Question

How do I determine the efficiency for photosynthesis?

For example: The sun supplies energy at the rate of about 1.0 kilowatts per square meter of surface area (1 watt = 1 J/s). The plants in an agricultural field produce the equivalent of 20. kg of sucrose(C12H22O11) per hour per hectare(1 ha = 10,000 m^2). Assume that sucrose is produced by the reaction given below.

12 CO2(g) + 11 H2O(l) => C12H22O11(s) + 12O2(g) H = -5640 kJ

Instructions:

1. I need a detailed step-by-step explanation because I would like to understand this.

2. Please include any common mistakes students make when solving for this type of problem so that I can understand correct and incorrect reasoning behind this problem.

Answer 1

Moles of sucrose = mass of sucrose/molecular weight

= (20kg x 1000g/kg) / (342.3 g/mol)

= [58.428 mol/(hr-ha)] x (1ha/10000m2) x (1hr/3600s)

= 1.623 x 10^-6 mol/(m2-s)

Energy required for photosynthesis = 5640 kJ/mol

Energy needed to produce 20 kg sucrose

= Energy required for photosynthesis x Moles of sucrose

= (5640 kJ/mol) x [1.623 x 10^-6 mol/(m2-s)]

= 0.00915372 kJ/(m2-s)

Energy supplied from sun = 1 kW/m2 = 1kJ/(m2-s)

Efficiency for photosynthesis

= (Energy needed to produce 20 kg sucrose) / (Energy supplied from sun)

= [0.00915372 kJ/(m2-s)] / [1kJ/(m2-s)]

= 0.00915 x 100

= 0.915 %

Points to avoid mistakes -

Proper unit conversion

Hour into seconds and Hectare into m2

Convert mass into moles for the chemical reaction

Write with proper units as per the question given.