Question

The sun supplies about \(1.0\) kilowatt of energy for each square meter of surface area \(\left(1.0 \mathrm{~kW} / \mathrm{m}^{2}\right.\), where a watt \(=1 \mathrm{~J} / \mathrm{s}\) ). Plants produce the equivalent of about \(0.20 \mathrm{~g}\) of sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) per hour per square meter. Assuming that the sucrose is produced as follows, calculate the percentage of sunlight used to produce sucrose. $$ \begin{aligned} 12 \mathrm{CO}_{2}(g)+11 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}+12 \mathrm{O}_{2}(g) \\ \Delta H=5645 \mathrm{~kJ} \end{aligned} $$

Short answer

The plants use approximately \(0.0914\%\) of the sunlight to produce sucrose.

Step-by-step solution

Convert mass of sucrose to moles

First, find the molar mass of sucrose (C12H22O11) by adding up the molar masses of each of its elements: Molar mass of C = 12.01 g/mol Molar mass of H = 1.01 g/mol Molar mass of O = 16.00 g/mol Molar mass of sucrose (C12H22O11) = 12(12.01) + 22(1.01) + 11(16.00) = 342.30 g/mol Now, convert the mass of sucrose (0.20 g) to moles: Moles of sucrose = mass (g) / molar mass (g/mol) Moles of sucrose = 0.20 g / 342.30 g/mol ≈ 0.000584 mol

Calculate the energy used to produce sucrose

Use the enthalpy change (ΔH = 5645 kJ) to find the energy used in producing 0.000584 mol of sucrose: Energy used (kJ) = moles of sucrose × ΔH Energy used (kJ) = 0.000584 mol × 5645 kJ/mol ≈ 3.29 kJ Now, convert this energy used to produce sucrose per hour to watts (J/s): Energy used (J) = 3.29 kJ × 1000 J/kJ = 3290 J 1 hour = 3600 seconds Energy used (J/s) = 3290 J / 3600 s ≈ 0.914 W

Calculate the percentage of sunlight used

Now that we have the amount of energy used by the plants (0.914 W), we can compare it with the energy supplied by the sun (1.0 kW/m^2) to find the percentage of sunlight used. Percentage of sunlight used = (Energy used by plants / Energy supplied by the sun) × 100 Percentage of sunlight used = (0.914 W / 1000 W) × 100 ≈ 0.0914 % So, the plants use approximately 0.0914% of the sunlight to produce sucrose.

Key concepts

Enthalpy Change in Photosynthesis

The key to understanding how plants convert light to chemical energy lies in the concept of enthalpy change during photosynthesis. Photosynthesis is a chemical process where plants, algae, and certain bacteria transform sunlight into chemical energy stored in glucose or, in this exercise, sucrose. The change in heat content for this reaction, known as the enthalpy change (abla H), is a measure of the energy absorbed or released when substances are converted from reactants to products.

In photosynthesis, sunlight provides the energy needed to drive the reaction of carbon dioxide and water to produce glucose, oxygen, and in our specific case, sucrose. The abla H value of 5645 kJ indicates that this amount of energy is absorbed from light to form one mole of sucrose. The enthalpy change is crucial in calculating the efficiency of photosynthesis, as it reflects the total energy captured by the plants and converted into chemical energy.

Molar Mass of Sucrose

Delving into the realm of chemistry, the molar mass is a fundamental concept representing the mass of one mole of a substance. It is calculated by summing the masses of the individual atoms making up the molecule. For sucrose, which has the chemical formula abla{C}_{12}abla{H}_{22}abla{O}_{11}, its molar mass is determined by adding the mass contributions from 12 carbon atoms, 22 hydrogen atoms, and 11 oxygen atoms. The precise molar mass calculation leads us to a value of 342.30 grams per mole.

Knowing the molar mass helps us translate between the mass of the substance in grams and the number of moles, which is essential in stoichiometry – the area of chemistry that deals with the relative quantities of reactants and products in chemical reactions. In this exercise, understanding the molar mass is required to convert the mass of produced sucrose into moles, connecting the dots between the physical substance and its participation in the photosynthetic reaction.

Conversion of Energy Units

Our understanding of energy transformations is not complete without the ability to convert between different energy units. To comprehend the process of photosynthesis thoroughly, it is essential to convert energy values from one unit to another. In the given problem, energy is provided in kilojoules (kJ), which is a unit of energy in the metric system, and measures how much energy is required to carry out a reaction. However, solar energy is provided in kilowatts per square meter (kW/m^2), where one watt is equal to one joule per second (abla W = abla J/s).

These conversions are essential for calculating the efficiency of photosynthesis. Converting the energy used in producing sucrose to watts allows us to compare it directly to the sunlight's power input. It is through these conversions, including converting an hour to seconds, that we can express the plant's energy use in terms of the percentage of available sunlight, revealing the actual efficiency of photosynthesis. The conversions highlight that despite the abundance of sunlight, plants utilize only a small fraction for sucrose production.