Respuesta :
Answer:
986.9 °C
Explanation:
To find the temperature we can use the Arrhenius equation:
[tex] k = Ae^{(-E_{a}/RT)} [/tex] (1)
Where:
k: is the reaction rate coefficient
A: is the pre-exponential factor
Ea: is the activation energy
R: is the gas constant
T: is the absolute temperature
With the values given:
T = 590 °C = 863 K; Ea = 117 kJ/mol; k = 0.00289 s⁻¹,
we can find the pre-exponential factor, A:
[tex]A = \frac{k}{e^{(-E_{a}/RT)}} = \frac{0.00289 s^{-1}}{e^{(-117 \cdot 10^{3} J*mol^{-1}/(8.314 J*K^{-1}*mol^{-1}*863 K))}} = 3.49 \cdot 10^{4} s^{-1}[/tex]
Now, we can find the temperature by solving equation (1) for T:
[tex] ln(\frac{k}{A}) = -\frac{E_{a}}{RT} [/tex]
[tex] T = -\frac{E_{a}}{R*ln(\frac{k}{A})} = -\frac{117 \cdot 10^{3} J*mol^{-1}}{8.314 J*K^{-1}*mol^{-1}*ln(\frac{0.492 s^{-1}}{3.49 \cdot 10^{4} s^{-1}})} = 1259.9 K = 986.9 ^{\circ} C [/tex]
Therefore, at 986.9 °C, the rate constant will be 0.492 s⁻¹.
I hope it helps you!
Answer:
The temperature is 1259.4 K (986.4°C)
Explanation:
Step 1: Data given
The activation energy Ea, = 117 kJ/mol
the rate constant for the reaction is 0.00289 s −1 at 590 °C
The new rate constant = 0.492 s −1
Step 2: Calculate the temperature
ln(kX / k 590°C) = Ea/R * (1/T1 - 1/T2)
⇒with kX is the rate constant at the new temperature = 0.492 /s
⇒with k 590 °C = the rate constant at 590 °C = 0.00289 /s
⇒with Ea = the activation energy = 117000 J/mol
⇒with R = the gas constant = 8.314 J/mol*K
⇒with T1 = 590 °C = 863 K
⇒ with T2 = the new temperature
ln (0.492 / 0.00289) = 117000/8.314 *(1/863 - 1/T2)
5.137 =14072.6 * (1/863 - 1/T2)
3.65*10^-4 = (1/863 - 1/T2)
3.65*10^-4 = 0.001159 - 1/T2
0,000794 = 1/T2
T2 = 1259.4 K
The temperature is 1259.4 K or 986.4 °C
