Respuesta :
Explanation:
The given data is as follows.
[tex]K_{1} = 2.57 M^{-1}s^{-1}[/tex] , [tex]T_{1}[/tex] = 701 K
[tex]K_{2}[/tex] = ?
[tex]T_{2}[/tex] = 525 K
[tex]E_{a} = 150 kJ/ mol \times \frac{1000 J}{1 kJ}[/tex]
= 150000 J / mol
Now, we will calculate rate constant as follows.
[tex]ln[\frac{K_{2}}{K_{1}}] = (\frac{E_{a}}{R}) \times [(\frac{1}{T_{1}}) - (\frac{1}{T_{2}})][/tex]
where, R = 8.314 J per mol K
Putting the given values into the above formula as follows.
[tex]ln[\frac{K_{2}}{K_{1}}] = (\frac{E_{a}}{R}) \times [(\frac{1}{T_{1}}) - (\frac{1}{T_{2}})][/tex]
[tex]ln [\frac{K_{2}}{2.57}] = \frac{150000 J/mol}{8.314 J/K mol} \times [(\frac{1}{701}) - (\frac{1}{525})][/tex]
[tex]ln [\frac{K_{2}}{2.57}][/tex] = -8.628
[tex]\frac{K_{2}}{2.57}[/tex] = anti ln[-8.628]
[tex]{K_{2}}{2.57}[/tex] = 0.000179
[tex]K_{2} = 0.000179 \times 2.57[/tex]
[tex]K_{2}[/tex] = 0.00046
or, = [tex]4.6 \times 10^{-4} L/mol s[/tex]
Thus, we can conclude that the rate constant at 525 K is [tex]4.6 \times 10^{-4} L/mol s[/tex].
