Question

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

Answer 1

Answer:

0.092 %

Explanation:

The equation of the reaction can be computed as :

$12CO_2_{(g)} + 11H_2O_{(l)} \to C_{12}H_{22}O_{11} + 12O_{2_(g)}$

$\Delta H = 5645 \ kJ$

recall that; the number of moles = $\dfrac{mass}{molar \ mass}$

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

$=\dfrac{0.20g \ of \ sucrose }{m^2 \ 3600 \ s}\times \dfrac{1 \ mol}{342.34 \ g}\times 5.645 kJ/mol$

$= 9.16 \times 10^{-4} \ kJ/m^2 s$

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

$1.0 \dfrac{kW}{m^2 }= 1.0 \dfrac{kJ}{m^2 s}$

Finally, the percentage of sunlight used to produce sucrose :

= $\dfrac{9.16 \times 10^{-4} \ kJ/m^2 \ s}{1.0 \ kJ/m^2 . s} \times 100\%$

= 0.092 %