. The uncoiling of the DNA double helix is a first-order process with an activat
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Question
. The uncoiling of the DNA double helix is a first-order process with an activation energy of about 420 kJ/mol. Because of the rather large activation energy, the reaction rate is extremely sensitive to temperature. At 50°C, the uncoiling proceeds fairly quickly, with a half-life of 2.0 minutes. What is the half-life at normal body temperature (37°C)? HINT: Recall that in class we showed that for a first-order reaction the half-life is related to the rate constant by the following equation. t1/2 = ln 2 k
Explanation / Answer
Arrhenius Equation: k=A e-Ea/RT
k = rate constant
T= Absolute temperature in kelvin
A = Pre-exponential factor
Ea = Activation energy
R = Universal gas constant = 8.314 J/mol K
Solution to the problem:
Ea = 420 kJ/mol (Ea is the activation energy)
t1/2 at 50oC is 2 minutes.
Conversion of temperature from celsius to kelvin
T1 = 273+50 = 323 K
T2 = 273+37 = 310 K
At 50oC temperature, First order t1/2 = ln2/k
2 min = ln2/k (ln2 = 0.693)
2min = 0.693/k
k = 0.693/2 min
K = 0.347 min-1
ln [k2/k1] = [Ea/R] [1/T1 - 1/T2]
As activation energy is in kJ/mol gas constant value is written as 8.314*10-3 (conversion of joule to kilojoule).
8.314 J = 0.008314 kJ and can be written as 8.314*10-3
ln [k37/0.347] = [420/8.314*10-3] [1/323 - 1/310]
= [420/0.008314][1/323-1/310]
k37 = 4.92 * 10-4 min-1
t1/2 = ln2/k
ln2 = 0.693
t1/2 = ln2/4.92*10-4
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