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CHMA11H3 Lecture Notes - Lecture 5: Gas Constant, Gibbs Free Energy, Reaction Quotient

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greywoodchuck686
22 Oct 2017
School
UTSC
Department
Chemistry
Course
CHMA11H3
Professor
Nirusha Thavarajah

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CHMA11H3 Full Course Notes
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CHMA11H3 Full Course Notes
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Document Summary

[(cid:2157)](cid:2183)[(cid:2158)](cid:2184: r is universal gas constant (8. 314 j/k*mol, t is temperature in k, for gasses, pressure in terms of bar, for solutes, in terms of m, for reactions with solids, concentration is not included in expressions for. Q or k: for reactions in aqueous, molar concentration of water is not included in, reaction quotient is comparison of the progress of a reaction to the equilibrium state of expressions for q or k the reaction. If q < k, reaction will proceed in forward direction. If q > k, reaction will proceed in reverse direction. If q = k, there is no change, reaction is at equilibrium.

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Related Questions

4. a. Given that the pKa for phenol is 10.0, estimate the pH values at which 1/3 and 2/3 of the phenol present in a 1 mM aqueous solution is deprotonated.

b. What is the value for the free energy change at the biochemists’ standard state ( ∆G°′ ) for the dissociation reaction of phenol? Remember that the biochemists’ standard state just means to use pH=7 instead of assuming 1 M H+.

The answers are below: Can someone explain this out in more detail, and tell me where the numbers are coming from?

a. When 1/3 of the phenol is protonated, then [phenol]/[phenolate] = 1⁄2 . Given that the pKa is 10.0, Ka =[phenolate][H+]/[phenol], so [H+] = Ka [phenol]/[phenolate] = 1⁄2 Ka = 5×10-11.

Therefore pH = -log [H+] = -log (5×10-11) ≠10.3
When 2/3 of the phenol is protonated, then [phenol]/[phenolate] = 2.

[H+] = Ka [phenol]/[phenolate] = 2 Ka = 2×10-10. Therefore pH = -log [H+] = -log (2×10-10) ≠9.7

Ka = 10-10.

b. ∆G°′ = ∆G° + RT ln Q where Q represents the reaction quotient with non-standard concentrations. ∆G°=-RTlnK. So∆G°′=-RTlnK+RTlnQ=RTln(Q/K).
For the dissociation reaction of phenol, Q = [phenolate][H+]/[phenol].
At the biochemists’ standard state of pH=7, [H+] = 1×10-7 while the other species still have a 1 M standard state.

So Q = 1×10-7 .

Therefore ∆G°′ = RT ln(Q/K) = RT ln(1×10-7 /1×10-10) = RT ln 1000 = (8.314 J mol-1 K-1)(298 K) (6.91) ∆G°′ = +17,100 J mol-1 = +17.1 kJ mol-1

The positive value for ∆G°′ means that the dissociation reaction is not favored at pH = 7. In fact, phenol is barely (about 1/1000, or 0.1%) dissociated at that pH.

Hi, So I collected the data and just need answers for these questions. not sure where to start and don't understand them. Please, answer all or don't answer. Thanks in advance and I appreicate it. I'll rate it if you do answer all, and I appreciate your help!Conclusions:

1. Create a data table that displays the voltage of each trial (Ecell) and the temperature of each trial. Then, calculate and display as a calculated column ?G for the cell at each of the temperatures, using ?G = -n F E.

2. Calculate a ?G0 value for each temperature, given that ?G = ?G0 + RT ? ln Q, and include a calculated column for this in your data table.

3. Calculate and include a column for an E0 value for 25 0C, given your E0 value and the relationship ?G0 = -nFE0 . Using the standard reduction potential table, calculate an accepted E0 value at 250C. Then, calculate your percentage of error. Offer explanation for why these values are different!

4. Determine the equilibrium constant K for this reaction from your value of ?G0 at 250C, given that ?G0 = - RT ? ln K. Include a calculated column in your data table for K. Assume 250C for the standard free energy in order to determine the equilibrium constant.

5. What is the relationship between temperature and cell potential, E?

6. What is the relationship between cell potential, E, and ?G? Explain why this makes sense!

7. Define the Faraday constant, F.

8. Are most oxidation/reduction reactions exothermic, endothermic, or both, do you think? Explain why you think this?

9. Should your reaction be exothermic or endothermic? Explain in terms of your calculated energy and/or voltage values!

10. From your value of ?G0 at 250C, and an accepted value of ?S0 , calculate the ?H0 for this reaction at standard temperature conditions. Is this consistent with your answer to number 9? Explain!