Problem 24 - Chemistry: Principles and Reactions

Question

In photosynthesis, the following reaction takes place: \(6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 6 \mathrm{O}_{2}(g)+\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s) \quad \Delta H=2801 \mathrm{~kJ}\) (a) Calculate \(\Delta H\) when one mole of \(\mathrm{CO}_{2}\) reacts. (b) How many kilojoules of energy are liberated when \(15.00 \mathrm{~g}\) of glucose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\), is burned in oxygen?

Step-by-Step Solution

Answer

(a) The enthalpy change when one mole of CO2 reacts is 467 kJ. (b) 38.85 kJ of energy are liberated when 15.00 g of glucose is burned in oxygen.

Step 1: Calculate ΔH when one mole of CO2 reacts

Given that the balanced equation is as follows: $6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 6 \mathrm{O}_{2}(g)+\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)$ and its corresponding enthalpy change is: \(\Delta H=2801 \mathrm{~kJ}\) Since the enthalpy change is given for 6 moles of CO2, we can calculate the enthalpy change for one mole of CO2 by dividing by 6: \(\Delta H_{1 \mathrm{~mol} \mathrm{CO}_2} = \frac{2801 \mathrm{~kJ}}{6}\)

Step 2: Solve for ΔH per one mole of CO2

\(\Delta H_{1 \mathrm{~mol} \mathrm{CO}_2} = \frac{2801 \mathrm{~kJ}}{6} = 467 \mathrm{~kJ}\) (a) So, the enthalpy change when one mole of CO2 reacts is 467 kJ.

Step 3: Calculate the energy liberated when 15.00 g of glucose is burned

We are given the mass of glucose and need to calculate the energy liberated. First, convert the mass of glucose to moles by dividing by its molar mass. The molar mass of glucose is 180.16 g/mol. \(\mathrm{Moles~of~glucose} = \frac{15.00 \mathrm{~g}}{180.16 \mathrm{~g/mol}}\)

Step 4: Solve for moles of glucose

\(\mathrm{Moles~of~glucose} = \frac{15.00 \mathrm{~g}}{180.16 \mathrm{~g/mol}} = 0.08327 \mathrm{~mol}\) Now use the stoichiometry of the balanced equation to find the energy liberated per mole of glucose burned: Energy liberated = moles of glucose × ΔH per mole of glucose

Step 5: Solve for energy liberated

Since we have the enthalpy change when one mole of CO2 reacts (Step 2), and one mole of glucose is produced for every mole of CO2 reacted, the energy liberated per mole of glucose is also 467 kJ. Energy liberated = \(0.08327 \mathrm{~mol} × 467 \mathrm{~kJ/mol} = 38.85 \mathrm{~kJ}\) (b) Thus, 38.85 kJ of energy are liberated when 15.00 g of glucose is burned in oxygen.

Key Concepts

Understanding Enthalpy Change Calculation

In the context of chemical reactions, enthalpy change, designated as \( \Delta H \), represents the heat absorbed or released. When speaking about photosynthesis or combustion, these processes involve energy changes that can be quantified with enthalpy changes.

To calculate the enthalpy change for a single molecule or mole of a reactant or product, it's necessary to begin with a balanced chemical equation. For instance, in our photosynthesis example, the reaction has an overall enthalpy change of 2801 kJ. This quantity corresponds to the conversion of six moles of carbon dioxide (\( \mathrm{CO}_2 \) gas) into glucose (\( \mathrm{C}_6\mathrm{H}_{12}\mathrm{O}_6 \) solid) and oxygen (\( \mathrm{O}_2 \) gas). To find the enthalpy change associated with the reaction of one mole of \( \mathrm{CO}_2 \) we divide the total enthalpy change by the stoichiometric coefficient of \( \mathrm{CO}_2 \) in the balanced equation, which is six in this case.

Calculating the \( \Delta H \) for a single mole then becomes a straightforward division: \( \Delta H_{1 \mathrm{~mol} \mathrm{CO}_2} = \frac{2801 \mathrm{~kJ}}{6} \) yielding 467 kJ per mole of \( \mathrm{CO}_2 \) reacted. This provides a clearer understanding of just how much energy is involved when photosynthesis occurs at the molecular level.

Stoichiometry in Chemical Reactions

Stoichiometry is the section of chemistry that involves using balanced equations to determine the proportions of reactants and products. It's like a recipe that tells you how much of each ingredient you need for your desired outcome.

In our example of photosynthesis, stoichiometry tells us that six molecules of \( \mathrm{CO}_2 \) react with an equal number of water molecules to produce a glucose molecule and six molecules of oxygen. The coefficients in the balanced equation (6,6,6,1) directly inform us about these ratios. When the exercise asks to calculate energy liberated from burning glucose, we rely on stoichiometry to first convert mass to moles, and then apply the enthalpy change per mole to find the total energy.

The stoichiometric conversion involves dividing the mass of glucose by its molar mass, leading us to the amount in moles: \( \mathrm{Moles~of~glucose} = \frac{15.00 \mathrm{~g}}{180.16 \mathrm{~g/mol}} \) which is essential for identifying the amount of energy released during the reaction. This is a pivotal step in understanding any chemical process and crucial for effectively performing these calculations.

Energy Liberation in Reactions

Chemical reactions are often accompanied by energy changes; some absorb energy while others release it. In the case of burning glucose (combustion), energy is released into the surroundings, an exothermic process.

Using the principles of stoichiometry, we've deduced that 38.85 kJ of energy is liberated when 15.00 g of glucose is burned. This value is found by multiplying the number of moles of glucose, 0.08327, by the enthalpy change per mole of glucose, 467 kJ. Thus, for every mole of glucose combusted, 467 kJ of energy are released in the form of heat.

This concept of energy liberation is not only important in biochemical processes like photosynthesis and respiration but also in various industries, where exothermic reactions are harnessed for energy production. Understanding how to calculate this energy release is essential for scientists and engineers who work on developing sustainable and efficient methods of energy generation.