Question

1. Entropy (ΔS°), Enthalpy (ΔH°) and Gibb's Free Energy (ΔG°) follow Hess's law meaning their value for a reaction is the sum of the products minus the sums of the reactants. Consider the following Reaction at 25.0°C

Reaction	2Fe2O3(s)	+3C(s)	-->	4Fe(s)	+3CO2(g)
ΔH° (KJ/mol)	-824.2	0		0	-393.5
ΔS° (J/molK)	87.4	5.74		27.23	213.6

A. Calculate ΔG° where ΔG° = ΔH° - TΔS° for this reaction ___________________________

B. If 6 electrons are transferred in the balanced redox reaction above, Calculate E° for the reaction

C. Using the Nernst Equation (E = E° - (RT/nF) ln Q), what is the E value of the reaction if the partial pressure of CO₂ drops to 0.0346 atm?

D. Explain in complete sentences why this reaction is or is not spontaneous using both ΔG° , E° and E calculated above.

Answer 1

Calculations steps

A.

1 . Calculate H^o  by plugging in the given values of H⁰ in the expression -

_r  H^o   = ( sum of enthalpies of products ) - (sum of enthalpies of reactants )

..................= [ 4(0) + 3 (-393.5 ) ] - [ 2 ( - 824.2 ) + 3(0) ]

...................= 467.9 kJ / mol

2. Similarly -

_r S^o   = [ 3( 213.6 ) + 4 (27.23) ] - [ 87.4 (2) + 3 ( 5.74) ]

..................= 557.7 J /mol

............or = 0.5577 kJ / mol.

3. Substituting the values in expression

........................... G^o_r = H^o_r   - T S^o_r

.....we get ,.......................= 467.9 - 298 ( 0.5577 )

..........................................= 301. 7054 kJ

B. Calculate E^o for the reaction using equation ,............. G^o_r = -nF E^o  _(cell)

......................................................................or, ...........E^o_(cell)  = - G^o_r / n F

......................................................................................................= - 301.7054 / ( 6 x 96.487)

......................................................................................................= - 0.5211 V

( where , n is given as = 6 , F is Faraday's constant = 96.487 kJ per Volt gram equivalent , T = 25⁰ C or 298 K )

D. Thus ,

(i) we are getting positive values of G^o , H^o , S^o

(ii) Mathematically when G^o , H^o , S^o , have positive values reaction is nonspontaneous at low temperatures (ie . at 25 C or 298 K )

(iii) Again , since the calculated E^o   & E values are found to be negative the reaction is not

feasible / spontaneous.

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Please post the fourth part of this question ( C ) separately as fresh question for detailed answer.. Glad to help.

However , a hint may be useful -

Substitute the calculated value of E^o & partial pressure of CO₂ in the given Nernst equation .

E = - 0.0506 V