Q22.74CP - Chemistry: Molecular Nature Of Matter And Change

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 ${\mathbf{\text{CO}}}_{\mathbf{\text{2}}}$ ?

Step-by-Step Solution

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 $26.7 L$ .

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

Step 1: 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${\text{nCO}}_{\text{2}}{\text{(g) + nH}}_{\text{2}}\text{O(l)}\to {\text{(CH}}_{\text{2}}{\text{O)}}_{\text{n}}{\text{ + nO}}_{\text{2}}\text{(g)}$  .

Step 2: 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 –

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

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 $pa$  –

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

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

Step 3: 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 –

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

Applying the ideal gas law is –

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

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