Arrhenius Rate Equation
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Arrhenius Rate Equation

The activation energy for a process is 55,000 cal/mole. The rate of this process is known at 400oC.
What is the incremental temperature needed to double the rate?

The strings:

S7P5A52 (Change - Chemical Change)

The math:
Pj Problem of Interest is of type change (chemical change). The rate of the chemical reaction is being changed by changing the temperature.

Temperature is a primary influencer of the rate of a chemical reaction. Concentrations of reactants is also a primary influencer of chemical reactions. In biochemical systems, enzymes are important influencer of biochemical reactions.

In 1893 A.D. Svante Arrhenius proposed an equation (now known as Arrhenius equation) that relates the rate of a chemical reaction with temperature. Arrhenius was also a pioneer in modern electrolyte theory.

Arrhenius Rate Equation
φ = Ae (-Ea/kT) -------------(1)

where A is a constant, e is natural logarithm,
Ea is activation energy
k is Boltzman's constant = 3.30 x 10-24 (cal/oK)/atom.
T is temperature degrees Kelvin.
If the activation energy is expressed in cal/mole, k is replaced by the universal gas constant R = 1.987 (cal/oK)/mole.

Now, 400oC = 673oK
So, at 400oC, rate = Ae (-Ea/673R)
At temperature T, rate is doubled

So, 2Ae (-Ea/673R) = Ae (-Ea/RT)
Taking natural logarithm:
So, ln2 - Ea/673R = -Ea/RT
So, ln2 = (Ea/R)[1/673 - 1/T]
so, Rln2/Ea = [1/673 - 1/T]

So, (1.987 x 0.693)/55000 = 0.001486 - 1/T
So, 0.0000250 = 0.0014858 - 1/T
So, 1/T = 0.0014858 - 0.0000250 = 0.001461
So, T = 684.5oK

So, rate at 400oC will be doubled if temperature is raised by 11.5oC.

Blessed are they that have not seen, and yet have believed. John 20:29