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.

Step by step solution

01

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: \[E_\text{used} = n_\text{sucrose} \times \Delta H = 58.4\ \text{mol/h} \times 5640\ \text{ kJ/mol} = 329\:360\ \text{kJ/h}\]

02

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: \[E_\text{sun} = 3.6\ \text{MJ/h/m}^2 \times 10\:000\ \text{m}^2 = 36\:000\:000\ \text{kJ/h}\]

03

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%.

Unlock Step-by-Step Solutions & Ace Your Exams!

• Full Textbook Solutions

  Get detailed explanations and key concepts

• Unlimited Al creation

  Al flashcards, explanations, exams and more...

• Ads-free access

  To over 500 millions flashcards

• Money-back guarantee

  We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!