Answer:
[tex]t_{1/2}=375.5min[/tex]
Explanation:
Hello!
In this case, since this problem refers to two different temperatures, it is possible to compute the rate constant at 652 K given the half-life at such temperature:
[tex]k=\frac{ln(2)}{58.0 min}=0.0120min^{-1}[/tex]
Next, by using the T-variable version of the Arrhenius equation, we can compute the rate constant at 623 K:
[tex]ln(\frac{k_2}{k_1} )=-\frac{Ea}{R}(\frac{1}{T_2}-\frac{1}{T_1} ) \\\\ln(\frac{k_2}{k_1} )=-\frac{218000J/mol}{8.3145\frac{J}{mol*K}}(\frac{1}{623K}-\frac{1}{652K})\\\\ln(\frac{k_2}{k_1} )=-1.872\\\\k_2=0.0120min^{-1}exp(-1.872)\\\\k_2=0.00185min^{-1}[/tex]
Finally, the half-life at 623 K turns out to be:
[tex]t_{1/2}=\frac{ln(2)}{0.00185min^{-1}} \\\\t_{1/2}=375.5min[/tex]
Best regards!
