This problem is asking for the equilibrium constant at two different temperatures by describing the chemical equilibrium between gaseous nitrogen, oxygen and nitrogen monoxide at 25 °C and 1496 °C as the room temperature and the typical temperature inside the cylinders of a car's engine respectively:
N₂(g) + O₂(g) ⇄ 2 NO(g)
Thus, the calculated equilibrium constants turned out to be 6.19x10⁻³¹ and 9.87x10⁻⁵ at the aforementioned temperatures, respectively, according to the following work:
There is a relationship between the Gibbs free energy, enthalpy and entropy of the reaction, which leads to the equilibrium constant as shown below:
[tex]\Delta _rG=\Delta _rH-T\Delta _rS\\\\\Delta _rG=-RT ln(K)[/tex]
Which means we can calculate the enthalpy and entropy of reaction and subsequently the Gibbs free energy and equilibrium constant. In such a way, we calculate these two as follows, according to the enthalpies of formation and standard entropies of N₂(g), O₂(g) and NO(g) since these are assumed constant along the temperature range:
[tex]\Delta _rH=2*90.25 kJ/mol - (0 kJ/mol+0 kJ/mol)=180.5kJ/mol\\\\\Delta _rS=2*(0.211 kJ/mol*K)-(0.192kJ/mol*K+0.205kJ/mol*K)=0.025kJ/mol*K[/tex]
Then, we calculate the Gibbs free energy of reaction at both 25 °C and 1496 °C:
[tex]\Delta _rG_{25\°C}=180.5-(25+298.15)*0.025=172.42kJ/mol\\\\\Delta _rG_{1496\°C}=180.5-(1496+298.15)*0.025=135.65kJ/mol[/tex]
And finally, the equilibrium constants derived from the general Gibbs equation and Gibbs free energies of reaction:
[tex]K=exp(-\frac{\Delta _rG}{RT} )\\\\K_{25\°C}=exp[-\frac{172420 J/mol}{(8.3145\frac{J}{mol*K})(298.15K)} ]=6.19x10^{-31}\\\\K_{1496\°C}=exp[-\frac{135650J/mol}{(8.3145\frac{J}{mol*K})(1769K)} ]=9.87x10^{-5}[/tex]
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