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At the end of 2012 , global population was about \(7.0\) billion people. What mass of glucose in kg would be needed to provide \(1500 \mathrm{cal} /\) person/day of nourishment to the global population for one year? Assume that glucose is metabolized entirely to \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l)\) according to the following thermochemical equation: $$ \begin{aligned} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) \longrightarrow 6 \mathrm{CO}_{2}(g)+& 6 \mathrm{H}_{2} \mathrm{O}(l) \\ \Delta H^{\circ}=-2803 \mathrm{~kJ} \end{aligned} $$

Short Answer

Expert verified
The mass of glucose needed to provide 1500 cal/person/day of nourishment to the global population for one year is approximately \(1.03 * 10^{12}\) kg.

Step by step solution

01

Calculate total daily caloric requirement for the global population

To calculate the total daily caloric requirement for the global population, multiply the population by the daily caloric requirement per person: Total daily calories = Global population × Daily caloric requirement per person = (7.0 * 10^9) × 1500 cal
02

Convert caloric requirement to kilojoules

Convert the total daily caloric requirements from calories to kilojoules using the conversion factor 1 cal = 4.184 J: Total daily kilojoules = Total daily calories × (4.184 J/cal) = (7.0 * 10^9 × 1500) × 4.184 J/cal ≈ 4.39 * 10^13 J
03

Determine the total amount of glucose needed to provide daily kilojoules

By studying the thermochemical equation, we can see that the reaction of one mole of glucose with oxygen releases -2803 kJ of energy. So we can calculate the amount of glucose in moles needed to generate the required daily energy for the global population: Moles of glucose = Total daily kilojoules / -ΔH° = (4.39 * 10^13 J) / (-2803 * 10^3 J/mol) ≈ 15.67 * 10^9 mol
04

Calculate the mass of glucose per day

We can calculate the mass of glucose needed daily by multiplying the number of moles by the molar mass of glucose (180.156 g/mol): Mass of glucose per day = (15.67 * 10^9 mol) × (180.156 g/mol) ≈ 2.82 * 10^12 g
05

Convert mass of glucose to kilograms per day

Convert the mass of glucose per day from grams to kilograms: Mass of glucose per day = 2.82 * 10^12 g × (1 kg / 10^3 g) = 2.82 * 10^9 kg
06

Calculate mass of glucose needed for one year

Multiply the mass of glucose needed per day by the number of days in one year: Mass of glucose per year = Mass of glucose per day × 365 = 2.82 * 10^9 kg/day × 365 days ≈ 1.03 * 10^12 kg So, the mass of glucose needed to nourish the global population for one year is approximately \(1.03 * 10^{12}\) kg.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Global Caloric Intake Calculation
Understanding and determining the global caloric intake requirement involves an intriguing application of chemistry and mathematics. The process begins with calculating the total daily caloric requirement for the entire global population. This is achieved by multiplying the population count by the individual daily caloric requirement. For instance, if the global population is around 7 billion and the daily requirement is 1500 calories per person, the total daily requirement becomes a multiplication of these two figures.

To make these calculations relevant and accurate, it's important to note the unit conversions involved. Caloric values need to be converted from calories to kilojoules since scientific energy calculations are commonly expressed in joules or kilojoules. This is where the conversion factor of 1 calorie being equivalent to 4.184 joules comes into play. Once the caloric requirements are expressed in joules, we can move towards understanding how these energy requirements are met through metabolism – in this case, the metabolism of glucose.
Thermochemical Equation Metabolism
The metabolism of glucose serves as an excellent exemplar of chemical energy conversion within organisms, and the process can be represented by a thermochemical equation. This equation not only shows the substances involved in the reaction, but also the energy change associated with the reaction, denoted as \( \Delta H^\circ \).

In the given example, we see glucose (\( \mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6 \) in its solid form) reacting with oxygen gas (\( \mathrm{O}_2 \) gas) to yield carbon dioxide (\( \mathrm{CO}_2 \) gas) and water (\( \mathrm{H}_2\mathrm{O} \) liquid), with an energy release of -2803 kJ per mole of glucose. This negative sign indicates that the reaction is exothermic, meaning it releases energy, which is a central concept in metabolism where food substances are broken down to release energy. Understanding this thermochemical equation is crucial in calculating the amount of glucose required to meet the global caloric intake because it provides the direct relationship between the mass of glucose and the amount of energy it can yield.
Glucose Metabolism Stoichiometry
Diving deeper into the realm of biochemistry, the stoichiometry of glucose metabolism quantitatively connects the mass of glucose consumed to the energy produced. By examining the thermochemical equation provided, you can calculate the moles of glucose required to produce a certain amount of energy expressed in kilojoules, using the reaction’s \( \Delta H^\circ \) value.

The calculation usually begins with the total energy requirement, from which you determine the moles needed based on the energy yield per mole. In our example, one mole of glucose produces -2803 kJ of energy. Thus, you can calculate the total moles of glucose necessary by dividing the total daily kilojoules by this energy yield per mole. The final step then involves converting moles to mass, where the molar mass of glucose (180.156 g/mol) is used. This stoichiometry is not just an abstract concept; it has real-world implications for understanding how much raw material (in this case, glucose) is needed to sustain energy levels – an important consideration in fields like nutrition, environmental science, and global health.

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Most popular questions from this chapter

5.104 We can use Hess's law to calculate enthalpy changes that cannot be measured. One such reaction is the conversion of methane to ethylene: $$ 2 \mathrm{CH}_{4}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2}(g) $$ Calculate the \(\Delta H^{\circ}\) for this reaction using the following thermochemical data: $$ \begin{array}{ll} \mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l) & \Delta H^{\circ}=-890.3 \mathrm{~kJ} \\ \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g) & \Delta H^{\circ}=-136.3 \mathrm{~kJ} \\ 2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) & \Delta H^{\circ}=-571.6 \mathrm{~kJ} \\ 2 \mathrm{C}_{2} \mathrm{H}_{6}(g)+7 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) & \Delta H^{\circ}=-3120.8 \mathrm{~kJ} \end{array} $$

For the following processes, calculate the change in internal energy of the system and determine whether the process is endothermic or exothermic: (a) A balloon is heated by adding 850 J of heat. It expands, doing $382 \mathrm{~J}\( of work on the atmosphere. (b) A \)50-g$ sample of water is cooled from \(30^{\circ} \mathrm{C}\) to \(15^{\circ} \mathrm{C}\), thereby losing approximately \(3140 \mathrm{~J}\) of heat. (c) A chemical reaction releases \(6.47 \mathrm{~kJ}\) of heat and does no work on the surroundings.

(a) What are the units of molar heat capacity? (b) What are the units of specific heat? (c) If you know the specific heat of copper, what additional information do you need to calculate the heat capacity of a particular piece of copper pipe?

(a) What is meant by the term state function? (b) Give an example of a quantity that is a state function and one that is not. (c) Is the volume of a system a state function? Why or why not?

From the enthalpies of reaction $$ \begin{array}{rlr} \mathrm{H}_{2}(g)+\mathrm{F}_{2}(g) \longrightarrow 2 \mathrm{HF}(g) & \Delta H=-537 \mathrm{~kJ} \\ \mathrm{C}(s)+2 \mathrm{~F}_{2}(g) \longrightarrow \mathrm{CF}_{4}(g) & \Delta H=-680 \mathrm{~kJ} \\ 2 \mathrm{C}(s)+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{4}(g) & \Delta H=+52.3 \mathrm{~kJ} \end{array} $$ calculate \(\Delta H\) for the reaction of ethylene with \(\mathrm{F}_{2}\) : $$ \mathrm{C}_{2} \mathrm{H}_{4}(g)+6 \mathrm{~F}_{2}(g) \longrightarrow 2 \mathrm{CF}_{4}(g)+4 \mathrm{HF}(g) $$

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