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Use the table below to answer the questions that follow. Thermodynamic Quantitie

ID: 968380 • Letter: U

Question

Use the table below to answer the questions that follow.

            Thermodynamic Quantities for Selected Substances at 298.15 K (25°C)

                                                                                                        

            Substance        H°f (kJ/mol) G°f (kJ/mol) S (J/K-mol)

            Carbon                                             

               C (s, diamond)    1.88                   2.84                2.43

               C (s, graphite)     0                        0                     5.69

               C2H2 (g)        226.7                 209.2              200.8

               C2H4 (g)          52.30                 68.11            219.4

               C2H6 (g)         -84.68               -32.89            229.5

               CO (g)           -110.5               -137.2              197.9

               CO2 (g)         -393.5               -394.4              213.6

            Hydrogen               

               H2( g)                 0                        0                 130.58

            Oxygen                   

               O2 (g)                 0                        0                 205.0

               H2O (l)          -285.83             -237.13              69.91

                                                                                                     

The value of S° for the combustion of ethane to carbon dioxide and water vapor,

            2C2H6 (g) + 7 O2 (g) 4CO2 (g) + 6H2O (g)

is __________ J/K mol. The value of H for this reaction is__________ kJ/mol.

Explanation / Answer

2C2H6 (g) + 7 O2 (g) 4CO2 (g) + 6H2O (l)

S° for reaction = 2* entropy of C2H8(g) + 7 * entropy of O2(g) - 4* entropy of CO2(g) - 6 * entropy of H2O (g)

= 2* 229.5 + 7 * 205.0 - 4* 213.6 - 6* 69.91   = 620.14J/K -mol

H for this reaction = 2* H°f (kJ/mol)(C2H6(g) + 7 * H°f (kJ/mol)(O2(g)) - 4 * H°f (kJ/mol)(CO2(g) -                                                                                                                     - 6*    H°f (kJ/mol) (H2O)(g)

   = 2(-84.68) + 7(0) -4(-393.5) -6(-285.83) = 3119.62 kJ /mol

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