Question

Under anaerobic conditions, glucose is broken down in muscle tissue to form lactic acid according to the reaction \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} \rightarrow 2 \mathrm{CH}_{3} \mathrm{CHOHCOOH}\). Thermodynamic data at \(T=298 \mathrm{K}\) for glucose and lactic acid are given in the following table: $$\begin{array}{lccc} & \Delta \boldsymbol{H}_{f}^{\circ}\left(\mathbf{k J} \mathbf{~ m o l}^{-1}\right) & C_{P, m}\left(\mathbf{J} \mathbf{K}^{-1} \mathbf{m o l}^{-\mathbf{1}}\right) & S_{m}^{\circ}\left(\mathbf{J} \mathbf{K}^{-\mathbf{1}} \mathbf{m o l}^{-\mathbf{1}}\right) \\ \hline \text { Glucose } & -1273.1 & 219.2 & 209.2 \\ \text { Lactic } & -673.6 & 127.6 & 192.1 \end{array}$$ Calculate, \(\Delta S\) for the system, the surroundings, and the universe at \(T=325 \mathrm{K}\). Assume the heat capacities are constant \\[ \text { between } T=298 \mathrm{K} \text { and } T=330 . \mathrm{K} \\]

Short answer

The entropy changes at \(T = 325K\) are: - For the system: \(211\,J/(mol\cdot K)\) - For the surroundings: \(228\,J/(mol\cdot K)\) - For the universe: \(439\,J/(mol\cdot K)\)

Step-by-step solution

1. Calculate the Standard Enthalpy Change

To calculate the standard enthalpy change \(\Delta H\) for the given reaction, we'll use the standard enthalpies of formation \(\Delta H_{f}^{\circ}\) for the products and the reactants: \(\Delta H = \Sigma n_{products} \Delta H_{f,products}^\circ - \Sigma m_{reactants} \Delta H_{f,reactants}^\circ\) For this reaction, \(\Delta H = 2\Delta H_{f,LacticAcid}^\circ - 1\Delta H_{f,Glucose}^\circ\) Using the given data from the table: \(\Delta H = 2(-673.6\,kJ/mol) - 1(-1273.1\, kJ/mol)\) \(\Delta H = -1347.2 + 1273.1 = -74.1\,kJ/mol\)

2. Calculate the Standard Entropy Change at 298K

To calculate the standard entropy change \(\Delta S\) for the reaction at 298K, we'll use the standard molar entropies \(S_{m}^{\circ}\) for the products and the reactants: \(\Delta S^{\circ}_{298} = \Sigma n_{products} S_{m,products}^{\circ} - \Sigma m_{reactants} S_{m,reactants}^{\circ}\) For this reaction, \(\Delta S^{\circ}_{298} = 2S_{m,LacticAcid}^{\circ} - 1S_{m,Glucose}^{\circ}\) Using the given data from the table: \(\Delta S^{\circ}_{298} = 2(192.1\,J/(mol\cdot K)) - 1(209.2\, J/(mol\cdot K))\) \(\Delta S^{\circ}_{298} = 384.2 - 209.2 = 175\,J/(mol\cdot K)\)

3. Calculate Entropy Change for the System at 325K

To calculate the entropy change for the system at 325K, we need to account for both the standard entropy change \(\Delta S^{\circ}_{325}\) and the change in entropy due to the temperature change from 298K to 325K, using the heat capacities \(C_{P,m}\): \(\Delta S_{system} = \Delta S^{\circ}_{298} + \Delta S(T)\) The temperature-dependent term is given by: \(\Delta S(T) = \Sigma n_{products} C_{P,m,products}(T) - \Sigma m_{reactants} C_{P,m,reactants}(T)\) For this reaction, we can use the heat capacities \(C_{P, m}\) given in the table: \(\Delta S(T) = 2(127.6\, J/(mol\cdot K)) - 219.2\, J/(mol\cdot K)\) \(\Delta S(T) = 36\, J/(mol\cdot K)\) Therefore, the entropy change for the system at \(T = 325K\) is: \(\Delta S_{system} = 175 + 36 = 211\,J/(mol\cdot K)\)

4. Calculate Entropy Change for the Surroundings at 325K

The entropy change for the surroundings is given by the heat released in the reaction divided by the temperature: \(\Delta S_{surroundings} = -\frac{\Delta H}{T}\) Using the calculated standard enthalpy change \(\Delta H\) and the given temperature \(T = 325K\): \(\Delta S_{surroundings} = -\frac{-74.1\,kJ/mol}{325\,K}\) \(\Delta S_{surroundings} = 0.228\,kJ/(mol\cdot K) = 228\,J/(mol\cdot K)\)

5. Calculate Entropy Change for the Universe at 325K

The entropy change for the universe is the sum of the entropy changes for the system and the surroundings: \(\Delta S_{universe} = \Delta S_{system} + \Delta S_{surroundings}\) Using the calculated values from steps 3 and 4: \(\Delta S_{universe} = 211\,J/(mol\cdot K) + 228\,J/(mol\cdot K)\) \(\Delta S_{universe} = 439\,J/(mol\cdot K)\) So, the entropy changes at \(T = 325K\) are: - For the system: \(211\,J/(mol\cdot K)\) - For the surroundings: \(228\,J/(mol\cdot K)\) - For the universe: \(439\,J/(mol\cdot K)\)

Key concepts

Enthalpy Change

Enthalpy change, denoted as \( \Delta H \), is a central concept in thermodynamics that represents the heat content during a chemical reaction under constant pressure. This value can be thought of as the total energy change, accounting for both internal energy and the energy associated with pressure and volume. In chemical reactions, the enthalpy change tells us how much heat is absorbed or released. \( \Delta H \) is calculated using the formula: \[ \Delta H = \Sigma n_{products} \Delta H_{f,products}^{\circ} - \Sigma m_{reactants} \Delta H_{f,reactants}^{\circ} \] where \( \Delta H_{f}^{\circ} \) indicates the standard enthalpies of formation for the respective substances at standard conditions (usually 298 K and 1 atm). A negative \( \Delta H \) indicates an exothermic process, where energy is released, while a positive value suggests an endothermic process, involving the absorption of energy.

In our exercise involving the conversion of glucose into lactic acid, the calculation of \( \Delta H \) revealed that the reaction is exothermic, releasing 74.1 kJ/mol at standard conditions, indicating that heat is being released to the surroundings as glucose is converted to lactic acid.

Entropy Change

Entropy, symbolized by \( S \), is a measure of disorder or randomness in a system. The change in entropy, \( \Delta S \) is a key factor in determining the spontaneity of a process. The formula for calculating the standard entropy change is: \[ \Delta S^{\circ}_{298} = \Sigma n_{products} S_{m,products}^{\circ} - \Sigma m_{reactants} S_{m,reactants}^{\circ} \] It involves subtracting the sum of entropies of reactants from the sum of entropies of products. A positive \( \Delta S \) indicates an increase in disorder, whereas a negative value means the system has become more ordered.

In the case of the process we're looking at, the entropy change for the system increased, suggesting that the products (lactic acid molecules) are less ordered than the reactants (glucose molecules). The step-by-step solution shows how to calculate the entropy change considering not just the reaction conditions but also the effect of a temperature change from the standard value.

Gibbs Free Energy

Gibbs free energy, represented by \( G \), combines the concepts of enthalpy (\( H \) and entropy (\( S \) into a single value that predicts the spontaneity of a reaction. The change in Gibbs free energy, \( \Delta G \) is given by the equation: \[ \Delta G = \Delta H - T\Delta S \] where \( T \) is the temperature in kelvins. A negative \( \Delta G \) indicates that a reaction is spontaneous, whereas a positive value suggests it is non-spontaneous under the given conditions. If \( \Delta G \) is zero, the system is at equilibrium.

While the exercise did not ask for a \( \Delta G \) calculation, understanding Gibbs free energy is crucial as it integrates the heat exchange with the surroundings (\( \Delta H \) and the change in system disorder (\( \Delta S \) to fully evaluate the feasibility of a reaction.

Chemical Reactions

Chemical reactions involve the transformation of reactants into products. These reactions are governed by the laws of thermodynamics, which dictate that energy is conserved, and often involve changes in enthalpy and entropy. The direction and extent to which a chemical reaction proceeds depend on various factors, including temperature, pressure, concentration of reactants and products, and the overall change in Gibbs free energy. Understanding the thermodynamic principles behind these reactions allows chemists to predict whether a reaction will occur and under what conditions it will be most favorable.

For example, the conversion of glucose to lactic acid that was analyzed in our exercise is a common biological reaction occurring under anaerobic conditions where organisms extract energy from glucose in the absence of oxygen.

Heat Capacity

Heat capacity is a term we encounter in the study of thermodynamics, referring to the amount of heat required to increase the temperature of a substance by a certain amount, usually one degree Celsius or one kelvin. The heat capacity at constant pressure (\( C_{P} \) is particularly relevant in chemical reactions occurring in open systems, where pressure remains constant. The formula to adjust entropy for temperature changes, utilizing \( C_{P} \) is as follows: \[ \Delta S(T) = \Sigma n_{products} C_{P,m,products} (T) - \Sigma m_{reactants} C_{P,m,reactants} (T) \] This allows us to understand how the entropy of a system evolves as the temperature changes. With a constant heat capacity assumption, it makes the calculation more straightforward as the same value applies across a temperature range—demonstrated in the thermodynamic calculation of the glucose-to-lactic acid reaction.