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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 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
Approximately \(1.02977 * 10^9\) kg of glucose is needed to provide 1500 Cal/person/day of nourishment to the global population for one year.

Step by step solution

01

Convert Cal to Joules

As there are 4.184 Joules in 1 Calorie, we can convert 1500 Cal/person/day to Joules/person/day: 1500 Cal/person/day * 4.184 J/Cal = 6276 J/person/day
02

Calculate daily Joules for global population

To find the amount of energy required for 7.0 billion people in one day, multiply the Joules/person/day by the population: 6276 J/person/day * 7.0 * 10^9 people = 4.3932 * 10^13 J/day
03

Calculate yearly Joules for global population

Now, we need to find the amount of energy required for a year. There are 365 days in a year, so we multiply the daily energy by 365: 4.3932 * 10^13 J/day * 365 days/year = 1.603318 * 10^16 J/year #Phase2Solution#:
04

Calculate amount of glucose needed for one year

We know that 2803 kJ of energy are released per mole of glucose. First, convert the enthalpy change to Joules: -2803 kJ/mol * 1000 J/kJ = -2.803 * 10^6 J/mol Now, divide the total energy needed for a year by the amount of energy released per mole of glucose to find the amount of glucose needed: (1.603318 * 10^16 J/year) / (-2.803 * 10^6 J/mol) = 5.720958 * 10^9 mol glucose/year
05

Calculate the mass of glucose needed for one year

The molar mass of glucose is approximately 180 g/mol. To find the mass of glucose needed for a year, multiply the amount of glucose needed in moles by the molar mass: (5.720958 * 10^9 mol glucose/year) * (180 g/mol) = 1.02977 * 10^12 g glucose/year Finally, convert grams to kilograms: 1.02977 * 10^12 g/year * (1 kg/1000 g) = 1.02977 * 10^9 kg glucose/year Therefore, approximately 1.02977 * 10^9 kg of glucose is needed to provide 1500 Cal/person/day of nourishment to the global population for one year.

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Key Concepts

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

Calorie to Joule Conversion
When working with energy values in different units, conversion is essential. In thermochemical calculations, often we have to convert calories (Cal) to joules (J). A calorie is a unit of energy specifically defined as the amount of heat needed to raise the temperature of one gram of water by one degree Celsius. In contrast, a joule is the SI unit for energy or work and is defined as the amount of work done when an applied force of one newton moves the object by one meter.

The conversion factor between these units is fixed: 1 calorie equals 4.184 joules. This means that to convert calories into joules, we multiply the number of calories by 4.184. For instance, if a person requires 1500 Cal per day, we multiply 1500 by 4.184 to determine how much energy this represents in joules. This precise conversion is crucial in exercises that bridge biology (which commonly uses calories) and physics or chemistry (which use joules).
Global Energy Requirements
Understanding global energy requirements involves large figures that can be hard to conceptualize. Energy consumed by humans can come from various food sources, often evaluated in terms of calorific intake. In our case, we consider the glucose required to fulfill the recommended daily calorie intake. Globally, this becomes a vast number when considering billions of people.

It's crucial to aggregate individual needs to the total required energy to support life on a planetary scale. To do this, we multiply the daily energy requirement in joules by the global population. Furthering to an annual need requires considering the number of days in a year. Such exercises not only test our mathematical prowess but also provide a profound perspective on our collective needs and the scale of human consumption of energy.
Molar Enthalpy Change
Molar enthalpy change is the heat released or absorbed when a reaction takes place at constant pressure, per mole of reactant or product. It's indicated by \( \Delta H^\circ \) and expressed in joules per mole (J/mol). Positive \( \Delta H^\circ \) indicates endothermic reactions, where heat is absorbed, and negative \( \Delta H^\circ \) signifies exothermic reactions, like our glucose metabolism example, where heat is released.

To apply this concept to real-world problems, such as determining the amount of a substance required to produce a specific amount of energy, we relate molar enthalpy change to the total energy needed. Dividing the total energy by the molar enthalpy gives us the moles of substance required, and further multiplying by the molar mass provides the mass in grams, often then converted to kilograms for practicality. This step is vital in calculating the required fuel or food (like glucose) needed to meet energy demands.

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Most popular questions from this chapter

The standard enthalpies of formation of gaseous propyne \(\left(\mathrm{C}_{3} \mathrm{H}_{4}\right),\) propylene \(\left(\mathrm{C}_{3} \mathrm{H}_{6}\right),\) and propane \(\left(\mathrm{C}_{3} \mathrm{H}_{8}\right)\) are \(+185.4,+20.4,\) and \(-103.8 \mathrm{kJ} / \mathrm{mol}\) , respectively.(a) Calculate the heat evolved per mole on combustion of each substance to yield \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(g) .\) (b) Calculate the heat evolved on combustion of 1 \(\mathrm{kg}\) of each substance. (c) Which is the most efficient fuel in terms of heat evolved per unit mass?

(a) Why is the change in enthalpy usually easier to measure than the change in internal energy? (b) \(H\) is a state function, but \(q\) is not a state function. Explain. (c) For a given process at constant pressure, \(\Delta H\) is positive. Is the process endothermic or exothermic?

For each of the following compounds, write a balanced thermochemical equation depicting the formation of one mole of the compound from its elements in their standard states and then look up \(\Delta H^{\circ} f\) for each substance in Appendix C. (a) \(\mathrm{NO}_{2}(g),(\mathbf{b}) \mathrm{SO}_{3}(g),(\mathbf{c}) \mathrm{NaBr}(s),(\mathbf{d}) \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}(s) .\)

Write balanced equations that describe the formation of the following compounds from elements in their standard states, and then look up the standard enthalpy of formation for each substance in Appendix C: (a) \(\mathrm{H}_{2} \mathrm{O}_{2}(g),(\mathbf{b}) \mathrm{CaCO}_{3}(s)\) (c) \(\mathrm{POCl}_{3}(l),(\mathbf{d}) \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l) .\)

Given the data $$\begin{aligned} \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}(g) & \Delta H=+180.7 \mathrm{kJ} \\ 2 \mathrm{NO}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}_{2}(g) & \Delta H=-113.1 \mathrm{kJ} \\ 2 \mathrm{N}_{2} \mathrm{O}(g) \longrightarrow 2 \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) & \Delta H=-163.2 \mathrm{kJ} \end{aligned}$$ use Hess's law to calculate \(\Delta H\) for the reaction $$\mathrm{N}_{2} \mathrm{O}(g)+\mathrm{NO}_{2}(g) \longrightarrow 3 \mathrm{NO}(g)$$

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