CHEM 201 Lecture Notes - Lecture 8: Pre-Exponential Factor, Rate Equation, Reaction Rate Constant

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7 Apr 2015

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How to speed up or slow down a process. Experimentally measure concentration [a] (or any quantity [a], V (cid:1) rt for a gas) as a function of time t. such as pressure pa = (cid:0)na. Fig k1 get equil value [a]eq at t = . De ne rate or velocity v of reaction (a +ve quantity) by slope at t rate v = d[a] dt could also measure [b] or [c]: note stoichiometry is important v = d[a] dt. In general, no way to predict rates of chemical reactions - must be determined experimentally. Many organic and inorganic reactions take minutes to hours. Experiment involves measuring concentration (or a related quantity) as function of time. For faster reactions: volume of gas, change in optical rotation, change in colour intensity (beer"s law, change in total p = pi pi at constant v for gas-phase reaction (must then relate p to p (one species, etc.

Related Questions

âPolymer beads 800 micrometer in diameter are used for timed released drug delivery. Initial uniform concentration of the drug in the beads is 5 micrograms/cm^3 and the diffusion coefficient for the drug through the polymer is 1x10^-8 cm^2/s. The beads are ingested and pass through the GI tract in your body in a period of 12 hours. Assume the concentration of the drug is zero in the GI tract.

a. Use divergence theorem to derive the partial differential equation that describes the drug concentration in a polymer sphere as a function of position and time.

b. What are the initial and boundary conditions for the equation in part a?

c. What is the drug concentration at the center of the bead after 4 hours?

d. How long does it take to release 50% of the drug?

e. What fraction of the drug is released during the 12-hour ingestion period?

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!

Please provide step by step solution.

ATP contains two high energy phosphates and one whose energy is not high. Many biochemical reactions require more free energy than hydrolysis of a single high energy phosphate can provide. In these cases the ATP hydrolysis removes both high energy phosphates as pyrophosphate, which is then hydrolyzed to make two ordinary inorganic phosphates, Pi. This harnesses double the single phosphate free energy. Pyrophosphate hydrolysis is according to the following reaction, which is catalyzed by the enzyme pyrophosphatase:

=O3PâOâPO3= + H2O + 2Pi. (Pi represents whatever forms of phosphate anion are present in the solution, e.g. H2PO4-, HPO4=, PO4-3 depending on the pH)

kcat = 257 per sec Km = 1.45 x 10-4 M

a) If the enzyme concentration were 7 nM, what would be the maximum hydrolysis rate? (You may use seconds rather than minutes as your time unit if you wish.)

b) If the pyrophosphate concentration were 1.07 x 10-6 M, what would be the reaction rate?

c) Aspartic acid 115 is part of the active site. When it is mutated to glutamic acid, kcat falls to 21 sec-1, and Km falls to 0.6 Î¼M. If the mutant enzyme is present at the same concentration as in (a) above, what percent of the wild type maximum rate would the mutant enzyme achieve?

d) If the substrate concentration is the same as in (b) above, what will be the reaction velocity of the mutant enzyme and what percentage of the activity at this substrate concentration of the wild type enzyme will this be?

CHEM 201 Lecture Notes - Lecture 8: Pre-Exponential Factor, Rate Equation, Reaction Rate Constant

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