Question

The sun supplies energy at a rate of about \(1.0\) kilowatt per square meter of surface area ( 1 watt \(=1 \mathrm{~J} / \mathrm{s}\) ). 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 ( \(1 \mathrm{ha}=10,000 \mathrm{~m}^{2}\) ). 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 sucrosethat is, determine the efficiency of photosynthesis.

Short answer

The efficiency of photosynthesis is approximately \(0.914\% \), meaning that only about \(0.914\% \) of the sunlight energy is converted into sucrose by the plants.

Step-by-step solution

Calculate the energy stored in the sucrose produced

First, we need to find the energy stored in the sucrose produced by the plants. We are given the enthalpy change for the reaction: \(\Delta H = 5640 \: \text{kJ}\). And we know that the plants produce \(20 \: \text{kg}\) of sucrose per hour per hectare. Given that the molar mass of sucrose (\(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\)) is \(342.3 \: \text{g/mol}\), we can find the number of moles in the produced sucrose: \[n_\text{sucrose} = \frac{20 \: \text{kg}}{342.3 \: \text{g/mol}} = 58.4 \text{mol}\] Now, we can calculate the energy stored in the produced sucrose: \[E_\text{sucrose} = \Delta H \times n_\text{sucrose} = 5640 \: \text{kJ} \times 58.4 \: \text{mol} = 328896 \: \text{kJ}\]

Calculate the energy provided by the sun

Next, we need to find the energy provided by the sun per hectare and per hour. We are given that the sun supplies energy at a rate of \(1.0 \: \text{kW/m}^2\), and that the area of one hectare is \(10,000 \: \text{m}^2\). The energy provided by the sun per square meter per hour can be found by multiplying the energy rate by the number of seconds in an hour: \[E_\text{Sun,m} = 1.0 \: \text{kW/m}^2 \times 3600 \: \text{s} = 3600 \: \text{kJ/m}^2\] Now, we can find the total energy provided by the sun per hectare: \[E_\text{Sun,ha} = E_\text{Sun,m} \times 10,000 \: \text{m}^2 = 3600 \: \text{kJ/m}^2 \times 10,000 \: \text{m}^2 = 36000000 \: \text{kJ}\]

Calculate the efficiency of photosynthesis

The efficiency of photosynthesis is the ratio between the energy converted into sucrose and the total energy provided by the sun: \[\text{Efficiency} = \frac{E_\text{sucrose}}{E_\text{Sun,ha}} = \frac{328896 \: \text{kJ}}{36000000 \: \text{kJ}}\] Now, we can calculate the efficiency as a percentage: \[\text{Efficiency (\%)} = \frac{E_\text{sucrose}}{E_\text{Sun,ha}} \times 100 = \frac{328896 \: \text{kJ}}{36000000 \: \text{kJ}} \times 100 \approx 0.914 \% \] The efficiency of photosynthesis is approximately \(0.914\% \).

Key concepts

Enthalpy Change

Enthalpy change (\(\Delta H\)) in a chemical reaction measures the heat absorbed or released at constant pressure. In photosynthesis, energy from sunlight is stored in chemical bonds when plants convert carbon dioxide and water into glucose or sucrose. The reaction's enthalpy change tells us how much energy is involved.For the production of sucrose, the enthalpy change was given as \(5640 \, \text{kJ}\) per mole. This high amount signifies that a lot of energy is stored within the sucrose molecules, showing how efficient plants are at capturing sunlight energy and converting it into chemical energy.

Sucrose Production

Plants convert sunlight into the chemical energy stored in sucrose through photosynthesis. The balanced chemical equation for sucrose production is:\[12 \text{CO}_2(g) + 11 \text{H}_2 \text{O}(l) \rightarrow \text{C}_{12} \text{H}_{22} \text{O}_{11}(s) + 12 \text{O}_2(g)\]Plants produce \(20 \, \text{kg}\) of sucrose per hour per hectare. To understand the scale of energy involvement, we calculate the moles of sucrose produced using its molar mass. Once we have the moles, we multiply by the enthalpy change to find the total energy stored. This illustrates the considerable amount of energy plants harness and store in carbohydrates.

Solar Energy Input

The sun provides vast amounts of energy. Its energy rate at Earth's surface is approximately \(1.0 \text{kW/m}^2\). To find the input energy, consider both time and area.

• Use the rate \(1.0 \text{kW/m}^2\)
• Convert kilowatts to kilojoules (1 kW = 1000 J/s).
• Multiply by seconds in an hour (3600) to get \(3600 \, \text{kJ/m}^2\) per hour.

Then, scale this to a hectare (\(10,000 \, \text{m}^2\)) to find the total solar energy input per hectare per hour, which equals \(36000000 \, \text{kJ}\). This immense energy input contrasts the relatively small percentage captured by plants.

Molar Mass Calculation

Molar mass helps us convert between grams and moles, crucial for energy calculations. For sucrose (\(\text{C}_{12} \text{H}_{22} \text{O}_{11}\)), its molar mass is calculated by adding the atomic masses of all constituent atoms:

• Carbon (\(C\)): \(12 \text{ atoms} \times 12.01 \text{ g/mol} = 144.12 \text{ g/mol}\)
• Hydrogen (\(H\)): \(22 \text{ atoms} \times 1.01 \text{ g/mol} = 22.22 \text{ g/mol}\)
• Oxygen (\(O\)): \(11 \text{ atoms} \times 16.00 \text{ g/mol} = 176.00 \text{ g/mol}\)

Total molar mass is \(342.3 \text{ g/mol}\). Knowing the molar mass lets us find the precise number of moles, a key step in determining the energy stored during photosynthesis.