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Ohio State University

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Problem 106

When heated, the DNA double helix separates into two random coil single strands. When cooled, the random coils reform the double helix: double helix $\Longrightarrow 2$ random coils.
(a) What is the sign of $\Delta S$ for the forward process? Why?
(b) Energy must be added to break $\mathrm{H}$ bonds and overcome dispersion forces between the strands. What is the sign of $\Delta G$ for the forward process when $T \Delta S$ is smaller than $\Delta H ?$
(c) Write an expression for $T$ in terms of $\Delta H$ and $\Delta S$ when the reaction is at equilibrium. (This temperature is called the melting temperature of the nucleic acid.)

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## Discussion

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## Video Transcript

were given a generic reversible reaction to model. How adding heat to the double helix structure of DNA causes the hydrogen bonds that form that structure to break and therefore caused the two strands to form two different random coils. Report A. We want to determine whether the sign of the change in entropy, where the forward direction of this reaction is positive or negative, and we want to explain how we know. So remember that the total entropy of a given system be correlated to the amount of disorder or randomness in that system, so that an increase in randomness or disorder, is equal to an increase in entropy. If we examine the the reaction that were given, we see that the word random even appears on on the product side. And when we go from left side to the right side in the forward direction, we start with one molecule of DNA double helix, two strands bonded together to two separate strands. So we increase the number of molecules in therefore increase molecular complexity and randomness. And that means that to go to go in the forward direction for this reaction, the increase entropy and that's why the change is going to be positive. Since we increase entropy, we have a higher ah, higher value for the entropy on the products than on the reactant. In the change in entropy is the entropy of the products, minus the entropy of the reactant. And if we have a higher value for the entropy of the products than the reactant stand, when we subtract them, Delta s comes out to be positive. And so that's the explanation as to how we know that the change in entropy Delta s for this. This generic reaction will be positive. And part B. We want to know, based on the equation, Delta G equals Delta H minus Tito s. We're told that this product of the temperature in the change in entropy is is smaller than the value for Delta H. And so because of that, we say we can plug in, we can plug in some numbers in order to qualitatively see whether Delta G will be positive or negative. We know that Delta H is greater than t Delta s. We're subtracting t Delta s from Delta H. So if Delta H were to equal to and Tito s would equal one. When we subtract them, we get a positive value of one. So since we know that Delta H is greater than T Delta s for any values we plug in for those the result meaning Delta G will always be positive, meaning that react, that reaction will always be non spontaneous under those conditions. So that comes out to a positive value. We're Delta G, meaning it's greater than zero. Finally, for part C, we want to arriving expression for the temperature. He in terms of Delta H and Delta s. When this reaction is at equilibrium member that when when a system is an equilibrium, a Delta G equals zero. When that's the case, we can get rid of this term Delta G instead of equal to zero and move over the product of Tea Delta s to the other side so that we can read out this equation now, as Delta H equals T Delta s. And now we just need to isolate T by dividing over Delta s and our final equation for he at equilibrium is still to each divided by Delta s