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
