Problem 91 - Chemistry and Chemical Reactivity

Question

Write a chemical equation for the oxidation of \(\mathrm{C}_{2} \mathrm{H}_{6}(\mathrm{g}) \mathrm{by} \mathrm{O}_{2}(\mathrm{g})\) to form \(\mathrm{CO}_{2}(\mathrm{g})\) and \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) Defining this as the system: (a) Predict whether the signs of \(\Delta S^{\circ}\) (system), \(\Delta S^{\circ}\) (surroundings), and \(\Delta S^{\circ}\) (universe) will be greater than zero, equal to zero, or less than zero. Explain your prediction. (b) Predict the signs of \(\Delta_{\mathrm{r}} H^{\circ}\) and \(\Delta_{\mathrm{r}} G^{\circ} .\) Explain how you made this prediction. (c) Will the value of \(K_{\mathrm{p}}\) be very large, very small, or near \(1 ?\) Will the equilibrium constant, \(K_{\mathrm{p}}\), for this system be larger or smaller at temperatures greater than 298 K? Explain how you made this prediction.

Step-by-Step Solution

Answer

The system entropy (\(\Delta S^{\circ}_{\text{system}}\)) and universe (\(\Delta S^{\circ}_{\text{universe}}\)) increase (>0), surroundings (\(\Delta S^{\circ}_{\text{surroundings}}\)) increase (>0), \(\Delta H^{\circ} < 0\), \(\Delta G^{\circ} < 0\); \(K_p\) is large and decreases with temperature.

Step 1: Write the Chemical Equation

The chemical equation for the oxidation of ethane \(C_2H_6(g)\) by oxygen \(O_2(g)\) to form carbon dioxide \(CO_2(g)\) and water \(H_2O(g)\) is:\[ 2C_2H_6(g) + 7O_2(g) \rightarrow 4CO_2(g) + 6H_2O(g) \]

Step 2: Predicting \(\Delta S^{\circ}\) for the System

In this reaction, the number of moles of gas changes from 9 (2 moles of \(C_2H_6\) and 7 moles of \(O_2\)) to 10 (4 moles of \(CO_2\) and 6 moles of \(H_2O\)). As the number of gas molecules increases, the entropy \(\Delta S^{\circ}\) for the system should be greater than zero because there is an increase in disorder.

Step 3: Predicting \(\Delta S^{\circ}\) for the Surroundings

The reaction is exothermic (it releases heat), which means that heat is released to the surroundings. This increases the entropy of the surroundings, making \(\Delta S^{\circ}_{\text{surroundings}}\) greater than zero as the energy spread increases in the environment.

Step 4: Predicting \(\Delta S^{\circ}\) for the Universe

Since both the system and surroundings experience an increase in entropy, \(\Delta S^{\circ}_{\text{universe}}\), which is the sum of \(\Delta S^{\circ}_{\text{system}}\) and \(\Delta S^{\circ}_{\text{surroundings}}\), should also be greater than zero, indicating the reaction is spontaneous.

Step 5: Predict \(\Delta_{\mathrm{r}} H^{\circ}\) and \(\Delta_{\mathrm{r}} G^{\circ}\)

The reaction is exothermic, releasing heat, so \(\Delta_{\mathrm{r}} H^{\circ} < 0\), indicating heat is released. As the reaction is both exothermic and increases the entropy of the universe (spontaneous), \(\Delta_{\mathrm{r}} G^{\circ}\) is also less than zero, meaning the reaction is thermodynamically favorable.

Step 6: Predicting the Nature of \(K_{\mathrm{p}}\)

Since \(\Delta_{\mathrm{r}} G^{\circ} < 0\), the reaction is spontaneous, indicating that \(K_{\mathrm{p}}\) will be very large because the equilibrium greatly favors the products. At temperatures higher than 298 K, due to Le Chatelier's principle, \(K_{\mathrm{p}}\) may decrease slightly because the exothermic reaction would partially shift to counterbalance the added heat.

Key Concepts

Entropy (ΔS)

Entropy, represented as \(\Delta S\), is a measure of disorder or randomness in a system. In chemical reactions, it reflects how the number of different ways molecules can be arranged changes. \(\Delta S^{\circ}\) for the system considers the change in entropy during a reaction. In our example, there is a transition from 9 moles of gases (ethane and oxygen) to 10 moles of gases (carbon dioxide and water). More moles indicate increased randomness, leading to a positive \(\Delta S^{\circ}_{\text{system}}\).

Moreover, when a reaction is exothermic, it releases heat into the surroundings, thus increasing the surroundings' entropy. This results in \(\Delta S^{\circ}_{\text{surroundings}}\) being positive. Since both the system and surroundings show increased entropy, \(\Delta S^{\circ}_{\text{universe}}\) also turns out positive, meaning the overall process is spontaneous.

Enthalpy (ΔH)

Enthalpy, denoted by \(\Delta H\), is the total heat content of a system. In chemistry, it provides insight into whether heat is absorbed or released. For our reaction, the oxidation of ethane is exothermic, indicating heat is released. Thus, \(\Delta_{\mathrm{r}} H^{\circ} < 0\), indicating the reaction decreases overall enthalpy and releases energy to the surroundings.

An exothermic process, like this one, means the products are energetically more stable than the reactants. When a process decreases enthalpy, the reaction pathway is energetically favorable. Consequently, such reactions occur naturally, supported by the accompanying increase in entropy of the universe.

Gibbs Free Energy (ΔG)

Gibbs Free Energy, \(\Delta G\), is a crucial component in determining the spontaneity of a chemical reaction. It combines the system's enthalpy and entropy changes. The relation between them is given by the formula: \(\Delta G = \Delta H - T\Delta S\), where \(T\) is the temperature in Kelvin.

In our reaction, both \(\Delta_{\mathrm{r}} H^{\circ} < 0\) and \(\Delta S^{\circ}_{\text{universe}} > 0\), implying \(\Delta_{\mathrm{r}} G^{\circ}\) will be negative \(\Delta G < 0\). Negative \(\Delta G\) indicates that the reaction proceeds spontaneously under standard conditions. Therefore, this chemical equation favors product formation when the conditions are right.

Equilibrium Constant (Kp)

The equilibrium constant, \(K_p\), determines the ratio of products to reactants when equilibrium is reached in a gaseous reaction. A negative \(\Delta_{\mathrm{r}} G^{\circ}\) suggests that \(K_p\) will be significantly greater than 1, favoring product formation.

For an exothermic reaction like the oxidation of ethane, an increase in temperature can shift the equilibrium position. According to Le Chatelier's principle, adding heat to an exothermic reaction will favor the reverse process, slightly decreasing \(K_p\) by favoring reactants.

Thus, while \(K_p\) is initially large at 298 K, higher temperatures may reduce it somewhat, as the system attempts to counteract the heat addition by shifting towards the reactants. This showcases how temperature impacts equilibrium.