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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 400^oC.
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 ^(-E_a/kT) -------------(1)

where A is a constant, e is natural logarithm,
E_a 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, 400^oC = 673^oK
So, at 400^oC, rate = Ae ^(-E_a/673R)
At temperature T, rate is doubled

So, 2Ae ^(-E_a/673R) = Ae ^(-E_a/RT)
Taking natural logarithm:
So, ln2 - E_a/673R = -E_a/RT
So, ln2 = (E_a/R)[1/673 - 1/T]
so, Rln2/E_a = [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 = 684.5^oK

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

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