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 \(\left(1 \mathrm{ha}=10,000 \mathrm{~m}^{2}\right)\). Assum- ing 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.

Since the field is 10,000 m² and the sun supplies energy at a rate of 1.0 kilowatt per square meter, we can calculate the total energy supplied to the field in an hour. Total energy per hour = Energy per m² * surface area * time in hours = 1.0 kW/m² * 10,000 m² * 1 hour Keep in mind,1 kW = 1000 J/s, and 1 hour = 3600 seconds. We'll convert the units to J/h: Total energy per hour = 1000 J/s * 10,000 * 3600 s = 36,000,000,000 J/h

We have the enthalpy change of the reaction (ΔH): 5640 kJ for one mole of sucrose produced. Let's find out the number of moles in 20 kg of sucrose produced per hour, based on the molecular weight of sucrose (C12H22O11): 12(12.01) + 22(1.01) + 11(16.00) = 342.3 g/mol. Number of moles = mass / molar mass Number of moles = 20,000 g / 342.3 g/mol = 58.44 mol Now, let's calculate the energy used in producing 20 kg (58.44 moles) of sucrose per hour: Energy used per hour = ΔH × number of moles of sucrose produced = 5640 kJ/mol × 58.44 mol = 329,319.36 kJ/h Convert to Joules: Energy used per hour = 329,319.36 kJ/h * 1000 J/kJ = 329,319,360 J/h

The efficiency of photosynthesis is the percentage of sunlight energy used to produce sucrose. We can find it using the energy used to produce sucrose and the total energy supplied by the sun: Efficiency = (Energy used per hour / Total energy per hour) * 100 = (329,319,360 J/h / 36,000,000,000 J/h) * 100 ≈ 0.915 % The efficiency of photosynthesis is approximately 0.915%.