Answer:
0.092 %
Explanation:
The equation of the reaction can be computed as :
[tex]12CO_2_{(g)} + 11H_2O_{(l)} \to C_{12}H_{22}O_{11} + 12O_{2_(g)}[/tex]
[tex]\Delta H = 5645 \ kJ[/tex]
recall that; the number of moles = [tex]\dfrac{mass}{molar \ mass}[/tex]
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
[tex]=\dfrac{0.20g \ of \ sucrose }{m^2 \ 3600 \ s}\times \dfrac{1 \ mol}{342.34 \ g}\times 5.645 kJ/mol[/tex]
[tex]= 9.16 \times 10^{-4} \ kJ/m^2 s[/tex]
Given that the sun supplies about 1.0 kilowatt, to KJ/m² s, we have:
[tex]1.0 \dfrac{kW}{m^2 }= 1.0 \dfrac{kJ}{m^2 s}[/tex]
Finally, the percentage of sunlight used to produce sucrose :
= [tex]\dfrac{9.16 \times 10^{-4} \ kJ/m^2 \ s}{1.0 \ kJ/m^2 . s} \times 100\%[/tex]
= 0.092 %
