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

Step-by-step explanation:

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

=
`(24 kg * 1000 g/kg)/(342.3 g/mol)`

=
`(70.114 mol)/(hr-ha)} * (1 ha)/(10000 m^(2)) * (1 hr)/(3600 s)`

=
`1.948 * 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) * [1.948 x 10^(-6) mol/(m^(2)s)]`

=
`0.01098 kJ/(m^(2)s)`

Energy supplied from sun =
`1 (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}}`

=
`(0.01098 kJ/(m^(2)s))/(1 kJ/(m^(2)s))`

=
`0.01098 * 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 %.