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Ch.19 - Chemical Thermodynamics
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All textbooksBrown 14th EditionCh.19 - Chemical ThermodynamicsProblem 99a

Chapter 19, Problem 99a

The oxidation of glucose (C6H12O6) in body tissue produces CO2 and H2O. In contrast, anaerobic decomposition, which occurs during fermentation, produces ethanol (C2H5OH) and CO2.

(a) Using data given in Appendix C, compare the equilibrium constants for the following reactions:

C6H12O6(s) + 6 O2(g) ⇌ 6 CO2(g) + 6 H2O(l)

C6H12O6(s) ⇌ 2 C2H5OH(l) + 2 CO2(g)

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Hello everyone. So in this video we're given two different reactions. The lactic acid fermentation of glucose and the anaerobic fermentation of glucose. So we're gonna go ahead and compare between the equilibrium constants. So first things first we need to go ahead and recognize that our delta G. Our delta G of the reaction is equal to the delta G of our product. So I just put P. R. O. D minus the delta G. Of our reactant. Now delta G. Is equal to negative R. Of T. Multiplied by L. N. K. So we're gonna go ahead and do some mathematical manipulation and solve for R. K. Term if we do so we get that K. Is equal to let's see E raised to the power of our of tea. So E raised to power Delta G over R. Of T. Okay, so and of course our tea at standard conditions is equal to 298 Kelvin's. So let's go ahead and proceed with the first reaction of the last electric acid fermentation. Alright, so we have again our delta G of reaction equaling times negative 711.62 units being killed, joules per mole minus negative 10.4. And these values I'm inputting is just given to us in the problem right over here. All right scrolling down again. So if we put these values into my calculator I'll get the value of negative 512.84 units being killing joules per mole Now for my key or my constant. We said that to be E to the negative delta G. Said that it's going to be negative 512.84. I want to convert this into jewels. How I can do this is simply Put times to the three because we just need to multiply that value the value by 1000 to give us jewels times A. C. R. Is r constant of negative 8.314 units being jewels per mole times kelvin. And then my T value is 2 98 kelvin's. So doing the race to this big old chunk here. Okay so of course I'm going to go ahead and use my calculator put in those values once I do so I get the value of 7.869 times 10 to the 89. Now doing it for our second reaction which is the anaerobic fermentation of glucose. Alright so we first do that. The delta G of reaction. Is it good to Two times negative 7 4.8 units being killed jules Permal. Now go ahead and add two times negative 9 4.4. Then we have the second delta G which is going to be of our reactant is 9 10. -9 10.4. So once I put all these values into my calculator I get the value of negative 2-8 kilo joules per mole. And of course we're gonna go ahead and solve our constant R. K value. So that's going to be E to the let's see again, we're gonna convert in the value into jewels. So do negative 2 to 8 times 10 to the third power jewels. And it's all over and see we have a constant of 8. units being jewels per mole times kelvin and then multiplied by R. T. Which is our temperature of 2 98 kelvin's. So putting all this into my calculator, then I get the K value being 9.2-5 times 10 to the 39. Now our equilibrium constant value for a lactic fermentation of glucose is way, way, way bigger than the value for the anaerobic fermentation of glucose. So we can say the K. Of the lactic is going to be greater than the K. Value for anaerobic. So then this right here is going to be my final answer for this problem. Thank you all so much for watching
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Textbook Question

Using the data in Appendix C and given the pressures listed, calculate Kp and ΔG for each of the following reactions: (c) N2H4(g) → N2(g) + 2 H2(g) PN2H4 = 0.5 atm, PN2 = 1.5 atm, PH2 = 2.5 atm

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(a) For each of the following reactions, predict the sign of ΔH° and ΔS° without doing any calculations. (i) 2 Mg(s) + O2 (g) ⇌ 2 MgO(s) (ii) 2 KI(s) ⇌ 2 K(g) + I2(g) (iii) Na2(g) ⇌ 2 Na(g) (iv) 2 V2O5(s) ⇌ 4 V(s) + 5 O2(g)

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(b) Based on your general chemical knowledge, predict which of these reactions will have K>1. (i) 2 Mg(s) + O2 (g) ⇌ 2 MgO(s) (ii) 2 KI(s) ⇌ 2 K(g) + I2(g) (iii) Na2(g) ⇌ 2 Na(g) (iv) 2 V2O5(s) ⇌ 4 V(s) + 5 O2(g)

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Textbook Question

The oxidation of glucose (C6H12O6) in body tissue produces CO2 and H2O. In contrast, anaerobic decomposition, which occurs during fermentation, produces ethanol (C2H5OH) and CO2.

(b) Compare the maximum work that can be obtained from these processes under standard conditions.

C6H12O6(s) + 6 O2(g) ⇌ 6 CO2(g) + 6 H2O(l)

C6H12O6(s) ⇌ 2 C2H5OH(l) + 2 CO2(g)

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Textbook Question

The conversion of natural gas, which is mostly methane, into products that contain two or more carbon atoms, such as ethane (C2H6), is a very important industrial chemical process. In principle, methane can be converted into ethane and hydrogen: 2 CH4(g) → C2H6(g) + H2(g) In practice, this reaction is carried out in the presence of oxygen: 2 CH4(g) + 12 O2(g) → C2H6(g) + H2O(g) (b) Is the difference in ΔG° for the two reactions due primarily to the enthalpy term (ΔH) or the entropy term (-TΔS)?

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Textbook Question

The conversion of natural gas, which is mostly methane, into products that contain two or more carbon atoms, such as ethane (C2H6), is a very important industrial chemical process. In principle, methane can be converted into ethane and hydrogen: 2 CH4(g) → C2H6(g) + H2(g) In practice, this reaction is carried out in the presence of oxygen: 2 CH4(g) + 1/2 O2(g) → C2H6(g) + H2O(g) (c) Explain how the preceding reactions are an example of driving a nonspontaneous reaction, as discussed in the 'Chemistry and Life' box in Section 19.7.

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