4. This problem will compare aerobic respiration and the efficiency of a diesel engine. You will need the following data: AcGo (glucose) = -2880 kJ mol-¹ ATP (aq) + H₂O (1) → ADP (aq) + P₁ (aq) + H3O+ (aq) AG+ = -31 kJ mol-¹ at 37°C (The difference in the superscripts reflects the different standard states.) Complete oxidation of 1 mol of glucose produces 38 moles of ATP. a) Calculate the efficiency of aerobic respiration under standard biochemical conditions. b) The following conditions are more typical of a biological system: P (CO₂) = 5.3 x 10-² atm, P (O₂) = 0.132 atm, [glucose] = 5.6 x 10-2 mol L-¹, [ATP] = [ADP] = [Pi] = 1.0 x 104 mol L-¹, pH = 7.4, T=310 K. Assuming that we can use concentrations and partial pressures for the activities, calculate the efficiency of aerobic respiration under these physiological conditions. c) A typical diesel engine operates between To = 873 K and Thi = 1923 K with an efficiency about 75% that of an ideal (Carnot) engine. Compare the efficiency of a diesel engine to aerobic respiration from part b. d) Why is biological energy conversion more or less efficient than a diesel engine?
4. This problem will compare aerobic respiration and the efficiency of a diesel engine. You will need the following data: AcGo (glucose) = -2880 kJ mol-¹ ATP (aq) + H₂O (1) → ADP (aq) + P₁ (aq) + H3O+ (aq) AG+ = -31 kJ mol-¹ at 37°C (The difference in the superscripts reflects the different standard states.) Complete oxidation of 1 mol of glucose produces 38 moles of ATP. a) Calculate the efficiency of aerobic respiration under standard biochemical conditions. b) The following conditions are more typical of a biological system: P (CO₂) = 5.3 x 10-² atm, P (O₂) = 0.132 atm, [glucose] = 5.6 x 10-2 mol L-¹, [ATP] = [ADP] = [Pi] = 1.0 x 104 mol L-¹, pH = 7.4, T=310 K. Assuming that we can use concentrations and partial pressures for the activities, calculate the efficiency of aerobic respiration under these physiological conditions. c) A typical diesel engine operates between To = 873 K and Thi = 1923 K with an efficiency about 75% that of an ideal (Carnot) engine. Compare the efficiency of a diesel engine to aerobic respiration from part b. d) Why is biological energy conversion more or less efficient than a diesel engine?
Human Anatomy & Physiology (11th Edition)
11th Edition
ISBN:9780134580999
Author:Elaine N. Marieb, Katja N. Hoehn
Publisher:Elaine N. Marieb, Katja N. Hoehn
Chapter1: The Human Body: An Orientation
Section: Chapter Questions
Problem 1RQ: The correct sequence of levels forming the structural hierarchy is A. (a) organ, organ system,...
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![4. This problem will compare aerobic respiration and the efficiency of a diesel engine. You will
need the following data:
A¢G° (glucose) =-2880 kJ mol-1
АТР (аq) + Н20 (I) —> ADP (aq) + Pr (аq) + Н:0* (аq)
(The difference in the superscripts reflects the different standard states.)
Complete oxidation of 1 mol of glucose produces 38 moles of ATP.
A-G+ = -31 kJ mol-1 at 37°C
a) Calculate the efficiency of aerobic respiration under standard biochemical conditions.
b) The following conditions are more typical of a biological system: P (CO2) = 5.3 x 102 atm,
P (O2) = 0.132 atm, [glucose] = 5.6 x 10 -² mol L', [ATP] = [ADP] = [Pi] = 1.0 x 10-4 mol L-',
pH = 7.4, T= 310 K. Assuming that we can use concentrations and partial pressures for the
activities, calculate the efficiency of aerobic respiration under these physiological conditions.
c) A typical diesel engine operates between Tio = 873 K and Thi = 1923 K with an efficiency
about 75% that of an ideal (Carnot) engine. Compare the efficiency of a diesel engine to
aerobic respiration from part b.
d) Why is biological energy conversion more or less efficient than a diesel engine?](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F120ef40c-ba61-489a-8f0e-beaff5d51a86%2F16976291-4236-4029-ba3e-eacea903114a%2Fbignayb_processed.png&w=3840&q=75)
Transcribed Image Text:4. This problem will compare aerobic respiration and the efficiency of a diesel engine. You will
need the following data:
A¢G° (glucose) =-2880 kJ mol-1
АТР (аq) + Н20 (I) —> ADP (aq) + Pr (аq) + Н:0* (аq)
(The difference in the superscripts reflects the different standard states.)
Complete oxidation of 1 mol of glucose produces 38 moles of ATP.
A-G+ = -31 kJ mol-1 at 37°C
a) Calculate the efficiency of aerobic respiration under standard biochemical conditions.
b) The following conditions are more typical of a biological system: P (CO2) = 5.3 x 102 atm,
P (O2) = 0.132 atm, [glucose] = 5.6 x 10 -² mol L', [ATP] = [ADP] = [Pi] = 1.0 x 10-4 mol L-',
pH = 7.4, T= 310 K. Assuming that we can use concentrations and partial pressures for the
activities, calculate the efficiency of aerobic respiration under these physiological conditions.
c) A typical diesel engine operates between Tio = 873 K and Thi = 1923 K with an efficiency
about 75% that of an ideal (Carnot) engine. Compare the efficiency of a diesel engine to
aerobic respiration from part b.
d) Why is biological energy conversion more or less efficient than a diesel engine?
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