Question

A certain reaction has an activation energy of 39.5 kJ/mol. As the temperature is increased from 25.0°C to a higher temperature, the rate constant increases by a factor of 5.90. Calculate the higher temperature (in °C).

Answer 1

To calculate the higher temperature, the Arrhenius equation is used to express the relationship between rate constants at different temperatures, activation energy, and the ideal gas constant. After substituting known values and solving for the unknown temperature, we convert the result from Kelvin to Celsius.

Step-by-step explanation:

The question pertains to calculating the higher temperature at which the rate constant of a chemical reaction increases by a factor of 5.90, given the reaction's activation energy of 39.5 kJ/mol. To solve this, we employ the Arrhenius equation which relates the rate constant k to the activation energy (Ea), the gas constant (R), and the temperature (T).

First, we write down the Arrhenius equation for both temperatures:

• k1 = A * exp(-Ea/(R*T1))
• k2 = A * exp(-Ea/(R*T2))

Since k2 = 5.90 * k1, we can write:

5.90 * A * exp(-Ea/(R*T1)) = A * exp(-Ea/(R*T2))

Dividing both sides by A * exp(-Ea/(R*T1)), we get:

5.90 = exp(Ea/R) * (1/T1 - 1/T2)

Next, we take the natural logarithm of both sides and solve for T2:

ln(5.90) = (Ea/R) * (1/T1 - 1/T2)

By substituting the known values (Ea = 39.5 * 10^3 J/mol, R = 8.314 J/Kmol, and T1 = 25.0 + 273.15 K), we can calculate T2, then convert back to °C.

Answer 2

Step-by-step explanation:

We shall apply Arrhenius equation which is given below .

`ln(k_2)/(k_1) = (E_a)/(R) [(1)/(T_1) -(1)/(T_2) ]`

K₂ and K₁ are rate constant at temperature T₂ and T₁ , Ea is activation energy .

Putting the given values

`ln(5.9)/(1) = (39500)/(8.3) [(1)/(298) -(1)/(T_2) ]`

`.000373= [(1)/(298) -(1)/(T_2) ]`

T₂ = 335.27 K

= 62.27 °C