Question

The sun supplies energy at a rate of about 1.0 kilowatt per square meter of surface area (1 watt=1 J/s). The plants in an agricultural field produce the equivalent of 20. kg of sucrose (C₁₂H₂₂O₁₁) per hour per hectare (1 ha=10,000 m²). Assuming that sucrose is produced by the reaction

$\mathbf{12}\mathit{C}{\mathit{O}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{+}\mathbf{11}{\mathit{H}}_{\mathbf{2}}\mathit{O}\mathbf{\left(}\mathit{I}\mathbf{\right)}\mathbf{\to }{\mathit{C}}_{\mathbf{12}}{\mathit{H}}_{\mathbf{22}}{\mathit{O}}_{\mathbf{11}}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{+}{\mathbf{120}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{∆}\mathit{H}\mathbf{=}\mathbf{5640}\mathit{k}\mathit{J}$

Calculate the percentage of sunlight used to produce the sucrose—that is, determine the efficiency of photosynthesis.

Short answer

The efficiency of photosynthesis is 0.92%.

Step-by-step solution

Determination of energy required for 20 kg sucrose

The given reaction is,

$\mathbf{12}\mathit{C}{\mathit{O}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{+}\mathbf{11}{\mathit{H}}_{\mathbf{2}}\mathit{O}\mathbf{\left(}\mathit{I}\mathbf{\right)}\mathbf{\to }{\mathit{C}}_{\mathbf{12}}{\mathit{H}}_{\mathbf{22}}{\mathit{O}}_{\mathbf{11}}\mathbf{\left(}\mathit{s}\mathbf{\right)}\mathbf{+}\mathbf{12}{\mathit{O}}_{\mathbf{2}}\mathbf{\left(}\mathit{g}\mathbf{\right)}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{∆}\mathit{H}\mathbf{=}\mathbf{5640}\mathit{k}\mathit{J}$

From the reaction, it is concluded that 1 mol of sucrose requires 5640 kJ energy.

Number of moles of sucrose produced are,

$$\begin{gathered} \mathit{M}\mathit{o}\mathit{l}\mathit{e}\mathit{s}\mathbf{}\mathbf{=}\frac{\mathbf{m}\mathbf{a}\mathbf{s}\mathbf{s}}{\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{a}\mathbf{r}\mathbf{}\mathbf{m}\mathbf{a}\mathbf{s}\mathbf{s}} \\ \mathit{m}\mathit{o}\mathit{l}\mathit{e}\mathit{s}\mathbf{=}\frac{\mathbf{20}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}\mathbf{g}}{\mathbf{342}\mathbf{g}\mathbf{m}\mathbf{o}{\mathbf{l}}^{\mathbf{-}\mathbf{1}}}\mathbf{58}\mathbf{.}\mathbf{48}\mathit{m}\mathit{o}\mathit{l} \end{gathered}$$

The energy absorbed by the production of sucrose per hour per hectare is,

$\frac{\mathbf{(}\mathbf{58}\mathbf{.}\mathbf{48}\mathbf{m}\mathbf{o}\mathbf{l}\mathbf{)}\mathbf{\times }\mathbf{(}\mathbf{5640}\mathbf{\times }{\mathbf{10}}^{\mathbf{3}}\mathbf{J}\mathbf{)}}{\mathbf{3600}\mathbf{s}\mathbf{\times }\mathbf{10}\mathbf{,}\mathbf{000}{\mathbf{m}}^{\mathbf{2}}}\mathbf{=}\mathbf{9}\mathbf{.}\mathbf{161}\mathit{J}{\mathit{s}}^{\mathbf{-}\mathbf{1}}{\mathit{m}}^{\mathbf{-}\mathbf{1}}$

Determination of efficiency of photosynthesis

The amount of energy released by sun is 1kW/m² = 10³ Js^-1m^-2

Now, the efficiency of photosynthesis is,

$\begin{array}{rcl}\mathit{E}\mathit{f}\mathit{f}\mathit{i}\mathit{c}\mathit{i}\mathit{e}\mathit{n}\mathit{c}\mathit{y}& \mathbf{=}& \frac{\mathbf{O}\mathbf{u}\mathbf{t}\mathbf{p}\mathbf{u}\mathbf{t}\mathbf{}\mathbf{e}\mathbf{n}\mathbf{e}\mathbf{r}\mathbf{g}\mathbf{y}}{\mathbf{I}\mathbf{n}\mathbf{p}\mathbf{u}\mathbf{t}\mathbf{}\mathbf{e}\mathbf{n}\mathbf{e}\mathbf{r}\mathbf{g}\mathbf{y}}\mathbf{\times }\mathbf{100}\\ & \mathbf{=}& \frac{\mathbf{9}\mathbf{.}\mathbf{161}}{{\mathbf{10}}^{\mathbf{3}}}\mathbf{\times }\mathbf{100}\\ & \mathbf{=}& \mathbf{0}\mathbf{.}\mathbf{92}\mathbf{\%}\end{array}$

Thus, the efficiency of photosynthesis is 0.92%.