Problem 97 - Chemistry The Central Science In Si Units

Question

At the end of \(2012,\) global population was about 7.0 billion people. What mass of glucose in \(\mathrm{kg}\) 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 \(\mathrm{CO}_{2}(\mathrm{~g})\) and \(\mathrm{H}_{2} \mathrm{O}(I)\) according to the following thermochemical equation: $$ \begin{aligned} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g) \longrightarrow 6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(I) \\ \Delta H^{\circ} &=-2803 \mathrm{~kJ} \end{aligned} $$

Step-by-Step Solution

Answer

About 1.03 trillion kg of glucose is needed for one year.

Step 1: Convert Calories to Joules

The energy required per person per day is 1500 Calories. First, we need to convert Calories (food calories) to Joules. Remember that 1 Calorie is equivalent to 4184 Joules. So, for one person: \[1500 \text{ Cal} \times 4184 \text{ J/Cal} = 6,276,000 \text{ J/person/day} \]

Step 2: Calculate Energy for Global Population for One Day

Multiply the energy needed by one person per day by the global population to find the energy required for all people in one day: \[6,276,000 \text{ J/person/day} \times 7.0 \times 10^9 \text{ people} = 4.3932 \times 10^{16} \text{ J/day} \]

Step 3: Calculate Energy for Global Population for One Year

Multiply the daily energy requirement by the number of days in a year to find the annual energy requirement: \[4.3932 \times 10^{16} \text{ J/day} \times 365 \text{ days/year} = 1.603518 \times 10^{19} \text{ J/year} \]

Step 4: Calculate Moles of Glucose Required

Using the enthalpy of glucose combustion, find out how many moles are needed for the annual energy requirement. The reaction gives off \(-2803 \text{ kJ/mol}\), or \(-2.803 \times 10^6 \text{ J/mol}\). The number of moles required is:\[\frac{1.603518 \times 10^{19} \text{ J}}{2.803 \times 10^6 \text{ J/mol}} = 5.72 \times 10^{12} \text{ mol} \]

Step 5: Find Mass of Glucose Required

Finally, calculate the mass of glucose needed. The molar mass of glucose (\(\mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6\)) is about 180.18 g/mol. Convert this to kilograms: \[5.72 \times 10^{12} \text{ mol} \times 180.18 \text{ g/mol} \times \frac{1 \text{ kg}}{1000 \text{ g}} \approx 1.03 \times 10^{15} \text{ kg} \]

Step 6: Conclusion

To sustain the global population with 1500 Cal/day/person for one year, approximately 1.03 trillion kg of glucose is required.

Key Concepts

Calorimetry

Calorimetry is the science that measures the heat of chemical reactions or physical changes. It's a vital part of thermochemistry, pivotal for understanding energy transfer in reactions. For our exercise, the concept of calorimetry helps us determine how much energy is required to sustain a population using glucose.

• Calorimetry involves precise measurements of temperature change to deduce heat transfer.
• The heat of reaction is determined through experiments often using a calorimeter device.
• This process allows for the computation of energy changes, vital in determining how much glucose is necessary to supply energy needs globally.

Determining the energy required per person per day and converting it to joules is a fundamental application of calorimetry principles.

Enthalpy Change

Enthalpy change refers to the heat change that occurs at constant pressure during a chemical reaction. It is symbolized as \( \Delta H \)

• This value indicates whether a reaction is exothermic or endothermic.
• In the context of glucose combustion, \( \Delta H^{\circ} = -2803 \, \text{kJ/mol} \) signifies that the process releases energy, making it exothermic.

Understanding enthalpy change helps predict the heat energy released or absorbed, aiding in calculating the moles of glucose needed based on energy output.

Stoichiometry

Stoichiometry involves the calculation of reactants and products in chemical reactions. It's like mathematically mapping the route of a reaction.

• Based on balanced chemical equations, it allows the prediction of quantities.
• For every mole of glucose combusted, a specific amount of energy is released.
• The exercise converts energy needs into moles of glucose using stoichiometry, utilizing the known \( \Delta H \) value.

This concept is a cornerstone of chemistry, ensuring the accurate scaling of reactants used in reactions.

Unit Conversion

Unit conversion is the process of converting values from one set of units to another. It ensures consistency and accuracy across calculations.

• In the exercise, converting Calories to Joules is the first essential step.
• This step uses the equivalence of \(1\, \text{Cal} = 4184\, \text{J}\).
• Converting grams to kilograms when finding the mass of glucose is another critical conversion.

Mastering unit conversion allows for the seamless translation of measurements and is crucial for precise scientific calculations.

Molar Mass Calculation

Molar mass is the mass of one mole of a substance, usually expressed in grams per mole (g/mol). This calculation is key for converting between moles and mass.

• Glucose has the molecular formula \( \text{C}_6\text{H}_{12}\text{O}_6 \).
• The molar mass of glucose is calculated as the sum of the atomic masses of all its atoms, approximately 180.18 g/mol.
• This value assists in transitioning from moles to mass when determining the needed glucose for the global population.

Understanding molar mass calculation bridges the gap between chemical measures in moles and practical measures in mass, essential for practical applications in the real-world.