Question

For the following reaction:



3H₂SO₄(g) + 2Fe(OH)₃(s) → Fe₂(SO₄)₃(s) + 6H₂O(l)







1. Calculate ΔH°_rx (in kJ) for this reaction. Report your answer to two decimal places in standard notation (i.e. 123.45 kJ).





2. Calculate ΔS°_rx (in J/K) for this reaction. Report your answer to two decimal places in standard notation (i.e. 123.45 J/K).





3. Calculate ΔG°_rx (in kJ) at 511.71 K for this reaction. Report your answer to two decimal places in standard notation (i.e. 123.45 kJ).

Assume ΔH°_f and S° do not vary as a function of temperature.





4. Is this reaction spontaneous or not spontaneous at the temperature given in part 3?

Compound

ΔH°f (kJ mol-1)

S° (J mol-1 K-1)

H2SO4 (g)

-735.13

298.78

Fe(OH)3 (s)

-832.62

104.56

Fe2(SO4)3 (s)

-2583.00

307.46

H2O (l)

-285.83

69.91

Answer 1

1) H_f = H_f products - H_f reactants.

H_f =[ (-2583)+6(-285.83)] - [2(-832.62) +3(-735.13)]

Note: Hf value for individual molecule given in the question and will multiply with its mole number.

Hf = [-2528-1714.98] - [-1665.24-2205.39]

Hf = [-4242.98]-[-3870.63]

Hf = -4242.98+3870.63

Hf = -372.35 KJ mol^-1.

2) S = S_products - S _rectants

Note: Hf value for individual molecule given in the question and will multiply with its mole number.

₌[ (307.46)+6(69.91)] - [2(104.56) +3(298.78)]

=726.92 - 1105.46

= -378.54 Jmol^-1K^-1. (or) -0.37854 KJ mol^-1K^-1

3) G = H - T S

   = (-372.35) - 511.71(-0.37854)

   =-372.35 + 193.702

   = -178.648 KJ.

4)The reaction is spontaneous because the G = -Ve (negative value).