The oxidation of glucose (C 6 H 12 O 6 ) in body tissue produces CO 2 and H 2 O. In contrast, anaerobic decomposition, which occurs during fermentation, produces ethanol (C 2 H 5 OH) and CO2. (a) Using data given in Appendix C, compare the equilibrium constants for the following reactions: C 6 H 12 O 6 (s) + 6 O 2 (g) ⇌ 6 CO 2 (g) + 6 H 2 O(l) C 6 H 12 O 6 (s) ⇌ 2 C 2 H 5 OH(l) + 2 CO 2 (g)

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

Ch.19 - Chemical Thermodynamics

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Brown 14th Edition

Ch.19 - Chemical Thermodynamics

Problem 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)

Verified step by step guidance

Identify the reactions for which you need to compare the equilibrium constants: the oxidation of glucose and the fermentation of glucose.

Use the standard Gibbs free energy change (\( \Delta G^\circ \)) for each reaction, which can be found in Appendix C, to calculate the equilibrium constant \( K \) using the relation \( \Delta G^\circ = -RT \ln K \).

For each reaction, calculate \( \Delta G^\circ \) by using the formula \( \Delta G^\circ = \sum \Delta G^\circ_{\text{products}} - \sum \Delta G^\circ_{\text{reactants}} \).

Substitute the \( \Delta G^\circ \) values into the equation \( \Delta G^\circ = -RT \ln K \) to solve for \( K \), where \( R \) is the gas constant (8.314 J/mol·K) and \( T \) is the temperature in Kelvin.

Compare the calculated \( K \) values for both reactions to determine which reaction is more favorable under standard conditions.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Equilibrium Constant (K)

The equilibrium constant (K) is a numerical value that expresses the ratio of the concentrations of products to reactants at equilibrium for a given chemical reaction. It provides insight into the extent of a reaction; a large K indicates that products are favored, while a small K suggests reactants are favored. Understanding how to calculate and interpret K is essential for comparing different reactions, such as those involving glucose oxidation and fermentation.

Oxidation and Reduction Reactions

Oxidation and reduction (redox) reactions involve the transfer of electrons between substances. In the context of glucose metabolism, oxidation refers to the loss of electrons from glucose as it is converted to carbon dioxide and water, while reduction involves the gain of electrons by other species. Recognizing these processes is crucial for understanding the energy transformations that occur during cellular respiration and fermentation.

Thermodynamics and Reaction Spontaneity

Thermodynamics studies the relationships between heat, work, and energy in chemical processes. The spontaneity of a reaction is determined by changes in Gibbs free energy (ΔG), where a negative ΔG indicates a spontaneous reaction. This concept is important when comparing the thermodynamic favorability of aerobic versus anaerobic processes, as it helps explain why one pathway is preferred under certain conditions.

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Spontaneity of Processes

Related Practice

Textbook Question

(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

(c) In each case, indicate whether K should increase or decrease with increasing temperature. (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)

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)

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)?

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.