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.

Step by step solution

01

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: \[E_\text{sucrose} = \Delta H \times n_\text{sucrose} = 5640 \: \text{kJ} \times 58.4 \: \text{mol} = 328896 \: \text{kJ}\]

02

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: \[E_\text{Sun,ha} = E_\text{Sun,m} \times 10,000 \: \text{m}^2 = 3600 \: \text{kJ/m}^2 \times 10,000 \: \text{m}^2 = 36000000 \: \text{kJ}\]

03

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 \(0.914\% \).

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!