Answer: The rate constant at 296 K is [tex]1.51\times 10^{-3}min^{-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 266 K = [tex]1.35\times 10^{-5}min^{-1}[/tex]
[tex]K_2[/tex] = rate constant at 296 K = ?
[tex]Ea[/tex] = activation energy for the reaction = 103 kJ/mol = 103000 J/mol
R = gas constant = 8.314 J/mole.K
[tex]T_1[/tex] = initial temperature = 266 K
[tex]T_2[/tex] = final temperature = 296 K
Now put all the given values in this formula, we get
[tex]\log (\frac{K_2}{1.35\times 10^{-5}})=\frac{103000}{2.303\times 8.314J/mole.K}[\frac{1}{266}-\frac{1}{296}][/tex]
[tex]\log (\frac{K_2}{1.35\times 10^{-5}})=2.049[/tex]
[tex]K_2=1.51\times 10^{-3}min^{-1}[/tex]
Thus the rate constant at 296 K is [tex]1.51\times 10^{-3}min^{-1}[/tex]
