Answer:
[tex]\boxed{\text{39 }^{\circ}\text{C}}[/tex]
Explanation:
We can use the Arrhenius equation:
[tex]\ln(\frac{k_2 }{k_1}) = \frac{E_{a} }{R}(\frac{1}{T_1} - \frac{1 }{T_2 })[/tex]
Data:
Eā = 40.1 kJĀ·molā»Ā¹
kā = 0.0160 sā»Ā¹; Ā Ā Ā Ā Ā Ā Ā kā = 0.0320 sā»Ā¹
Tā = 26 Ā°C = 299.15 K; Tā = ? Ā
Calculations:
[tex]\ln(\frac{0.0320}{0.0160}) = \frac{40 100 }{8.314}(\frac{ 1}{299.15}} - \frac{1 }{T_{2} })\\\\\ln2 = 4823(\frac{ 1}{299.15}} - \frac{1}{T_{2} }) \\\\\ln2 - 16.12 = -\frac{4823}{T_{2} }\\\\-15.43 = -\frac{4823}{T_{2} }\\\\T_{2} = \frac{4823}{15.43} =\textbf{312.6 K}[/tex]
Tā = 312.6 K = 39 Ā°C
The reaction will go twice as fast at [tex]\boxed{\text{39 }^{\circ}\text{C}}[/tex].
