Question

You are working in an alternative fuels laboratory and have been asked by your supervisor to determine the efficiency of photosynthesis for a plant that is being considered as a source of biofuels. You know that the plant produces the equivalent of 24 kg of sucrose per hectare per hour. You also know that sun supplies energy at a rate of 1.0 kilowatt per square meter of surface area. Assume that sucrose is produced by the reaction: 12 CO3 (g) + 11 H2O (g) ? C12H22O11 (s) + 12O2 (l) ? H = -5640 kJ

Answer 1

Explanation:

Moles of sucrose = $\frac{\text{mass of sucrose}}{\text{molecular weight}}$

                       = $\frac{24 kg \times 1000 g/kg}{342.3 g/mol}$

                        = $\frac{70.114 mol}{hr-ha}} \times \frac{1 ha}{10000 m^{2}} \times \frac{1 hr}{3600 s}$

                        = $1.948 \times 10^{-6} mol/(m^{2}s)$

Energy required for photosynthesis = 5640 kJ/mol

Energy needed to produce 24 kg sucrose

         = Energy required for photosynthesis x Moles of sucrose

          = $(5640 kJ/mol) \times [1.948 x 10^{-6} mol/(m^{2}s)]$

          = $0.01098 kJ/(m^{2}s)$

Energy supplied from sun = $1 \frac{kW}{m^{2}}$

                                            = 1 $kJ/(m^{2}s)$

Efficiency for photosynthesis = $\frac{\text{Energy needed to produce 24 kg sucrose}}{\text{Energy supplied from sun}}$

                              = $\frac{0.01098 kJ/(m^{2}s)}{1 kJ/(m^{2}s)}$

                             = $0.01098 \times 100$

                              = 1.098%

Thus, we can conclude that efficiency of photosynthesis for a plant that is being considered as a source of biofuels is 1.098 %.