Question

I need someone to show me how to solve thus problem, I think I have the right idea, but I may be plugging in something wrong.

What is ΔS° at 298 K for the following reaction?

Fe2O3(s)+ 3CO(g)→ 3CO2(g) + 2Fe(s)

Substance ΔG°f(kJ/mol) ΔH°f(kJ/mol)
Fe2O3(s). -741.0 -822.2

CO(g) -137.2 -110.5

CO2(g) -394.4 -393.5

Thanks

Answer 1

Consider reaction, Fe₂O₃ (s)+ 3 CO (g)$\rightarrow$  3 CO₂ (g) + 2 Fe (s)

The standard enthalpy change of reaction is given as , r H ⁰ = H ⁰_f ( products ) -  H ⁰_f ( reactants )

H ⁰_f ( products ) = 3 H ⁰_f CO₂ (g) + 2 H ⁰_f Fe (s)

H ⁰_f ( products ) = [ 3( - 393.5 ) + 2 (0) ] k J / mol = - 1180.5 k J / mol

H ⁰_f ( reactants ) = H ⁰_f Fe₂O₃ (s) + 3 H ⁰_f CO (g)

H ⁰_f ( reactants ) = [ ( -822.2 ) + 3 ( - 110.5 ) ] k J / mol = - 1153.7 k J / mol

r H ⁰ = ( - 1180.5 k J / mol ) - ( - 1153.7 k J / mol ) = - 26.8 k J / mol

For above reaction, G ⁰ reaction = G ⁰ (products) - G ⁰ (reactants)

G ⁰ (products) = 3 f G ⁰ CO₂ (g) + 2 f G ⁰ Fe (s)

G ⁰ (products) = [ 3 ( - 394.4 ) + 2 ( 0) ] k J / mol = - 1183.2 k J / mol

G ⁰ (reactants) = f G ⁰ Fe₂O₃ (s) + 3 f G ⁰ CO (g)

G ⁰ (reactants) = [ ( -741.0 ) + 3 ( - 137.2 ) ] k J / mol = - 1152.6 k J / mol

G ⁰ reaction = ( - 1183.2 k J / mol) - ( - 1152.6 k J / mol ) = -30.6 k J / mol

We have relation, G ⁰ reaction = r H ⁰ - T r S ⁰

We have, G ⁰ reaction = - 30.6 k J / mol ,   r H ⁰ = - 26.8 k J / mol & T = 298 K

$\therefore$ - 30.6 k J / mol = - 26.8 k J / mol - 298 K r S ⁰

- 298 K r S ⁰ = - 30.6 k J / mol + 26.8 k J / mol

- 298 K r S ⁰ = - 3.8 k J / mol = - 3800 J / mol

r S ⁰ = - 3800 J / mol / ( - 298 K )

r S ⁰ = 12.75 J K ^-1 mol ^-1

ANSWER : Standard entropy change of reaction at 298 K is 12.75 J K ^-1 mol ^-1