Question

The specific reaction rate of a reaction at 492k is 2.46 second inverse and at 528k 47.5 second inverse.find activation energy and ko.

Answer 1

Answer:

Ea = 177x10³ J/mol

ko = $1.52x10^{19}$ J/mol

Explanation:

The specific reaction rate can be calculated by Arrhenius equation:

$k = koxe^{-Ea/RT}$

Where k0 is a constant, Ea is the activation energy, R is the gas constant, and T the temperature in Kelvin.

k depends on the temperature, so, we can divide the k of two different temperatures:

$\frac{k1}{k2} = \frac{koxe^{-Ea/RT1}}{koxe^{-Ea/RT2}}$

$\frac{k1}{k2} = e^{-Ea/RT1 + Ea/RT2}$

Applying natural logathim in both sides of the equations:

ln(k1/k2) = Ea/RT2 - Ea/RT1

ln(k1/k2) = (Ea/R)x(1/T2 - 1/T1)

R = 8.314 J/mol.K

ln(2.46/47.5) = (Ea/8.314)x(1/528 - 1/492)

ln(0.052) = (Ea/8.314)x(-1.38x$10^{-4}$

-1.67x$10^{-5}$xEa = -2.95

Ea = 177x10³ J/mol

To find ko, we just need to substitute Ea in one of the specific reaction rate equation:

$k1 = koxe^{-Ea/RT1}$

$2.46 = koxe^{-177x10^3/8.314x492}$

$1.61x10^{-19}ko = 2.46$

ko = $1.52x10^{19}$ J/mol