Answer: [tex]4.3\times 10^{-13}s^{-1}[/tex]
Explanation:
According to the Arrhenius equation,
[tex]K=A\times e^{\frac{-Ea}{RT}}[/tex]
or,
[tex]\log (\frac{K_2}{K_1})=\frac{Ea}{2.303\times R}[\frac{1}{T_1}-\frac{1}{T_2}][/tex]
where,
[tex]K_1[/tex] = rate constant at [tex]600.0K[/tex] = [tex]6.1\times 10^{-8}s^{-1}[/tex]
[tex]K_2[/tex] = rate constant at [tex]775.0[/tex] = [tex]?[/tex]
[tex]Ea[/tex] = activation energy for the reaction = 262 kJ/mol = 262000J/mol
R = gas constant = 8.314 J/mole.K
[tex]T_1[/tex] = initial temperature = [tex]600.0K[/tex]
[tex]T_2[/tex] = final temperature = [tex]775.0K[/tex]
Now put all the given values in this formula, we get
[tex]\log (\frac{6.1\times 10^{-8}}{K_2})=\frac{262000}{2.303\times 8.314J/mole.K}[\frac{1}{600.0K}-\frac{1}{775.0K}][/tex]
[tex]\log (\frac{6.1\times 10^{-8}s^}{K_2})=5.150[/tex]
[tex](\frac{6.1\times 10^{-8}}{K_2})=141253.8[/tex]
Therefore, the value of the rate constant at 775.0 K is [tex]4.3\times 10^{-13}s^{-1}[/tex]
