Question

A particular reaction has an activation energy, Ea, of 117 kJ/mol. If the rate constant for the reaction is 0.00289 s −1 at 590 °C, at what temperature(in°C) would the rate constant be 0.492 s −1?

Answer 1

The temperature is 1259.4 K (986.4°C)

Step-by-step 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

Answer 2

986.9 °C

Step-by-step explanation:

To find the temperature we can use the Arrhenius equation:

`k = Ae^{(-E_(a)/RT)}` (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:

`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)`

Now, we can find the temperature by solving equation (1) for T:

`ln((k)/(A)) = -(E_(a))/(RT)`

`T = -(E_(a))/(R*ln((k)/(A))) = -(117 \cdot 10^(3) J*mol^(-1))/(8.314 J*K^(-1)*mol^(-1)*ln((0.492 s^(-1))/(3.49 \cdot 10^(4) s^(-1)))) = 1259.9 K = 986.9 ^(\circ) C`

Therefore, at 986.9 °C, the rate constant will be 0.492 s⁻¹.

I hope it helps you!