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# Biochemistry Questions

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1. Adenosine triphosphate can be hydrolyzed to give adenosine diphosphate:
ATP-4(aq) + H2O(l) ↔ ADP-3(aq) + H3O+(l) + HPO-24

Calculate the fraction of adenosine that exists in each form (ATP & ADP) at pH = 7 at 298 K and 310 K. Assume the buffer keeps the pH @ 7 no matter how much ATP is hydrolyzed. HPO4 concentration is at 0.006 M.

Solution: pH = 7

At 298 K, ΔGorxn = -RTlnKeq = -30.5 x 103 J/mol (standard value for ATP hydrolysis)

Hence -8.314 J/mol K x 298K x ln[ADP][HPO42-]/[ATP] = -30.5 x 103 J/mol (I've ignored H+ as you have indicated it as being a pure liquid and not aqueous)

Solving this leads to,

[ADP]/[ATP] = 36.6 x 106 (Let me know if you need the ratio when H+ is aqueous)

Now, ΔHorxn = -20 x 103 J/mol (standard value for ATP hydrolysis)

ln K310 = lnK298 - (ΔHorxn / R)(1/ 310 - 1/298)

K310 = 15.9 x 104

[ADP][HPO42-]/[ATP] = 15.9 x 104

[ADP]/[ATP] = 26 x 106

This makes sense as the reaction is exothermic, the equilibrium constant will decrease with increase in temperature. Hence the ratio of ADP to ATP should also decrease.

2. What is the total entropy change for mixing 855 mL of tea @ 85⃝C with 15 mL of lemon juice at 3⃝C? Be sure to consider everything that changes. Assum the pressure is constant and state any additional assumptions that you make.
Solution: I'm assuming that both tea and lemon juice are mostly water. You can assume lemon juice is mostly citric acid which will lead to different calculations. On mixing these solutions, I'm assuming the solutions come to equilibrium. When a hot body comes in contact with a colder body, heat will be transferred to the colder body. So assuming these are ideal solutions, we can calculate the heat transferred from tea to lemon juice and thus find the entropy as well. (If the temperature was the same for both solutions, we can use other equations to ...

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The solution assists with answering the given example biochemistry questions.

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