Question

A key part of the carbon cycle is the fixation of ${\mathbf{\text{CO}}}_{\mathbf{\text{2}}}$by photosynthesis to produce carbohydrates and oxygen gas.

(a) Using the formula$\mathbf{(}\mathit{C}{\mathit{H}}_{\mathbf{2}}\mathit{O}{\mathbf{)}}_{\mathbf{\text{n}}}$ to represent a carbohydrate, write a balanced equation for the photosynthetic reaction.

(b) If a tree fixes$\mathbf{\text{48 g}}$ of $\mathit{C}{\mathit{O}}_{\mathbf{2}}$per day, what volume of ${\mathit{O}}_2$gas measured at $\mathbf{\text{1.0 atm}}$and${\mathbf{\text{78}}}^{\mathbf{\text{o}}}\mathbf{\text{F}}$ does the tree produce per day?

(c) What volume of air ( ${\mathbf{\text{0.035 mol \% CO}}}_{\mathbf{\text{2}}}$) at the same conditions contains this amount of${\mathbf{\text{CO}}}_{\mathbf{\text{2}}}$ ?

Short answer

(a) The balanced equation for the photosynthetic reaction is .

${\text{nCO}}_{\text{2}}{\text{(g) + nH}}_{\text{2}}\text{O(l)}\to {\text{(CH}}_{\text{2}}{\text{O)}}_{\text{n}}{\text{+ nO}}_{\text{2}}\text{(g)}$

(b) The volume of oxygen produced by the tree per day is $26.7L$.

(c) The volume of air that contains carbon dioxide is $7.6\times {10}^{4}L$.

Step-by-step solution

Balanced Equation

(a)

Photosynthesis can be described with the following reaction –

${\text{nCO}}_{\text{2}}{\text{(g) + nH}}_{\text{2}}\text{O(l)}\to {\text{(CH}}_{\text{2}}{\text{O)}}_{\text{n}}{\text{+ nO}}_{\text{2}}\text{(g)}$

Plants fixate carbon dioxide and produce carbohydrates and oxygen using light energy and converting it into chemical energy stored in the carbohydrates.

Therefore, the reaction is obtained as${\text{nCO}}_{\text{2}}{\text{(g) + nH}}_{\text{2}}\text{O(l)}\to {\text{(CH}}_{\text{2}}{\text{O)}}_{\text{n}}{\text{+ nO}}_{\text{2}}\text{(g)}$ .

Calculation for Volume

(b)

By looking at the chemical equation it is observed that the mol ratio of oxygen to carbon dioxide is. So, calculate the chemical amount of${\text{CO}}_{\text{2}}$that the tree fixes per day –

$$\begin{gathered} \text{m}\left({\text{CO}}_{\text{2}}\right)\text{= 48g} \\ \frac{{\displaystyle \text{M}\left({\text{CO}}_{\text{2}}\right){\text{= 44.009gmol}}^{\text{- 1}}}}{{\displaystyle \text{n}\left({\text{CO}}_{\text{2}}\right)\text{=}\frac{\text{m}\left({\text{CO}}_{\text{2}}\right)}{\text{M}\left({\text{CO}}_{\text{2}}\right)}}} \\ \text{n}\left({\text{CO}}_{\text{2}}\right)\text{=}\frac{48g}{{\displaystyle {\text{4.009gmol}}^{\text{- 1}}}} \\ \text{n}\left({\text{CO}}_{\text{2}}\right)\text{=1.09mol} \end{gathered}$$

Next, we can calculate the volume of oxygen produced by using ideal gas law. Before that, we have toconvert the temperature from${}^{°}F$to$K$and pressure from$atm$to$pa$–

$$\begin{gathered} \text{T = 78}{}^{°}F\text{=}\left(\text{(78 - 32)×}\frac{\text{5}}{\text{9}}\text{+ 273.15}\right)\text{K= 298.71K} \\ \text{p = 1 atm = 101325 Pa} \\ \text{n}\left({\text{O}}_{\text{2}}\right)\text{= n}\left({\text{CO}}_{\text{2}}\right) \\ \text{n}\left({\text{O}}_{\text{2}}\right)\text{=1.09mol} \\ \text{pV = nRT} \\ \text{V =}\frac{\text{nRT}}{\text{p}} \end{gathered}$$

Therefore, the value for volume is obtained as$26.7L$ .

Calculation for Volume

(c)

If$\text{0.035 mol\%}$of air is$C{O}_2$, we can calculate the volume of air which contains$48g$of carbon dioxide byusing the formula for mole fraction$x_i$and ideal gas law –

$$\begin{gathered} \text{m}\left({\text{CO}}_{\text{2}}\right)\text{=48g} \\ n\left({\text{CO}}_{\text{2}}\right)\text{=1.09mol} \\ {\text{x}}_{{\text{CO}}_{\text{2}}}\text{=}\frac{\text{n}\left({\text{CO}}_{\text{2}}\right)}{\text{n( total )}}\text{×100\%} \\ \text{n(total) =}\frac{\text{n}\left({\text{CO}}_{\text{2}}\right)}{{\text{x}}_{{\text{CO}}_{\text{2}}}}\text{×100\%} \\ \text{n(total) =}\frac{\text{1.09mol}}{\text{0.035\%}}\text{×100\%} \\ \text{n(total) = 3114mol} \end{gathered}$$

Applying the ideal gas law is –

role="math" localid="1663330032456" $$\begin{gathered} \text{n(air) = 3114mol} \\ \text{p = 101325Pa} \\ \text{T = 298.71K} \\ \text{pV = nRT} \\ \text{V =}\frac{\text{nRT}}{\text{p}} \\ \text{V(air) =}\frac{{\displaystyle {\text{3114mol×8.314 JK}}^{\text{- 1}}{\text{mol}}^{\text{- 1}}\text{×298.71K}}}{{\displaystyle \text{101325Pa}}} \\ {\text{V(air) = 76.3m}}^{\text{3}} \\ ={\text{7.6×10}}^{\text{4}}\text{L} \end{gathered}$$

Therefore, the value for volume is obtained as${\text{7.6×10}}^{\text{4}}\text{L}$ .