Question

The sun supplies energy at a rate of about 1.0 kilowatt per square meter of surface area \((1 \text { watt }=1 \mathrm{Js} \text { ). The plants in an }\) agricultural field produce the equivalent of \(20 . \mathrm{kg}\) 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 $$ \begin{aligned} 12 \mathrm{CO}_{2}(g)+11 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(s)+& 12 \mathrm{O}_{2}(g) \\ & \Delta H=5640 \mathrm{kJ} \end{aligned} $$ calculate the percentage of sunlight used to produce the sucrose-that is, determine the efficiency of photosynthesis.

Short answer

The efficiency of photosynthesis in producing sucrose in the given agricultural field is approximately 0.915%.

Step-by-step solution

Calculate the energy used to produce sucrose

We are given that the plants produce 20 kg of sucrose per hour, and the reaction has a \(\Delta H = 5640\) kJ. First, let's determine the molar mass of sucrose absolutely: \[M_\text{sucrose} = (12 \times 12.01) + (22 \times 1.01) + (11 \times 16.00) = 342.3\ \text{g/mol}\] Now, we need to determine the number of moles of sucrose produced per hour: \[n_\text{sucrose} = \frac{20\ \text{kg}}{342.3\ \text{g/mol}} \times 1000\ \text{g/kg} \approx 58.4\ \text{mol/h}\] Next, we'll calculate the energy used to produce this amount of sucrose: \[E_\text{used} = n_\text{sucrose} \times \Delta H = 58.4\ \text{mol/h} \times 5640\ \text{ kJ/mol} = 329\:360\ \text{kJ/h}\]

Determine the total energy supplied by the sun

We are given that the sun supplies energy at a rate of 1.0 kilowatt per square meter. First, we need to convert kW/m^2 to J/h (since the energy used was found in J/h): \[1\ \text{kW/m}^2 = 1\:000\ \text{J/s/m}^2\times 3\:600\ \text{s/h} = 3.6\: \text{MJ/h/m}^2\] Since the agricultural field has an area of 1 hectare, which is equivalent to 10,000 m^2, we can calculate the total energy supplied by the sun per hour as follows: \[E_\text{sun} = 3.6\ \text{MJ/h/m}^2 \times 10\:000\ \text{m}^2 = 36\:000\:000\ \text{kJ/h}\]

Calculate the efficiency of photosynthesis

Finally, we can calculate the efficiency of photosynthesis by dividing the energy used to produce sucrose by the total energy supplied by the sun: \[\text{Efficiency} = \frac{E_\text{used}}{E_\text{sun}} = \frac{329\:360\ \text{kJ/h}}{36\:000\:000\ \text{kJ/h}}\] \[\text{Efficiency} \approx 0.00915 \times 100\%\] \[\text{Efficiency} \approx 0.915\%\] The efficiency of photosynthesis in producing sucrose in the given agricultural field is approximately 0.915%.

Key concepts

Solar Energy Conversion

Solar energy conversion refers to the process by which sunlight is transformed into usable forms of energy. The sun emits abundant energy, reaching the Earth mostly in the form of light waves. Plants leverage this energy through a process called photosynthesis. Here's how it works:

The sun's energy is captured by chlorophyll, which is the green pigment in plants. Chlorophyll absorbs sunlight and uses its energy to split water molecules, releasing oxygen and converting carbon dioxide and water into glucose and other carbohydrates needed by the plant.

This energy transformation is crucial not only for the survival of plants but also for the entire ecosystem as it forms the foundation of food chains. Harnessing solar power for human use has similarities, such as converting sunlight into electricity using solar panels, demonstrating the universal importance and potential of solar energy.

Sucrose Production

In the context of photosynthesis, sucrose production is the creation of a sugar used for energy and growth in plants. Sucrose, a type of carbohydrate, is formed naturally in many plants during photosynthesis:

• Sucrose has the chemical formula \( \text{C}_{12}\text{H}_{22}\text{O}_{11} \).
• Plants produce it to store energy derived from the sun.

Sucrose serves as an energy source for plant cells and is an essential building block for other organic compounds. Its production involves combining simpler sugars (like glucose and fructose) in the chloroplasts, which are special plant cell structures.

This sugar is then transported throughout the plant to provide energy wherever needed - from cell growth to flower formation. Understanding these processes helps in evaluating solar energy conversion into chemical energy, which is a step towards assessing the efficiency of photosynthesis.

Thermodynamic Calculations

Thermodynamic calculations in photosynthesis involve assessing energy inputs and outputs, particularly focusing on changes in enthalpy (\( \Delta H \)). This helps in measuring the energy efficiency of conversions:

In the given exercise, the production of 20 kg of sucrose with a \( \Delta H \) of 5640 kJ reflects the energy requirement to drive this chemical reaction. Calculating energy efficiency necessitates comparing this required energy against the solar energy absorbed per hectare of the agricultural field.

Let's recapitulate this calculation step. The required energy for sucrose production is determined by multiplying the number of moles of sucrose by the enthalpy change, which is obtained using the molar mass of sucrose to first find how many moles are produced from 20 kg. Comparatively, the sun supplies a substantial amount of energy per square meter each hour, converted into total joules based on the field area.

These calculations provide insight into how much of the sun's vast energy supply is effectively converted into usable sugar by the plant, allowing for a better understanding of the efficiency of photosynthesis.

Chemical Reactions in Photosynthesis

Chemical reactions in photosynthesis are complex processes transforming sunlight into chemical energy stored in glucose and other carbohydrates. The core reaction for sucrose involves carbon dioxide and water, which are converted into sucrose and oxygen:

• Equation: \( 12 \ \text{CO}_{2} + 11 \ \text{H}_{2}\text{O} \rightarrow \text{C}_{12}\text{H}_{22}\text{O}_{11} + 12 \ \text{O}_{2} \)
• This chemical reaction indicates that carbon fixation occurs, wherein carbon dioxide is converted into sugars.

Photosynthesis can be divided into two main stages: light-dependent reactions and light-independent reactions (Calvin cycle). In the first stage, sunlight energizes electrons in chlorophyll, leading to the splitting of water (photolysis) and the release of oxygen.

The second stage involves the Calvin cycle, where carbon is fixed, leading to the creation of glucose and ultimately sucrose. The energy carrier molecules ATP and NADPH produced in the light-dependent reactions are used to power this cycle.

Understanding these chemical reactions is essential in grasping how solar energy is converted to biochemical energy, further enabling the calculation of photosynthesis's efficiency in an agricultural context.