Answer:
Rate constant at 725 K is [tex]5.2\times 10^{-4}s^{-1}[/tex]
Explanation:
According to Arrhenius equation for a reaction-
[tex]ln(\frac{k_{2}}{k_{1}})=\frac{E_{a}}{R}(\frac{1}{T_{1}}-\frac{1}{T_{2}})[/tex]
where [tex]k_{2}[/tex] and [tex]k_{1}[/tex] are rate constants of reaction at [tex]T_{2}[/tex] and [tex]T_{1}[/tex] temperatures (in kelvin) respectively.
[tex]E_{a}[/tex] is activation energy of reaction.
Here [tex]T_{1}[/tex]= 600 K , [tex]k_{1}[/tex]= [tex]6.1\times 10^{-8}s^{-1}[/tex]
[tex]T_{2}[/tex]= 725 K, [tex]E_{a}[/tex]= 262 kJ/mol and R = 8.314 J/(mol.K)
So plugin all the values in the above equation-
[tex]ln(\frac{k_{2}}{6.1\times 10^{-8}s^{-1}})=\frac{262\times 10^{3}J/mol}{8.314J/(mol.K)}\times (\frac{1}{600K}-\frac{1}{725K})[/tex]
So, [tex]k_{2}[/tex] = [tex]5.2\times 10^{-4}s^{-1}[/tex]
Hence rate constant at 725 K is [tex]5.2\times 10^{-4}s^{-1}[/tex]