Question

A key part of the carbon cycle is the fixation of \(\mathrm{CO}_{2}\) by photosynthesis to produce carbohydrates and oxygen gas. (a) Using the formula \(\left(\mathrm{CH}_{2} \mathrm{O}\right)_{n}\) to represent a carbohydrate, write a balanced equation for the photosynthetic reaction. (b) If a tree fixes \(48 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) per day, what volume of \(\mathrm{O}_{2}\) gas measured at 1.0 atm and \(78^{\circ} \mathrm{F}\) does the tree produce per day? (c) What volume of air ( \(0.040 \mathrm{~mol} \% \mathrm{CO}_{2}\) ) at the same conditions contains this amount of \(\mathrm{CO}_{2} ?\)

Short answer

The photosynthesis equation is \( \text{CO}_2 + \text{H}_2 \text{O} \rightarrow \text{(CH}_2 \text{O}) + \text{O}_2 \). The volume of \(\text{O}_2\) produced is 26.5L. The volume of air containing 48g \(\text{CO}_2\) is 66,729L.

Step-by-step solution

Write the Balanced Photosynthesis Equation

Photosynthesis converts \(\text{CO}_2\) and water into carbohydrates and oxygen. The general equation is: \[n \text{CO}_2 + n \text{H}_2 \text{O} \rightarrow (\text{CH}_2 \text{O})_n + n \text{O}_2.\] Using 1 mole of carbohydrate as \((\text{CH}_2 \text{O})_n\), we get: \[ \text{CO}_2 + \text{H}_2 \text{O} \rightarrow \text{(CH}_2 \text{O}) + \text{O}_2.\]

Calculate the Volume of \(\text{O}_2\) Produced

Each mole of \(\text{CO}_2\) fixed produces 1 mole of \(\text{O}_2\). Fixing 48g of \(\text{CO}_2\) corresponds to: \[\frac{48 \text{g}}{44 \text{g/mol}} = 1.09 \text{ mol of CO}_2\] (since the molar mass of \(\text{CO}_2\) is 44g/mol). Using the Ideal Gas Law \[V = nRT/P\] with R = 0.0821 L atm/mol K, T = 78°F = 298 K, and P = 1 atm, we find: \[V = (1.09 \text{ mol} \times 0.0821 \text{ L atm/mol K} \times 298 \text{ K}) / 1 \text{ atm} = 26.5 \text{ L of O}_2.\]

Calculate the Volume of Air Containing \(48 g \text{CO}_2\)

Since air is 0.040 mol% \(\text{CO}_2\), use the relation that 1 mol% means 1 mol in 100 mol of air. So, 0.040 mol% means \[ 0.040 \text{ mol CO}_2 = 1 \text{ mol air} / 100 = 1/2500 \text{ mol CO}_2 per mol air (approx.)\]. The fixed \(\text{CO}_2\) is 1.09 mol, thus: \[ 1.09 \text{ mol CO}_2 \times 2500 \text{ mol air/ mol CO}_2 = 2725 \text{ mol air.}\] Convert mol to volume: \[ V = nRT/P = (2725 \text{ mol} \times 0.0821 \text{ L atm/mol K} \times 298 \text{ K})/1 \text{ atm} = 66,729 \text{ L air.}\]

Key concepts

Photosynthesis Equation

Photosynthesis is an essential process in the carbon cycle that turns carbon dioxide (\text{CO}_2) and water (\text{H}_2O) into carbohydrates and oxygen (\text{O}_2). The general equation for photosynthesis can be written as: \[ n \text{CO}_2 + n \text{H}_2 \text{O} \rightarrow (\text{CH}_2 \text{O})_n + n \text{O}_2 \] Here, \text{(CH}_2\text{O})_n represents a general carbohydrate. Essentially, each mole of \text{CO}_2 fixed by a plant will produce one mole of carbohydrate and one mole of \text{O}_2. Therefore, understanding this process helps us grasp how plants contribute to oxygen production and the overall carbon cycle.

Oxygen Production

Oxygen production is a crucial outcome of photosynthesis. Each fixed mole of \text{CO}_2 through photosynthesis results in a mole of \text{O}_2 being released. For example, if a tree fixes 48g of \text{CO}_2 per day, we first calculate the moles of \text{CO}_2: \[ \frac{48 \text{g}}{44 \text{g/mol}} = 1.09 \text{ mol CO}_2 \] Because one mole of fixed \text{CO}_2 yields one mole of \text{O}_2, the tree will produce 1.09 mol of \text{O}_2 per day. This conversion highlights the direct relationship between \text{CO}_2 fixation and \text{O}_2 production, emphasizing the importance of plants in maintaining our atmosphere.

Ideal Gas Law

The Ideal Gas Law is used to relate the volume, temperature, and pressure of a gas sample. The formula is: \[ V = nRT/P \] where

• \text{V} is the volume
• \text{n} is the number of moles
• \text{R} is the gas constant (0.0821 L atm / mol K)
• \text{T} is temperature in Kelvin
• \text{P} is pressure in atm

To find the volume of oxygen produced at 1 atm and 78°F (298 K), we use: \[ V = (1.09 \text{ mol} \times 0.0821 \text{ L atm/mol K} \times 298 \text{ K}) / 1 \text{ atm} = 26.5 \text{ L of O}_2 \] This calculation demonstrates how we use the Ideal Gas Law to determine the volume of gases involved in photosynthesis and air mixtures.

Molar Mass Calculation

Molar mass plays a vital role in calculating the moles of a substance from its mass. For \text{CO}_2, the molar mass is calculated as the sum of the atomic masses of its constituent atoms:

• 1 carbon atom (C) = 12g/mol
• 2 oxygen atoms (O) = 2x16g/mol = 32g/mol

Therefore, the molar mass of \text{CO}_2 is: \[ 12 \text{g/mol} + 32 \text{g/mol} = 44 \text{g/mol} \] Using this value, we can accurately determine the moles of \text{CO}_2 when given a mass, such as 48 grams. The conversion goes as: \[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} = \frac{48 \text{g}}{44 \text{g/mol}} = 1.09 \text{ mol} \] This step is essential in quantifying how much \text{CO}_2 is fixed and subsequently how much \text{O}_2 is produced.

Carbon Dioxide Fixation

Carbon dioxide fixation is a key process in the carbon cycle where \text{CO}_2 is converted into organic molecules by plants during photosynthesis. This process reduces atmospheric \text{CO}_2 levels and provides oxygen. For instance, with 48g of \text{CO}_2 being fixed, the number of moles fixed is: \[ \frac{48 \text{g}}{44 \text{g/mol}} = 1.09 \text{ mol} \] Knowing that air contains 0.040 mol% \text{CO}_2, we need to find the volume of air containing 1.09 mol \text{CO}_2. This means: \[ 1.09 \text{ mol CO}_2 \times 2500 \text{ mol air / mol CO}_2 = 2725 \text{ mol air} \] Using the Ideal Gas Law again to convert moles to volume: \[ V = (2725 \text{ mol} \times 0.0821 \text{ L atm/mol K} \times 298 \text{ K}) / 1 \text{ atm} = 66,729 \text{ L air} \] This calculation highlights how much air is involved in fixing the given amount of \text{CO}_2, showcasing the broader impact on environmental scales.