Consider the following reaction:
2NO(g)+O2(g)→2NO2(g)
Estimate ΔG∘ for this reaction at each of the following temperatures and predict whether or not the reaction will be spontaneous. (Assume that ΔH∘ and ΔS∘ do not change too much within the give temperature range.) I need to find the temperature are 298K and 702K. For 298K It is simple because at standard temperature
ΔG∘ = DG(products)- DG(reactants).

Answer:

A. [tex]\mathbf{\Delta G^0 = -72.6 \ kJ/mol}[/tex] ; as such the reaction is said to be spontaneous since the value of [tex]\mathbf{\Delta G^0 }[/tex] is negative.

B.  [tex]\mathbf{\Delta G^0_{702 \ K} = -13.29 \ kJ/mol.K}}[/tex] and the reaction is spontaneous

Explanation:

The equation for this chemical reaction is :

[tex]2NO_{(g)} +O_{2(g)} \to 2NO_{2(g)}[/tex]

Using the following relation to calculate [tex]\Delta G^0[/tex];

[tex]\Delta G^0 = [2(\Delta G^0_{NO_{2(g)}}] - [1(\Delta G^0_{O_{2(g)}})+ 2(\Delta G^0_{NO_{g}})][/tex]

At 298 K; the standard Gibbs Free Energy for the formation are as follows:

[tex]\Delta G^0_{NO_{2(g)}} = 51.2 \ kJ/mol[/tex]

[tex]\Delta G^0_{O_{2(g)}} = 0[/tex]

[tex]\Delta G^0_{NO_{g}}= 87.6 \ kJ/mol[/tex]

Replacing them into the above equation;

[tex]\Delta G^0 = [2(51.2 \ kJ/mol}] - [1(0)+ 2(87.6 \ kJ/mol})][/tex]

[tex]\Delta G^0 = [102.4 \ kJ/mol}] - [175.2 \ kJ/mol})][/tex]

[tex]\mathbf{\Delta G^0 = -72.6 \ kJ/mol}[/tex]

Thus; [tex]\mathbf{\Delta G^0 = -72.6 \ kJ/mol}[/tex] ; as such the reaction is said to be spontaneous since the value of [tex]\mathbf{\Delta G^0 }[/tex] is negative.

B.

Using the same above chemical equation;

The relation used for calculating [tex]\mathbf{\Delta G^0}[/tex] of the reaction when the temperature is 702 K is:

[tex]\Delta G^0_{702 \ K} = \Delta H^0_{xn} - T \Delta S^0_{rxn}[/tex]

where;

[tex]\Delta G^0_{702 \ K} =[/tex] Gibbs free energy of the reaction at 702 K

[tex]\Delta H^0_{xn}[/tex] = standard enthalpy of the reaction = -116.2 kJ/mol

[tex]\Delta S^0_{rxn}[/tex] = standard entropy of the reaction = -146.6 J/mol/K

Temperature T = 702 K

[tex]\Delta G^0_{702 \ K} = -1162. \ kJ/mol - 702 \ K ( -146.6 \ J/mol. K (\dfrac{1 \ kJ }{1000 \ J})[/tex]

[tex]\Delta G^0_{702 \ K} = -1162. \ kJ/mol - 702 \ K ( 0.1466 \ kJ/mol.K})[/tex]

[tex]\Delta G^0_{702 \ K} = -13.2868 \ kJ/mol.K}[/tex]

[tex]\mathbf{\Delta G^0_{702 \ K} = -13.29 \ kJ/mol.K}}[/tex]

Thus [tex]\mathbf{\Delta G^0_{702 \ K} = -13.29 \ kJ/mol.K}}[/tex] and the reaction is spontaneous

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