The following is a quote from an article in the August 18, 1998, issue of The Ne

Question

The following is a quote from an article in the August 18, 1998, issue of The New York Times about the breakdown of cellulose and starch: "A drop of 18 degrees Fahrenheit [from 77 ^circ { m F} to 59 ^circ { m F}] lowers the reaction rate six times; a 36-degree drop [from 77 ^circ { m F} to 41 ^circ { m F}] produces a fortyfold decrease in the rate." Assuming the value of E_a calculated from the 36-degree drop and assuming that the rate of breakdown is first order with a half-life at 25 ^circ { m C} of 2.7 years, calculate the half-life for breakdown at a temperature of -15 ^circ { m{C}}.

Explanation / Answer

A)
Half-life and rate constant for a 1st order order reaction are related as:
t½ = ln(2)/k

By changing the temperature the rate constant changes. From relation above one can find a relation of two half-lifes in terms of the rate constant ratio:
(t½) / (t½) = k/k
=>
(t½) = (t½) (k/k)

From Arrhenius equation it can be derived that the ratio of two rate constants is given by:
k/k = Ae^{ -Ea/(RT) } / Ae^{ -E/(RT) }
= e^{ (Ea/R) (1/T - 1/T) }

This relation can also be used to determine the activation energy.
ln( k/k ) = e^{ (Ea/R) (1/T - 1/T) }
<=>
Ea = Rln( k/k ) / (1/T - 1/T)

for the drop of 36°F
T = 77°F = 25°C = 298K
T = 41°F = 5°C = 278K
k/k = 40
=>
Ea = 0.0083145kJ/molK ln( 40 ) / (1/278K - 1/298K)
= 127 kJ/mol

By changing the temperature from T=25°C to T=-5°C the half-life changes to:
(t½) = (t½) (k/k)
= (t½) e^{ (Ea/R) (1/T - 1/T) }
= 2.7y e^{ (127 kJ/mol//0.0083145kJ/molK) (1/258K - 1/298K) }
= 2.7y e^{ 7.94968 }
= 7653.6 y

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