 8.314 x 10-3 x 298 x ln k

negative sign could be canceled out as it is on both sides..

ln k = 242.0/(8.314 x 10-3 x 298)

ln k = 97.68

k = e97.68

k = 2.64 x 1042

(b) 3NO2(g) + H2O(l) <-----> 2HNO3(aq) + NO(g)

deltta G0 for the reaction = [2(-111.25) + 86.55] - [3(51.31) + (-237.13)]

deltta G0 for the reaction = (-222.50 + 86.55) - (153.93 - 237.13)

deltta G0 for the reaction = ( -135.95) - ( - 83.2)

deltta G0 for the reaction = -135.95 + 83.2

deltta G0 for the reaction = -52.75 kJ/mol

-52.75 = - 8.314 x 10-3 x 298 x ln k

ln k = 52.75/(8.314 x 10-3 x 298)

ln k = 21.29

k = e21.29

k = 1.76 x 109

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