https://en.wikipedia.org/wiki/Cori_cycleUnder anaerobic conditions, glucose is broken down in muscle tissue to form lactic acid according to thefollowing reaction:C6H12O6(s) → 2CH3HOHCOOH (aq)Room temperature is considered to be 25ºC, body temperature is 32°C, and a severe fever isclassified as anything above 40.6 °C.The following thermodynamic data was collected at 25°CAH°f (kJ/mol)Cpm (J/Kmol)S°m (J/K)Glucose-1273.1219.2209.2Lactic Acid-673.6127.6192.1 Group ACalculate AGo, at 25°C and at body temperature. What assumptions will you have to make? Whatcan you say about the two values?Hint: You will need to find AH, and AS; at 25°CHint: There are a couple different methods to solving for AGº, (T2), you can calculate AH°, and ASº,for the new temperature or use the equation for when you know AG°-(T1) and want AG°, (T2), butwhat assumptions will you need to make and are they reasonable for this process?Hint: compare the two values, what conclusion can you draw about the formation of lactic acid atbody temp vs room temp?

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Under anaerobic conditions, glucose is Following reaction: C6H Room temperature is considered to b lassified as anything above 40.6 °C. he following thermodynamic data w AH°F (kJ/m Glucose -1273.1 actic Acid -673.6

Under anaerobic conditions, glucose is Following reaction: C6H Room temperature is considered to b lassified as anything above 40.6 °C. he following thermodynamic data w AH°F (kJ/m Glucose -1273.1 actic Acid -673.6

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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