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:

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:

Then, we calculate the Gibbs free energy of reaction at both 25 °C and 1496 °C:

And finally, the equilibrium constants derived from the general Gibbs equation and Gibbs free energies of reaction:
![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}](https://tex.z-dn.net/?f=K%3Dexp%28-%5Cfrac%7B%5CDelta%20_rG%7D%7BRT%7D%20%29%5C%5C%5C%5CK_%7B25%5C%C2%B0C%7D%3Dexp%5B-%5Cfrac%7B172420%20J%2Fmol%7D%7B%288.3145%5Cfrac%7BJ%7D%7Bmol%2AK%7D%29%28298.15K%29%7D%20%5D%3D6.19x10%5E%7B-31%7D%5C%5C%5C%5CK_%7B1496%5C%C2%B0C%7D%3Dexp%5B-%5Cfrac%7B135650J%2Fmol%7D%7B%288.3145%5Cfrac%7BJ%7D%7Bmol%2AK%7D%29%281769K%29%7D%20%5D%3D9.87x10%5E%7B-5%7D)
Learn more:
Answer: The rate of the loss of
is 0.52M/s
Explanation:
Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.
The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.

Rate of disappearance of
=![-\frac{1d[O_3]}{2dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1d%5BO_3%5D%7D%7B2dt%7D)
Rate of formation of
=![+\frac{1d[O_2]}{3dt}](https://tex.z-dn.net/?f=%2B%5Cfrac%7B1d%5BO_2%5D%7D%7B3dt%7D)
![-\frac{1d[O_3]}{2dt}=+\frac{1d[O_2]}{3dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1d%5BO_3%5D%7D%7B2dt%7D%3D%2B%5Cfrac%7B1d%5BO_2%5D%7D%7B3dt%7D)
Rate of formation of
= 
Thus Rate of disappearance of
=![\frac{2d[O_2]}{3dt}=\frac{2}{3}\times 7.78\times 10^{-1}M/s=0.52M/s](https://tex.z-dn.net/?f=%5Cfrac%7B2d%5BO_2%5D%7D%7B3dt%7D%3D%5Cfrac%7B2%7D%7B3%7D%5Ctimes%207.78%5Ctimes%2010%5E%7B-1%7DM%2Fs%3D0.52M%2Fs)
V = maximum capacity of human lung = 3 liter = 3 x 0.001 m³ = 0.003 m³ (Since 1 liter = 0.001 m³)
P = pressure of oxygen = 21.1 kilo pascal = 21.1 x 1000 = 21100 Pa (since 1 kilo = 1000)
T = temperature of air = 295 K
n = number of moles of oxygen
Using the ideal gas equation
PV = n RT
inserting the above values in the equation
(21100) (0.003) = n (8.314) (295)
n = 0.026 moles