Respuesta :
Explanation:
It is given that energy to transfer one water molecule is [tex]2.208 \times 10^{-20}[/tex] J/molecule
As it is known that in 1 mole there are [tex]6.022 \times 10^{23}[/tex] atoms.
So, energy in 1 mole = [tex]2.208 \times 10^{-20} \times 6.022 \times 10^{23}[/tex] J/mol
= 13.3 kJ/mol
As, [tex]log \frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303 \times 8.314 atm L/mol K} \times \frac{T_{2} - T_{1}}{T_{1}T_{2}}[/tex]
Putting the given values in the above formula as follows.
[tex]log \frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303 \times 8.314 atm L/mol K} \times \frac{T_{2} - T_{1}}{T_{1}T_{2}}[/tex]
[tex]log \frac{k_{2}}{k_{1}} = \frac{13.3 kJ/mol}{2.303 \times 8.314 atm L/mol K} \times \frac{293 K - 283 K}{283K \times 293 K}[/tex]
[tex]log \frac{k_{2}}{k_{1}}[/tex] = 0.08377
[tex]\frac{k_{2}}{k_{1}}[/tex] = 1.213 = [tex]\frac{Concentration_{2}}{Concentration_{1}}[/tex]
[tex]Concentration_{2}[/tex] = [tex]1.213 \times Concentration_{1}[/tex]
= [tex]1.213 \times 9 \times 10^{-4}[/tex]
= [tex]10.915 \times 10^{-4}[/tex] water molecules per oil molecule
Thus, we can conclude that the equilibrium concentration at 293 K, assuming that nothing else changes is [tex]10.915 \times 10^{-4}[/tex].
