Answer :
Explanation:
Let us assume that the value of [tex]K_{r}[/tex] = [tex]2.82 \times 10^{5} \times e^({\frac{-9835}{T}}) m^{6}/mol^{2}s[/tex]
Also at 1500 K, [tex]K_{r}[/tex] = [tex]2.82 \times 10^{5} \times e^({\frac{-9835}{1500}}) m^{6}/mol^{2}s[/tex]
[tex]K_{r} = 400.613 m^{6}/mol^{2}s[/tex]
Relation between [tex]K_{p}[/tex] and [tex]K_{c}[/tex] is as follows.
[tex]K_{p} = K_{c}RT[/tex]
Putting the given values into the above formula as follows.
[tex]K_{p} = K_{c}RT[/tex]
[tex]0.003691 = K_{c} \times 8.314 \times 1500[/tex]
[tex]K_{c} = 2.9 \times 10^{-7}[/tex]
Also, [tex]K_{c} = \frac{K_{f}}{K_{r}}[/tex]
or, [tex]K_{f} = K_{c} \times K_{r}[/tex]
= [tex]2.9 \times 10^{-7} \times 400.613[/tex]
= [tex]1.1617 \times 10^{-4} m^{6}/mol^{2}s[/tex]
Thus, we can conclude that the value of [tex]K_{f}[/tex] is [tex]1.1617 \times 10^{-4} m^{6}/mol^{2}s[/tex].