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The standard Gibbs energy change value (...

The standard Gibbs energy change value `(Delta_(r)G^(Theta))` at `1773K` are given for the following reactions:
`4Fe +3O_(2) rarr 2Fe_(2)O_(3), Delta_(r)G^(Theta) = - 1487 kJ mol^(-1)`
`4AI +3O_(2) rarr 2AI_(2)O_(3),Delta_(r)G^(Theta) =- 22500 kJ mol^(-1)`
`2CO +O_(2) rarr 2CO_(2),Delta_(r)G^(Theta) =- 515 kJ mol^(-1)`
Find out the possibility of reducing `Fe_(2)O_(3)` and `AI_(2)O_(3)` with `CO` at this temperature.

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To determine the possibility of reducing `Fe2O3` and `Al2O3` with `CO` at `1773 K`, we need to analyze the Gibbs free energy change (ΔG°) for the respective reduction reactions. A reaction is feasible if ΔG° is negative. ### Step-by-Step Solution: 1. **Write the Reduction Reactions:** - For `Fe2O3`: \[ \text{Fe}_2\text{O}_3 (s) + 3 \text{CO} (g) \rightarrow 2 \text{Fe} (s) + 3 \text{CO}_2 (g) ...
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On the basic of the following Delta_(r)G^(Theta) values at 1073K : S_(1)(s) +2O_(2)(g) rarr 2SO_(2)(g) Delta_(r)G^(Theta) =- 544 kJ mol^(-1) 2Zn(s) +O_(2)(g) rarr 2ZnO(s) Delta_(r)G^(Theta) =- 480 kJ mol^(-1) 2Zn(s) +S_(2)(s) rarr 2ZnS(s) Delta_(r)G^(Theta) =- 293 KJ mol^(-1) Show that roasting of zinc sulphide to zinc oxide is a spontaneous process.

On the basic of the following Delta_(r)G^(Theta) values at 1073K : S_(1)(s) +2O_(2)(g) rarr 2SO_(2)(g) Delta_(r)G^(Theta) =- 544 kJ mol^(-1) 2Zn(s) +O_(2)(g) rarr 2ZnO(s) Delta_(r)G^(Theta) =- 480 kJ mol^(-1) 2Zn(s) +S_(2)(s) rarr 2ZnS(s) Delta_(r)G^(Theta) =- 293 KJ mol^(-1) Show that roasting of zinc sulphide to zince oxide is a spontaneous process.

Calculated the equilibrium constant for the following reaction at 298K : 2H_(2)O(l) rarr 2H_(2)(g) +O_(2)(g) Delta_(f)G^(Theta) (H_(2)O) =- 237.2 kJ mol^(-1),R = 8.314 J mol^(-1) K^(-1)

Calculate the standard internal energy change for the following reactions at 25^(@)C : 2H_(2)O_(2)(l) rarr 2H_(2)O(l) +O_(2)(g) Delta_(f)H^(Theta) at 25^(@)C for H_(2)O_(2)(l) =- 188 kJ mol^(-1) H_(2)O(l) =- 286 kJ mol^(-1)

Calculate the standard Gibbs free energy change from the free energies of formation data for the following reaction: C_(6)H_(6)(l) +(15)/(2)O_(2)(g) rarr 6CO_(2)(g) +3H_(2)O(g) Given that Delta_(f)G^(Theta) =[C_(6)H_(6)(l)] = 172.8 kJ mol^(-1) Delta_(f)G^(Theta)[CO_(2)(g)] =- 394.4 kJ mol^(-1) Delta_(f)G^(Theta) [H_(2)O(g)] =- 228.6 kJ mol^(-1)

Consider the reaction: 4NH_(3)(g) +5O_(2)(g) rarr 4NO(g) +6H_(2)O(l) DeltaG^(Theta) =- 1010.5 kJ Calculate Delta_(f)G^(Theta) [NO(g)] if Delta_(f)G^(Theta) (NH_(3)) = -16.6 kJ mol^(-1) and Delta_(f)G^(Theta) [H_(2)O(l)] =- 237.2 kJ mol^(-1) .

Comment on the thermodynamic stability of NO(g) , given 1/2N_(2)(g)+1/2O_(2)(g)rarr NO(g), Delta_(r)H^(Θ)=90 kJ mol^(-1) NO(g)+1/2O_(2)(g) rarr NO_(2)(g), Delta_(r)H^(Θ)=-74 kJ mol^(-1)

Calculate in kJ for the following reaction : C(g) + O_(2)(g) rarr CO_(2)(g) Given that, H_(2)O(g) + C(g) + H_(2)(g) , Delta H = +131 kJ CO(g) + 1/2 O_(2)(g) rarr CO_(2)(g), " " Delta H = -242 kJ H_(2)(g) + 1/2 O_(2)(g) rarr H_(2)O(g), " "DeltaH = -242 kJ

Calculate the change in entropy for the following reaction 2CO(g) +O_(2)(g) rarr 2CO_(2)(g) Given: S_(CO)^(Theta)(g)=197.6 J K^(-1)mol^(-1) S_(O_(2))^(Theta)(g)=205.03 J K^(-1)mol^(-1) S_(CO_(2))^(Theta)(g)=213.6 J K^(-1)mol^(-1)

The standard Gibbs free energies for the reaction at 1773K are given below: C(s) +O_(2)(g) rarr CO_(2)(g), DeltaG^(Theta) =- 380 kJ mol^(-1) 2C(s) +O_(2)(g) hArr 2CO(g),DeltaG^(Theta) =- 500 kJ mol^(-1) Discuss the possibility of reducing Al_(2)O_(3) and PbO with carbon at this temperature, 4Al +3O_(2)(g) rarr 2Al_(2)O_(3)(s),DeltaG^(Theta) =- 22500 kJ mol^(-1) 2Pb +O_(2)(g) rarr 2PbO(s),DeltaG^(Theta) =- 120 kJ mol^(-1)

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