(28) (a) From a consideration of the following reactions, calculate H f ° for et

Question

(28)

(a) From a consideration of the following reactions, calculate Hf° for ethane, C2H6(g).

(1) CH3CHO(g) + 2H2(g) C2H6(g) + H2O(); H°(kJ) = -204

(2) 2H2(g) + O2(g) 2H2O(g); H°(kJ) = -484

(3) 2C2H5OH() + O2(g) 2CH3CHO(g) + 2H2O(); H°(kJ) = -348

(4) H2O() H2O(g); H° = 44

(5) 2C2H5OH() 4C(s) + 6H2(g) + O2(g); H° = 555

Hf° = _____kJ

(b) From appropriate entropy data, calculate S° for the formation of ethane from the elements.

2C(s) + 3H2(g) C2H6(g)

S° (C(s)) = 5.74 J/mol K
S° (H2(g)) = 130.6 J/mol K
S° (C2H6(g)) = 229.5 J/mol K

S° = _____J/K

(c) Calculate Gf° for ethane at 25 °C.

2C(s) + 3H2(g) C2H6(g)

Gf° = _____kJ/mol

Explanation / Answer

The calculations are as shown below

(a) We want,

2C(s) + 3H2(g) -----> C2H6(g)

To get this equation,

i. reverse 5 and divide by 2

ii. divide 3 by 2

iii. add 1

iv. multiply 4 by 2

v. reverse 2

Add, i to v,

we get,

delta Hf (C2H6) = -555/2 - 348/2 - 204 + 88 + 484 = -83.5 kJ

(b) delta So = sum of delta So for products - sum of delta So for reactants

                    = 229.5 - (2 x 5.74 + 3 x 130.6)

                    = -173.78 J/K

(c) delta Gfo = delta Gfo products - delta Gfo reactants

                     = -32.80 kJ/mol

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