Question

the activation energy of a bimolecular elementary reaction was obtained by measuring the rate constant and two different temperatures. if the rate constant at 298k was 1.28x10⁻⁶ m⁻¹s⁻¹ and the rate constant at 425k was 7.32x10⁻⁸ m⁻¹s⁻¹, what was the energy of activation? (in Kj)

Answer 1

The activation energy (Ea) is calculated using the Arrhenius equation relating rate constants at two different temperatures. After substituting the given rate constants and temperatures into the rearranged equation, the calculated Ea is then converted from joules to kilojoules.

Step-by-step explanation:

To calculate the activation energy (Ea) for the bimolecular elementary reaction in question, we use the Arrhenius equation in the form that relates two different temperatures (T1 = 298K and T2 = 425K) and the corresponding rate constants (k1 = 1.28x10^-6 M^-1s^-1 and k2 = 7.32x10^-8 M^-1s^-1). The equation is:

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

where R is the universal gas constant (8.314 J/mol K). By rearranging the equation, we can solve for Ea:

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

Plugging in the given values and performing the calculation gives us

Ea = ln(1.28x10^-6 / 7.32x10^-8) * (8.314 J/mol K / (1/425K - 1/298K))

Converting the result from joules to kilojoules (1 kJ = 1000 J) provides the activation energy in kilojoules.