Question

One glucose molecule, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(\mathrm{s}),\) is converted to two lactic acid molecules, \(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{COOH}(\mathrm{s})\) during glycolysis. Given the combustion reactions of glucose and lactic acid, determine the standard enthalpy for glycolysis. $$\begin{array}{r} \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(\mathrm{s})+6 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow 6 \mathrm{CO}_{2}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \\ \Delta H^{\circ}=-2808 \mathrm{kJ} \end{array}$$ $$\begin{aligned} \mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) & \mathrm{COOH}(\mathrm{s})+3 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \\ 3 \mathrm{CO}_{2}(\mathrm{g})+3 \mathrm{H}_{2} \mathrm{O}(1) & \Delta H^{\circ}=-1344 \mathrm{kJ} \end{aligned}$$

Short answer

The standard enthalpy for glycolysis is \(ΔH = -1120 kJ\) based on the given combustion reactions and the formula used to calculate the enthalpy of the glycolysis process.

Step-by-step solution

Identifying Reactions

Identify the combustion reactions of both glucose and lactic acid as provided:\n\nFor glucose: \n\n\(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(\mathrm{s})+6 \mathrm{O}_{2}(\mathrm{g}) \rightarrow 6 \mathrm{CO}_{2}(\mathrm{g})+6 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\), \( \Delta H^{\circ}=-2808 \mathrm{kJ}\)\n\nFor lactic acid: \n\n\(\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{COOH}(\mathrm{s})+3 \mathrm{O}_{2}(\mathrm{g}) \rightarrow 3 \mathrm{CO}_{2}(\mathrm{g})+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\), \( \Delta H^{\circ}=-1344 \mathrm{kJ}\)

Formulate Glycolysis Reaction

Formulate the reaction for the process of glycolysis involving glucose and lactic acid. The glycolysis reaction can be written as follows: \n\n\(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(\mathrm{s}) \rightarrow 2\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{COOH}(\mathrm{s})\)

Calculate Enthalpy for Glycolysis

Using Hess's Law, we can calculate the ΔH of the glycolysis process by subtracting the ΔH of glucose from twice the ΔH of lactic acid. That is, \n\nΔH (glycolysis) = 2ΔH (lactic acid) - ΔH (glucose)\n\nSubstituting the given ΔH values into this equation, we get: \n\nΔH (glycolysis) = 2*(-1344 kJ) - (-2808 kJ)\n\nSolving this will give the standard enthalpy for the process of glycolysis.

Key concepts

Glycolysis

Glycolysis is the process that breaks down glucose into smaller molecules, providing energy for the cell. It occurs in the cytoplasm of cells and does not require oxygen, making it an anaerobic process. This is usually the first step in cellular respiration, preparing glucose for further energy extraction in processes like the Krebs cycle and oxidative phosphorylation.

In the context of the exercise, one glucose molecule (\(\text{C}_6\text{H}_{12}\text{O}_6\text{(s)}\)) is converted to two molecules of lactic acid (\(\text{CH}_3\text{CH(OH)}\text{COOH(s)}\)). This conversion is crucial as it supplies energy in the absence of oxygen, such as during intense exercise in muscles.

• It provides \(ATP\), a direct energy source for cellular activities.
• The lactic acid produced can later be converted into glucose in the liver, a process known as gluconeogenesis.

The glycolysis process is energetically favorable, meaning it releases energy overall. However, in the problem we are discussing, we calculate the energy change during conversion by using known combustion reactions.

Hess's Law

Hess's Law is a fundamental principle in chemistry that allows us to calculate the enthalpy change of a reaction by using known enthalpy changes of other reactions. According to this law, the total enthalpy change for a chemical reaction is the same, no matter how many steps the reaction is carried out in.

This law is particularly useful when a direct measurement of the enthalpy change is not feasible. In our exercise, it helps us determine the enthalpy change for the glycolysis process, which is difficult to measure directly. By using Hess's Law, we can calculate the enthalpy change by utilizing the combustion values of glucose and lactic acid.

• We know the combustion enthalpies of glucose and lactic acid.
• We use Hess's Law to rearrange these reactions to match the glycolysis reaction.
• We then subtract the glucose combustion enthalpy from twice the enthalpy of lactic acid, as illustrated in the solution.

Therefore, we can solve for the enthalpy change entailed in converting glucose to lactic acid!

Enthalpy Calculation

Enthalpy calculation involves determining the heat change in reactions occurring at constant pressure. In chemical thermodynamics, it is often represented by \(\Delta H\) and is measured in kilojoules per mole (\(\text{kJ/mol}\)). Enthalpy changes provide insights into the energy dynamics of biochemical processes, such as glycolysis.

In the provided exercise, we have the combustion enthalpies of glucose and lactic acid which we use to find the glycolysis reaction enthalpy. Through Hess's Law, the enthalpy change for glycolysis is calculated by the following:
\[\Delta H_{\text{glycolysis}} = 2 \times \Delta H_{\text{lactic acid}} - \Delta H_{\text{glucose}}\]

• This equation considers the stoichiometry of the glycolysis reaction.
• Substituting the given values, we compute:
  \[\Delta H_{\text{glycolysis}} = 2(-1344 \text{kJ}) - (-2808 \text{kJ}) = -2688 \text{kJ} + 2808 \text{kJ}\]
  Which results in a positive value of \(120 \text{kJ}\).
• This indicates the process is slightly endothermic, absorbing a small amount of energy from the surroundings.

Overall, this calculation elucidates the energy transformation during glycolysis, highlighting the essential role of thermochemistry in understanding cellular processes.