Answer:
The temperature is 39.1 ⁰C
Explanation:
Applying Arrhenius equation which relates rate constants to the temperature
[tex]ln(\frac{K_2}{K_1}) = \frac{E_a}{R}[\frac{1}{T_1}-\frac{1}{T_2}][/tex]
where;
Ea is the activation energy = 66 kJ
R is gas constant = 8.314 J/k.mol
T₁ is the initial gas temperature = 35.4°C = 308.4K
T₂ is the final gas temperature = ?
K₁ is rate constant at T₁ = 0.531
K₂ is rate constant at T₂ = 0.724
[tex]ln(\frac{K_2}{K_1}) = \frac{E_a}{R}[\frac{1}{T_1}-\frac{1}{T_2}]\\\\ln(\frac{0.724}{0.531}) = \frac{66,000}{8.314}[\frac{1}{308.4}-\frac{1}{T_2}]\\\\0.31 = 7938.417[\frac{1}{308.4}-\frac{1}{T_2}]\\\\3.905 X 10^{-5} = \frac{1}{308.4}-\frac{1}{T_2}\\\\\frac{1}{T_2} = \frac{1}{308.4}- 3.905 X 10^{-5}\\\\\frac{1}{T_2} = 0.003243 - 0.00003905\\\\\frac{1}{T_2} = 0.003204\\\\{T_2} = 312.1 K[/tex]
T₂(°C) = 312.1 -273 = 39.1 ⁰C
Therefore, the temperature at which the rate constant is 0.724 s⁻¹ is 39.1°C
