Sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) is produced by plants as follows: $$ \begin{aligned} 12 \mathrm{CO}_{2}(g)+11 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}+12 \mathrm{O}_{2}(g) \\ \Delta H=5645 \mathrm{~kJ} \end{aligned} $$ About \(4.8 \mathrm{~g}\) of sucrose is produced per day per square meter of the earth's surface. The energy for this endothermic reaction is supplied by the sunlight. About \(0.1 \%\) of the sunlight that reaches the earth is used to produce sucrose. Calculate the total energy the sun supplies for each square meter of surface area. Give your answer in kilowatts per square meter \(\left(\mathrm{kW} / \mathrm{m}^{2}\right.\) where $\left.1 \mathrm{~W}=1 \mathrm{~J} / \mathrm{s}\right).$

Since we know the reaction of sucrose production, we can use stoichiometry to calculate the energy requirement. First, we must find the amount of energy required to produce 4.8 grams of sucrose per day. We are given the value of \(\Delta H = 5645 kJ\) for the entire reaction. The molar mass of sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) is: \(12(12.01) + 22(1.01) + 11(16.00) = 144.12 + 22.22 + 176 = 342.34 \mathrm{~g/mol}\) Now we find the moles of sucrose produced daily per square meter: \(\frac{4.8 \mathrm{~g}}{342.34 \mathrm{~g/mol}} = 0.014 \mathrm{~mol}\) Next, we calculate the energy required to produce 0.014 mol of sucrose per day per square meter using the given value of \(\Delta H\): Energy = \(\Delta H \times \mathrm{moles}\) Energy = \(5645 \mathrm{~kJ} \times 0.014 \mathrm{~mol} = 79.03 \mathrm{~kJ}\)

We are given that only 0.1% of the sunlight that reaches the earth is used to produce sucrose. Therefore, we can find the total energy supplied by the sun daily per square meter by dividing the energy used to produce sucrose (79.03 kJ) by the efficiency (0.1%): Total solar energy = \(\frac{Energy \, for \, sucrose \, production}{Percentage \, efficiency}\) Total solar energy = \(\frac{79.03 \mathrm{~kJ}}{0.001} = 79030 \mathrm{~kJ}\)

We have calculated the total solar energy supplied per day per square meter in kJ. Now, we need to convert this to kW/m². First, we should convert kJ to J: \(79030 \mathrm{~kJ} = 79030 \times 1000 = 79,030,000 \mathrm{~J}\) Since 1 second has 86400 seconds, now we can convert the energy to watts per square meter: Energy rate = \(\frac{Energy}{Time}\) Energy rate = \(\frac{79,030,000 \mathrm{~J}}{86400 \mathrm{~s}} = 914.583 \mathrm{~W/m}^2\) Finally, we can convert the energy rate to kilowatts per square meter: Total solar energy = \(914.583 \mathrm{~W/m}^2 \times \frac{1\, kW}{1000 \, W} = 0.9146 \mathrm{~kW/m}^2\) The total energy the sun supplies daily for each square meter of surface area is approximately \(0.9146 \mathrm{~kW/m}^2\).