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Question: The enthalpy change involved in oxidation of glucose is \( - 2880kJmo{l^{ - 1}} \) . \( 25\% \) of t...

The enthalpy change involved in oxidation of glucose is −2880kJmol−1- 2880kJmo{l^{ - 1}} . 25%25\% of this energy is available for muscular work. If 80kJ80kJ of muscular work is needed to walk one km, what is the maximum distance that a person will be able to walk after eating 120g120g of glucose?

Explanation

Solution

Hint : The oxidation of glucose is a biochemical process that takes place inside organisms and is exothermic in nature. The enthalpy released by this reaction is utilized in carrying out a variety of other processes by the body.

Complete Step By Step Answer:
The exergonic reaction of oxidizing glucose to pyruvic acid is an important source of energy in the body. The enthalpy change associated with the oxidation of a single mole of glucose is −2880kJ- 2880kJ .The amount of energy released on oxidation varies with the amount of glucose being oxidized.
The total amount of glucose involved in the process can be determined by calculating the number of moles of glucose using the given mass and molar mass.
The number of moles (n)(n) are simply calculated as the ratio of given mass and molar mass,
n=given massmolar mass=120g180gmol−1=0.67molesn = \dfrac{{{\text{given mass}}}}{{{\text{molar mass}}}} = \dfrac{{120g}}{{180gmo{l^{ - 1}}}} = 0.67moles
The product of enthalpy released per mole and the actual amount of glucose oxidized gives us the total amount of enthalpy released in the process.
net enthalpy=0.67moles×(−2880kJmol−1)=−1929.6kJmol−1{\text{net enthalpy}} = 0.67moles \times ( - 2880kJmo{l^{ - 1}}) = - 1929.6kJmo{l^{ - 1}}
Since, only one fourth i.e. 25%25\% of this energy is available for muscular work, the total energy available can be calculated as follows:
net enthalpy available=−1929.6kJmol−1×25100=482.4kJmol−1{\text{net enthalpy available}} = \dfrac{{ - 1929.6kJmo{l^{ - 1}} \times 25}}{{100}} = 482.4kJmo{l^{ - 1}}
Now, 80kJ80kJ of energy is needed to walk a distance of one kilometers, then the maximum distance can be calculated as follows:
distance=482.4kJ80kJkm−1=6.03km{\text{distance}} = \dfrac{{482.4kJ}}{{80kJk{m^{ - 1}}}} = 6.03km
Hence, a maximum distance of 6.03km6.03km can be walked using the energy available.

Note :
The amount of energy produced by the body through the oxidation of glucose is always higher than that consumed during different daily exercises or activities, this is because a large portion of energy is utilized in performing involuntary functions such as breathing, digestion and many other biological activities.