The average consumption of a footballer in one football match is about 2200 kJ of the body's energy supply. How much bread does he need to eat (it is supposed to be pure glycogen, glucose) to get this energy back into the body, assuming each loaf weighs 0.2 kg? The molecular weight of glucose is 180 g / mol. The standard free energy for burning glucose is -2800 kJ / mol. 50% efficiency should be calculated and differences in temperature (25 ° C compared to 37 ° C) should be overlooked.
The average consumption of a footballer in one football match is about 2200 kJ of the body's energy supply. How much bread does he need to eat (it is supposed to be pure glycogen, glucose) to get this energy back into the body, assuming each loaf weighs 0.2 kg? The molecular weight of glucose is 180 g / mol. The standard free energy for burning glucose is -2800 kJ / mol. 50% efficiency should be calculated and differences in temperature (25 ° C compared to 37 ° C) should be overlooked.
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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I have this wierd question, that is driving me nuts and does not make much of a sence to me! please help me out.
The average consumption of a footballer in one football match is about 2200 kJ of the body's energy supply. How much bread does he need to eat (it is supposed to be pure glycogen, glucose) to get this energy back into the body, assuming each loaf weighs 0.2 kg? The molecular weight of glucose is 180 g / mol. The standard free energy for burning glucose is -2800 kJ / mol. 50% efficiency should be calculated and differences in temperature (25 ° C compared to 37 ° C) should be overlooked.
I got the question rounded to something like this.
dG = 2200 kJ
Loaf of bread = 1,11 mol glukose
dG´°= -2800 kJ/mol
efficiency 50%
K = 298
R = 8,314 J / mol*K
However I am stuck, should the efficency be used to calculate the Keq?
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