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Chapter 11: Problem 63

The uncoiling of deoxyribonucleic acid (DNA) is a first-order reaction. Its activation energy is \(420 \mathrm{~kJ}\). At \(37^{\circ} \mathrm{C}\), the rate constant is \(4.90 \times 10^{-4} \mathrm{~min}^{-1}\). (a) What is the half-life of the uncoiling at \(37^{\circ} \mathrm{C}\) (normal body temperature)? (b) What is the half-life of the uncoiling if the organism has a temperature of \(40^{\circ} \mathrm{C}\left(\approx 104^{\circ} \mathrm{F}\right)\) ? (c) By what factor does the rate of uncoiling increase (per \({ }^{\circ} \mathrm{C}\) ) over this temperature interval?

Short Answer

Expert verified
Answer: The half-life of the uncoiling of DNA at 37°C is approximately 1414 minutes, and at 40°C, it is approximately 985 minutes. The rate of uncoiling increases by a factor of approximately 1.44 per °C over this temperature interval.

Step by step solution

01

Calculate the half-life at 37°C

First, we have k1 (rate constant) = 4.90 × 10^(-4) min^(-1). We can simply use the half-life equation for first-order reactions to calculate the half-life at 37°C: t_1/2 = ln(2) / k1 t_1/2 = ln(2) / (4.90 × 10^(-4) min^(-1)) ≈ 1414 min (approximately) So, the half-life of the uncoiling at 37°C is approximately 1414 minutes.
02

Calculate the rate constant at 40°C

First, we need to use the Arrhenius equation. We will find a constant A and then use it to calculate k2 (rate constant) at 40°C. We are given Ea = 420 kJ/mol. k1 = Ae^(-Ea/(R × T1)) k1 = 4.90 × 10^(-4) min^(-1) T1 = 37°C + 273 = 310 K R = 8.314 J/(mol·K) Plugging these values into the equation: 4.90 × 10^(-4) = A × e^(-420000/(8.314 × 310)) Now, you can solve for A. A = 2.61 × 10^(13) min^(-1) Now, we need to calculate k2 at 40°C. T2 = 40°C + 273 = 313 K. k2 = Ae^(-Ea/(R × T2)) k2 = 2.61 × 10^(13) × e^(-420000/(8.314 × 313)) k2 ≈ 7.04 × 10^(-4) min^(-1) The rate constant at 40°C is approximately 7.04 × 10^(-4) min^(-1).
03

Calculate the half-life at 40°C

Now, we can calculate the half-life at 40°C, using the rate constant k2: t_1/2 = ln(2) / k2 t_1/2 = ln(2) / (7.04 × 10^(-4) min^(-1)) ≈ 985 min (approximately) So, the half-life of the uncoiling at 40°C is approximately 985 minutes.
04

Calculate the rate of uncoiling increase per °C

Next, we need to find the factor by which the rate of uncoiling increases over the temperature interval of 37°C to 40°C. We can find the rate constant ratio k2/k1: Rate increase factor = k2 / k1 = (7.04 × 10^(-4)) / (4.90 × 10^(-4)) ≈ 1.44 Therefore, the rate of uncoiling increases by a factor of approximately 1.44 per °C over this temperature interval.

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Most popular questions from this chapter

In a first-order reaction, suppose that a quantity \(X\) of a reactant is added at regular intervals of time, \(\Delta t\). At first the amount of reactant in the system builds up; eventually, however, it levels off at a saturation value given by the expression $$\text { saturation value }=\frac{X}{1-10^{-a}} \quad \text { where } a=0.30 \frac{\Delta t}{t_{1 / 2}}$$ This analysis applies to prescription drugs, of which you take a certain amount each day. Suppose that you take \(0.100 \mathrm{~g}\) of a drug three times a day and that the half-life for elimination is \(2.0\) days. Using this equation, calculate the mass of the drug in the body at saturation. Suppose further that side effects show up when \(0.500 \mathrm{~g}\) of the drug accumulates in the body. As a pharmacist, what is the maximum dosage you could assign to a patient for an 8 -h period without causing side effects?

The decomposition of \(\mathrm{Y}\) is a zero-order reaction. Its half-life at \(25^{\circ} \mathrm{C}\) and \(0.188 M\) is 315 minutes. (a) What is the rate constant for the decomposition of Y? (b) How long will it take to decompose a \(0.219 \mathrm{M}\) solution of \(\mathrm{Y}\) ? (c) What is the rate of the decomposition of \(0.188 \mathrm{M}\) at \(25^{\circ} \mathrm{C}\) ? (d) Does the rate change when the concentration of \(\mathrm{Y}\) is increased to \(0.289 \mathrm{M}\) ? If so, what is the new rate?

Iodine-131 is used to treat tumors in the thyroid. Its first-order half-life is \(8.1\) days. If a patient is given a sample containing \(5.00 \mathrm{mg}\) of \(\mathrm{I}-131\), how long will it take for \(25 \%\) of the isotope to remain in her system?

The decomposition of phosphine, \(\mathrm{PH}_{3}\), to \(\mathrm{P}_{4}(g)\) and \(\mathrm{H}_{2}(g)\) is firstorder. Its rate constant at a certain temperature is \(1.1 \mathrm{~min}^{-1}\). (a) What is its half-life in seconds? (b) What percentage of phosphine is decomposed after \(1.25 \mathrm{~min}\) ? (c) How long will it take to decompose one fifth of the phosphine?

Ammonia is produced by the reaction between nitrogen and hydro- gen gases. (a) Write a balanced equation using smallest whole-number coefficients for the reaction. (b) Write an expression for the rate of reaction in terms of \(\Delta\left[\mathrm{NH}_{3}\right]\). (c) The concentration of ammonia increases from \(0.257 \mathrm{M}\) to \(0.815 \mathrm{M}\) in \(15.0 \mathrm{~min} .\) Calculate the average rate of reaction over this time interval.
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