The first part of the lecture on thermodynamics and kinetics is now available. Part 1 reviews thermodynamics, redox and half-cell reactions, kinetics, acid-base reactions, buffers, and hydrolysis. This section should be an overview.
Questions: 1) Just a notation question, but for representing Joules does it matter if an upper case or lower case j is used? I've always used J but noticed that you use j--is there a difference? Same for k vs K with equilibrium constants.
2) Do you have some sample problems (like a review sheet) that would help us review and practice some of these basic thermodynamic concepts?
3) On slide 2-10 you add the half-cell potentials from the Fe/Zn example problem. You said you added them because one was negative so you got 0.77 + 0.76 = 1.53 V. Do you add the absolute value of the potentials, or do you really subtract them and subtracting the negative made it positive?
4) On slide 2-21 what role does the Na play? Is it just a spectator ion?
Joule is J, not j. Changes made in the presentation.
We will review thermodynamic problems in class. The homework at the end of the second lecture is a prime example of what you need to know. Some other examples are presented in the second part of the lecture.
For the Fe/Zn example, the half cell potentials provided are for reduction. Based on the data, Fe3+ reduction with half cell potential of 0.77 V is favored. Since Fe3+ will reduce, Zn will oxidize. As the potentials provided are for reduction, we need to use a value for oxidation; which is the opposite of the reduction potential. Since Zn will oxidize we need to change the sign of the reduction potential. This is why the value for Zn is 0.76 V. For the overall potential we sum the values; 0.77 + 0.76 V.
Correct, Na+ is just a spectator ion. Another cation beside H+ is need.
Is it because it has an unstable nucleus of either odd-even or odd-odd configuration, or is there something more that makes Tc different from other elements that have at least one stable isotope?
The short answer is stable isotopes with A=97 or 99, the expected stable Tc isotopes, have lower energies than Tc. Therefore Tc will decay to 97Mo or 99Ru. The question that needs to be explored is why is this so.
Lecture 2, part 1- completed - IA
ReplyDeletepart 1 complete
ReplyDeletethanks for the update. Let me know if you have any questions.
ReplyDeleteLecture II, part I complete -- JN
ReplyDeleteQuestions:
1) Just a notation question, but for representing Joules does it matter if an upper case or lower case j is used? I've always used J but noticed that you use j--is there a difference? Same for k vs K with equilibrium constants.
2) Do you have some sample problems (like a review sheet) that would help us review and practice some of these basic thermodynamic concepts?
3) On slide 2-10 you add the half-cell potentials from the Fe/Zn example problem. You said you added them because one was negative so you got 0.77 + 0.76 = 1.53 V. Do you add the absolute value of the potentials, or do you really subtract them and subtracting the negative made it positive?
4) On slide 2-21 what role does the Na play? Is it just a spectator ion?
Joule is J, not j. Changes made in the presentation.
ReplyDeleteWe will review thermodynamic problems in class. The homework at the end of the second lecture is a prime example of what you need to know. Some other examples are presented in the second part of the lecture.
For the Fe/Zn example, the half cell potentials provided are for reduction. Based on the data, Fe3+ reduction with half cell potential of 0.77 V is favored. Since Fe3+ will reduce, Zn will oxidize. As the potentials provided are for reduction, we need to use a value for oxidation; which is the opposite of the reduction potential. Since Zn will oxidize we need to change the sign of the reduction potential. This is why the value for Zn is 0.76 V. For the overall potential we sum the values; 0.77 + 0.76 V.
Correct, Na+ is just a spectator ion. Another cation beside H+ is need.
Please let me know if you have any questions.
JD: Lecture 2 Part 1 completed
ReplyDeletethanks for the comments.
ReplyDeleteWhy is Tc naturally radioactive?
ReplyDeleteIs it because it has an unstable nucleus of either odd-even or odd-odd configuration, or is there something more that makes Tc different from other elements that have at least one stable isotope?
--JN and JD
The short answer is stable isotopes with A=97 or 99, the expected stable Tc isotopes, have lower energies than Tc. Therefore Tc will decay to 97Mo or 99Ru. The question that needs to be explored is why is this so.
ReplyDelete