Question

Suppose a layer of oil is on the top of a beaker of water. Water in many oils is slightly soluble, so its concentration is so low that we can treat it as an ideal solute. 1) Suppose that at 283 K, the equilibrium concentration is 9 × 10-4 water molecules per oil molecule, and it takes 2.208 × 10-20 J to transfer one water molecule into the oil. What is the equilibrium concentration at 293 K, assuming that nothing else changes

Answer 1

Explanation:

It is given that energy to transfer one water molecule is $2.208 \times 10^{-20}$ J/molecule

As it is known that in 1 mole there are $6.022 \times 10^{23}$ atoms.

So, energy in 1 mole = $2.208 \times 10^{-20} \times 6.022 \times 10^{23}$ J/mol

                                  = 13.3 kJ/mol

As,    $log \frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303 \times 8.314 atm L/mol K} \times \frac{T_{2} - T_{1}}{T_{1}T_{2}}$

Putting the given values in the above formula as follows.

                $log \frac{k_{2}}{k_{1}} = \frac{E_{a}}{2.303 \times 8.314 atm L/mol K} \times \frac{T_{2} - T_{1}}{T_{1}T_{2}}$

               $log \frac{k_{2}}{k_{1}} = \frac{13.3 kJ/mol}{2.303 \times 8.314 atm L/mol K} \times \frac{293 K - 283 K}{283K \times 293 K}$

                  $log \frac{k_{2}}{k_{1}}$ = 0.08377

                       $\frac{k_{2}}{k_{1}}$ = 1.213 = $\frac{Concentration_{2}}{Concentration_{1}}$

                 $Concentration_{2}$ = $1.213 \times Concentration_{1}$

                                             = $1.213 \times 9 \times 10^{-4}$

                                             = $10.915 \times 10^{-4}$ water molecules per oil molecule

Thus, we can conclude that the equilibrium concentration at 293 K, assuming that nothing else changes is $10.915 \times 10^{-4}$.