Respuesta :
Answer:
Ea=5.29 × 10⁴ J/mol
Explanation:
In going from 25 °C (298 K) to 35 °C (308 K), the rate of the reaction doubles. Since the rate of the reaction depends on the rate constant (k), this implies that the rate constant doubles. We can find the activation energy (Ea) using the two-point form of the Arrhenius equation.
[tex]ln\frac{k_{2}}{k_{1}} =\frac{-Ea}{R} .(\frac{1}{T_{2}}-\frac{1}{T_{1}})\\ln\frac{2k_{1}}{k_{1}}=\frac{-Ea}{8.314J/K.mol}.(\frac{1}{308K}-\frac{1}{298K} )\\Ea=5.29 \times 10^{4} J/mol[/tex]
The activation energy for the reaction has been 5.29 [tex]\rm \times\;10^4[/tex] J/mol.
The temperature of the reaction has been raised by 10 [tex]\rm ^\circ C[/tex], the rate of reaction doubles.
Let the initial rate = x
Final rate = 2x.
According to the Arrhenius equation for a two-point reaction:
[tex]\rm In\dfrac{Final\;Rate}{Initial\;rate}\;=\;-\dfrac{Activation\;energy}{Rydberg\;constant}\;\times\;(\dfrac{1}{Final\;temperature}\;-\;\dfrac{1}{Initial\;temperature})[/tex]
InitIal temperature = 25 [tex]\rm ^\circ C[/tex] = 298 K
Final temperature = 35 [tex]\rm ^\circ C[/tex] = 308 K.
In x = [tex]\rm \dfrac{-Ea}{8.314}\;\times\;(\dfrac{1}{308}\;-\;\dfrac{1}{298})[/tex]
Ea = 5.29 [tex]\rm \times\;10^4[/tex] J/mol.
The activation energy for the reaction has been 5.29 [tex]\rm \times\;10^4[/tex] J/mol.
For more information about the activation energy, refer to the link:
https://brainly.com/question/11334504
