The sun supplies about 1.0 kilowatt of energy for each square meter of surface area (1.0 kW/m^2 where a watt = 1 kJ/s) Plants produce the equivalent of about 0.20g of sucrose (C_12H_22O_11) per hour per square meter. Assuming that the sucrose is produced as follows, calculate the percentage of sunlight used to produce sucrose12CO2 (g) + 11H2O (I) --> C12H22O11 + 12O2(g) deltaH = 5645 kJ

0.092 %

Explanation:

The equation of the reaction can be computed as :

recall that; the number of moles =

By applying the method of enthalpy of combustion for sucrose at the same time changing the time from hours to seconds, we can determine the total energy output.

i.e

Given that the sun supplies about 1.0 kilowatt, to KJ/m² s, we have:

Finally, the percentage of sunlight used to produce sucrose :

=

= 0.092 %

wher is the rest of the question