Question

The sun supplies energy at a rate of about 1.0 kilowatt per square meter of surface area (1 watt = 1 J/s). The plants in an agricultural field produce the equivalent of \(20 .\) kg sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) per hour per hectare \(\left(1 \mathrm{ha}=10,000 \mathrm{m}^{2}\right) .\) Assuming that sucrose is produced by the reaction $$12 \mathrm{CO}_{2}(g)+11 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(s)+12 \mathrm{O}_{2}(g) \Delta H=5640 \mathrm{kJ}$$ calculate the percentage of sunlight used to produce the sucrose - that is, determine the efficiency of photosynthesis.

Short answer

The efficiency of photosynthesis can be calculated by finding the energy supplied by the sun per hour per hectare, the energy required to produce 20 kg of sucrose per hectare, and then determining the ratio of the energy required for sucrose production to the energy supplied by the sun. After performing these calculations, we find that the efficiency of photosynthesis is approximately 0.914%.

Step-by-step solution

Calculate the energy supplied by the sun per hour per hectare.

First, find the energy supplied by the sun per square meter per second, which is given to be 1.0 kilowatt, or 1000 J/s. To convert this to hourly energy, multiply by 3600 seconds (since there are 3600 seconds in an hour): \(1000 \text{ J/s} \times 3600 \text{ s/hour} = 3.6 \times 10^{6} \text{ J/hour}\). Now, calculate the energy per hectare (1 hectare = 10,000 square meters): \(3.6 \times 10^{6} \text{ J/h m}^2 \times 10,000 \text{ m}^2 = 3.6 \times 10^{10} \text{ J/hour}\).

Calculate the energy needed to produce 20 kg of sucrose per hectare.

The reaction tells us that the enthalpy change (\(\Delta H\)) is 5640 kJ for every 1 mole of sucrose. First, convert \(\Delta H\) to joules: \(5640\text{ kJ} × 10^3\text{ J/kJ} = 5.64 \times 10^6\text{ J/mole}\). Next, find the molar mass of sucrose (\(M\)): \(M(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}) = 12 \times 12 + 22 \times 1 + 11 \times 16 = 342\text{ g/mole}\). We are given that 20 kg (20,000 g) of sucrose is produced per hour per hectare. To find the number of moles of sucrose, divide 20,000 g by the molar mass: \(\frac{20,000\text{ g}}{342\text{ g/mole}} \approx 58.48\text{ moles}\). Now, multiply the number of moles by the enthalpy change per mole to find the total energy required for sucrose production: \(58.48\text{ moles} \times 5.64 \times 10^6\text{ J/mole} \approx 3.29 \times 10^8\text{ J}\).

Calculate the efficiency of photosynthesis.

Efficiency can be determined by dividing the energy required for sucrose production by the energy supplied by the sun, then multiplying by 100 to get the percentage: \(\frac{3.29 \times 10^8\text{ J}}{3.6 \times 10^{10}\text{ J}} = 0.00914\). Finally, multiply by 100 to get the efficiency: \(0.00914 \times 100 \approx 0.914\%\). Therefore, the efficiency of photosynthesis is approximately 0.914%.

Key concepts

Enthalpy Change of Sucrose Production

When learning about the enthalpy change of sucrose production, it's helpful to understand that enthalpy change, denoted as \( \Delta H \), represents the total heat content of a system. It is an important concept in thermochemistry and is directly associated with chemical reactions, corresponding to the heat absorbed or released under constant pressure.

In the context of sucrose production, the enthalpy change is the amount of energy required to convert carbon dioxide and water into sucrose and oxygen during photosynthesis. The given reaction in the exercise shows an enthalpy change of 5640 kJ, which is the energy needed for the formation of 1 mole of sucrose from the reactants. To calculate this on a larger scale, as seen in the exercise, you must consider the molar mass of sucrose and the amount produced in kilograms to work out the total energy involved. This type of chemistry calculation interfaces directly with the biological processes in plants, showcasing the vital role of energy transformations in nature.

Solar Energy Conversion in Photosynthesis

Photosynthesis is one of nature's most elegant examples of solar energy conversion. Through this process, plants absorb sunlight and convert it into chemical energy in the form of glucose or, for our particular interest, sucrose. This transformation isn't just essential for the survival of the plant but also serves as a foundational component of the entire food web.

The sun emits energy continuously, which plants capture using chlorophyll in their leaves. During photosynthesis, this solar energy powers the chemical reactions that assemble carbon dioxide and water into organic molecules. The efficiency of this conversion process is not 100%, which is a common misunderstanding. Most plants have relatively low photosynthetic efficiency due to energy losses at various stages of the conversion process. In our example, the efficiency is a critical measure that compares the energy captured and utilized by plants to the amount of energy delivered by sunlight.

Sucrose Production in Plants

Plants produce sucrose using the energy harnessed during photosynthesis. Sucrose, which is common table sugar, serves multiple purposes in plants. It acts as an energy source for plant cells and is critical for building the plant's structure. But how does this production actually take place on a molecular level?

The process begins with the conversion of energy from sunlight into chemical energy stored in adenosine triphosphate (ATP) and nicotinamide adenine dinucleotide phosphate (NADPH). These high-energy molecules are then used in the Calvin cycle to combine carbon dioxide and water, resulting eventually in the synthesis of sucrose. This example from the exercise provides an opportunity to peek into nature's detailed biochemical processes and learn about the mass-to-mole conversions and energy relationships in this fascinating chain of reactions.

Energy Calculations in Chemistry

Performing energy calculations is a crucial part of understanding chemical reactions, as seen in chemistry and biochemistry. It involves quantifying the energy changes associated with reactions, which is fundamental in fields ranging from environmental science to engineering and pharmacology.

In the context of the exercise, we dealt with converting units of energy from kilojoules to joules, understanding molar mass, and the concept of moles—all fundamental chemistry skills. Calculating the efficiency of photosynthesis provides a real-world application of these principles, showing the intricate balance between energy input and output. For students, understanding these calculations is essential for grasping why some reactions are energetically viable while others are not, and how this relates to broader ecological systems and technological innovations like solar panels.