Question

The two common sugars, glucose \(\left(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\right)\) and sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\), are both carbohydrates. Their standard enthalpies of formation are given in Table \(5.3\). Using these data, (a) calculate the molar enthalpy of combustion to \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l)\) for the two sugars; (b) calculate the enthalpy of combustion per gram of each sugar; (c) determine how your answers to part (b) compare to the average fuel value of carbohydrates discussed in Section \(5.8\).

Short answer

In summary, the molar enthalpy of combustion for glucose is -4176.8 kJ/mol and for sucrose is -8390.3 kJ/mol. The enthalpy of combustion per gram is -23.18 kJ/g for glucose and -24.51 kJ/g for sucrose. Both glucose and sucrose release more energy per gram when burned compared to the average fuel value of carbohydrates (16.7 kJ/g).

Step-by-step solution

a) Calculate the molar enthalpy of combustion

Firstly, we need to write down the combustion reactions for glucose and sucrose: For glucose: \(C_6H_{12}O_6(s) + 6O_2(g) \rightarrow 6CO_2(g) + 6H_2O(l)\) For sucrose: \(C_{12}H_{22}O_{11}(s) + 12O_2(g) \rightarrow 12CO_2(g) + 11H_2O(l)\) Now, let's use Hess's law to find the molar enthalpy of combustion: \(ΔH_{comb} = Σ_nΔH_f(products) - Σ_nΔH_f(reactants)\) Standard enthalpies of formation from Table 5.3: \(ΔH_{f,Glucose} = -1274 \, kJ/mol\) \(ΔH_{f,Sucrose} = -2220 \, kJ/mol\) \(ΔH_{f,CO_2} = -393.5 \, kJ/mol\) \(ΔH_{f,H_2O} = -285.8 \, kJ/mol\) For glucose: \(ΔH_{comb,Glucose} = [6(−393.5) + 6(−285.8)] - [1(-1274) + 6(0)]\) \(ΔH_{comb,Glucose} = -4176.8 \, kJ/mol\) For sucrose: \(ΔH_{comb,Sucrose} = [12(−393.5) + 11(−285.8)] - [1(-2220) + 12(0)]\) \(ΔH_{comb,Sucrose} = -8390.3 \, kJ/mol\)

b) Calculate enthalpy of combustion per gram of each sugar

To find the enthalpy of combustion per gram, we need to divide the molar enthalpy of combustion by the molar mass of each sugar. Molar mass of glucose: \(C_6H_{12}O_6\) : 6(12.01) + 12(1.01) + 6(16.00) = 180.18 \, g/mol Molar mass of sucrose: \(C_{12}H_{22}O_{11}\) : 12(12.01) + 22(1.01) + 11(16.00) = 342.34 \, g/mol For glucose: \(ΔH_{comb,Glucose (kJ/g)} = \frac{-4176.8}{180.18} = -23.18 \, kJ/g\) For sucrose: \(ΔH_{comb,Sucrose (kJ/g)} = \frac{-8390.3}{342.34} = -24.51 \, kJ/g\)

c) Compare to the average fuel value of carbohydrates

From Section 5.8, the average fuel value of carbohydrates is 4 Cal/g or 16.7 kJ/g. Compared to the average fuel value of carbohydrates (16.7 kJ/g), glucose has a higher enthalpy of combustion per gram (23.18 kJ/g) and sucrose has an even greater value (24.51 kJ/g). This means that both glucose and sucrose release more energy per gram when burned, compared to the average carbohydrate.

Key concepts

Standard Enthalpy of Formation

The standard enthalpy of formation, denoted as \( \Delta H_{f}^{\circ} \), is a crucial concept in thermodynamics and is defined as the change in enthalpy when one mole of a substance is formed from its elements in their most stable states under standard conditions (usually at a pressure of 1 bar and a temperature of 298 K). For example, the formation of glucose from carbon, hydrogen, and oxygen elements would have a certain standard enthalpy of formation value.

The standard enthalpy of formation for pure elements in their standard states is considered to be zero, because there is no change to the element itself. Conversely, for compounds, this value can be positive or negative depending on whether the formation process absorbs or releases energy, respectively. Understanding this concept is crucial in calculating reaction enthalpies, including combustion, which is our focus.

To illustrate, consider the formation of glucose (\(C_6H_{12}O_6(s)\)) referenced in the textbook exercise. Its standard enthalpy of formation is given as \( \Delta H_{f,Glucose} = -1274 \, kJ/mol \) in Table 5.3. This indicates that energy is released when glucose is formed from its constituent elements under standard conditions.

Hess's Law

Hess's law, which is also known as the law of constant heat summation, provides a pathway to calculating the overall enthalpy change (\( \Delta H \)) for a chemical reaction. It states that the total enthalpy change for a chemical reaction is the same, regardless of the number of intermediate steps or the route by which the reaction takes place, as long as the initial and final conditions are the same.

Applying Hess's law involves the principle of enthalpy conservation during a sequence of reactions. To find the enthalpy change for an overall reaction, you can sum the enthalpy changes of multiple steps that lead to the same final state. In the context of our exercise, the molar enthalpy of combustion for glucose and sucrose is calculated by summing up the standard enthalpies of formation of the products and reactants, representing the enthalpy change if we were to form these compounds directly from their constituent elements.

For glucose, the enthalpy of combustion calculation using Hess's law was shown as \( \Delta H_{comb,Glucose} = [6(\minus393.5) + 6(\minus285.8)] - [1(\minus1274) + 6(0)] \) resulting in \( \Delta H_{comb,Glucose} = \minus4176.8 \, kJ/mol\). This process demonstrates how Hess's law helps to derive the enthalpy of combustion from the enthalpies of formation.

Molar Enthalpy of Combustion

Molar enthalpy of combustion, denoted as \( \Delta H_{comb} \), is the amount of heat released when one mole of a substance completely burns in oxygen. It provides insight into the energy that can be obtained from fuel sources like carbohydrates (e.g., glucose and sucrose). The molar enthalpy of combustion is usually accompanied by the products, carbon dioxide (\(CO_2(g)\)) and water (\(H_2O(l)\)), the typical combustion outputs for organic compounds.

In the exercise, calculating the molar enthalpy of combustion involved using the chemical equations for the combustion of glucose and sucrose. The result is an indicator of how much heat is released per mole when these sugars combust. For example, glucose's calculation yielded \( \Delta H_{comb,Glucose} = \minus4176.8 \, kJ/mol \) indicating a significant release of energy.

This value is critical for various applications, such as dietary energy calculations or assessing potential fuels. Interestingly, when comparing molar enthalpy per gram, glucose and sucrose exceed the average fuel value for carbohydrates noted in the textbook, implying they are highly efficient energy sources.