Question

Based on the thermodynamic data below at 298 K, estimate the temperature, in CELSIUS, above which the following reaction would be non-spontaneous. FeO(s s) +Fe(s) +02(g)> Fe20,(s) ДН/ (kJ/mol) So (J/K-mol) FeO(s) -271.9 60.75 Fe(s) Fe203 (a) 618.1°C 27.15 205.0 89.96 0 -822.16 (b) 756.3°C (c) 1235°O (d) 2439°C (e) spontaneous at all temperatures

Answer 1

The basic equation to do this question is, ΔG⁰ =ΔH_rxn – T ΔS

Where,

ΔG⁰   - free energy

ΔH_rxn – enthalpy change in the reaction

T – Temperature in kelvin

ΔS – change in entropy

In order to proceed with the above reaction we have to find the ΔH_rxn

ΔH_rxn = ΔH_f of products - ΔH_f of reactants

From the table given.

ΔH_rxn = ( -822.16) – (0+ 0+(-271.9)) KJ/mol

= -542.26 KJ

ΔS= ΔS( products ) – ΔS(reactents)

   =(89.96) –(205+27.15+60.75) = -202.94 JK/ mol.

Now as we got all the unknowns , take the assumption that at the point of equilibrium ΔG⁰ =0

Then the equation will be, 0 = ΔH_rxn – T ΔS

So , =ΔH_rxn =T ΔS

Now find the temperature, using T = ΔH_rxn/ ΔS

T =(-542.26 * 1000 J/mol) /(-202.94 JK/ mol.) =2672 K

In Celsius scale 2672 -273 =2399 ⁰C

Now look at the reaction again,

ΔG⁰ =ΔH_rxn – T ΔS

As the ΔH_rxn and ΔS are negative the equation is like,

ΔG⁰ = T ΔS - ΔH_rxn.

Thus above all other temperature except temperature at equilibrium the reaction is non spontaneous.

The choice above 2399 ⁰C is option

c) 2439⁰C is the correct answer