Problem 79 - Chemistry and Chemical Reactivity

Question

Write a chemical equation for the oxidation of \(\mathrm{C}_{2} \mathrm{H}_{6}(\mathrm{g})\) 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 \(\Delta S_{\mathrm{m}}^{\circ}, \Delta S_{\text {surr }}^{\mathrm{o}},\) and \(\Delta S_{\text {univ }}^{\mathrm{e}}\) will be greater than zero, equal to zero, or less than zero. Explain your prediction. (b) Predict the signs of \(\Delta H^{\circ}\) and \(\Delta G^{\circ} .\) 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, \(K_{p},\) for this system be larger or smaller at temperatures greater than \(298 \mathrm{K} ?\) Explain how you made this prediction.

Step-by-Step Solution

Answer

The reaction is spontaneous, with \(K_p\) large at standard temperature; decreasing with higher temperature.

Step 1: Write the Balanced Chemical Equation

The oxidation of ethane (\(\mathrm{C}_2 \mathrm{H}_6\)) by oxygen (\(\mathrm{O}_2\)) produces carbon dioxide (\(\mathrm{CO}_2\)) and water (\(\mathrm{H}_2\mathrm{O}\)). The balanced chemical equation is:\[\mathrm{2C}_2 \mathrm{H}_6(\mathrm{g}) + 7\mathrm{O}_2(\mathrm{g}) \rightarrow 4\mathrm{CO}_2(\mathrm{g}) + 6\mathrm{H}_2\mathrm{O}(\mathrm{g})\] This equation ensures that the number of each type of atom is the same on both sides of the equation.

Step 2: Analyze Entropy Changes

(a) For the entropy of the system, \(\Delta S_{\mathrm{m}}^{\circ}\), the change will be less than zero because the number of gas molecules decreases from 9 to 10 (higher disorder to lower disorder).For the surroundings, \(\Delta S_{\text{surr}}^{\mathrm{o}}\), the entropy change will be greater than zero as the reaction releases heat to surroundings resulting in an increase in their entropy.The change in the universe's entropy, \(\Delta S_{\text{univ}}^{\mathrm{e}}\), will be greater than zero because the total entropy must increase for a spontaneous reaction.

Step 3: Analyze Enthalpy Changes

(b) The reaction releases heat (exothermic), so \(\Delta H^{\circ}\) will be less than zero.Since the reaction is exothermic and increases the entropy of the surroundings, \(\Delta G^{\circ}\) will also be less than zero because the reaction is spontaneous at standard conditions.

Step 4: Predict Equilibrium Constant Behavior

(c) The value of \(K_{\mathrm{p}}\) will be very large, indicating that the reaction strongly favors the formation of products. At temperatures greater than 298 K, the equilibrium constant \(K_p\) will tend to decrease because, according to Le Chatelier's principle, a higher temperature favors the endothermic direction, which is the reverse of this exothermic reaction.

Key Concepts

Chemical Equations

When we talk about chemical equations, we are referring to a symbolic representation of a chemical reaction. It includes reactants, products, and their stoichiometric coefficients, which are numbers that indicate the proportions of substances involved. A chemical equation provides vital information about the chemical species undergoing the reaction and the substances formed as a result.

For example, the oxidation of ethane (\(\mathrm{C}_2 \mathrm{H}_6\)) by oxygen (\(\mathrm{O}_2\)) can be expressed with the equation:\[\mathrm{2C}_2 \mathrm{H}_6(\mathrm{g}) + \mathrm{7O}_2(\mathrm{g}) \rightarrow \mathrm{4CO}_2(\mathrm{g}) + \mathrm{6H}_2\mathrm{O}(\mathrm{g})\]

• Reactants: \(\mathrm{C}_2 \mathrm{H}_6\), \(\mathrm{O}_2\)
• Products: \(\mathrm{CO}_2\), \(\mathrm{H}_2\mathrm{O}\)

This equation indicates that two molecules of ethane react with seven molecules of oxygen to produce four molecules of carbon dioxide and six molecules of water. Chemical equations help predict the amounts of products formed in a reaction.

Balancing Chemical Equations

In chemistry, balancing chemical equations is a fundamental skill. It involves ensuring that there is the same number of each type of atom on both sides of the equation. This balance obeys the Law of Conservation of Mass, which states that mass cannot be created or destroyed in a chemical reaction.

For the oxidation of ethane given above, balancing involves the following:

• List all reactants and products: \(\mathrm{C}_2 \mathrm{H}_6\), \(\mathrm{O}_2\), \(\mathrm{CO}_2\), and \(\mathrm{H}_2\mathrm{O}\)
• Count the number of each type of atom on both sides
• Adjust coefficients to have equal numbers on both sides
• Verify that the equation is balanced

In our equation, each side of the equation has:

• Carbon atoms: 4
• Hydrogen atoms: 12
• Oxygen atoms: 14

Achieving this perfect balance ensures the equation is correctly representing the chemical reaction process.

Thermodynamics in Chemistry

Thermodynamics in chemistry deals with the relationships between heat energy and physical reactions. It encompasses concepts such as entropy (\(\Delta S\)), enthalpy (\(\Delta H\)), and Gibbs free energy (\(\Delta G\)). These parameters help determine whether a chemical reaction will occur spontaneously and the reaction's energy profile.

**Entropy (\(\Delta S\))**:

• Represents disorder in the system
• For the oxidation reaction of ethane, \(\Delta S_{ ext{m}}^{\circ}\) is less than zero since gas molecules decrease indicating lower disorder.
• \(\Delta S_{ ext{surr}}^{\circ}\), the surroundings' entropy, increases due to heat release.
• Total entropy change, \(\Delta S_{ ext{univ}}^{\circ}\), will be positive because spontaneous reactions increase total entropy.

**Enthalpy (\(\Delta H\))**:

• The change in heat: this reaction is exothermic so \(\Delta H^{\circ}\) is negative.

**Gibbs Free Energy (\(\Delta G\))**:

• Combines both entropy and enthalpy to determine spontaneity.
• Negative \(\Delta G^{\circ}\) in this reaction indicates spontaneity at standard conditions.

Thermodynamics provides a comprehensive understanding of energy changes within chemical reactions.

Equilibrium Constant

The equilibrium constant (\(K_{p}\)) provides a quantitative measure that reflects the position of equilibrium in a chemical reaction at a given temperature. It is an essential concept in understanding reaction dynamics and predicting the extent of the reaction.

For the combustion of ethane, the equilibrium constant is expressed in terms of the partial pressures of the gases involved:\[K_{p} = \frac{(P_{\text{CO}_2})^4 (P_{\text{H}_2\mathrm{O}})^6}{(P_{\text{C}_2 \mathrm{H}_6})^2 (P_{\text{O}_2})^7}\]

• A large \(K_{p}\) value indicates the reaction heavily favors product formation, as is the case here.
• According to Le Chatelier’s Principle, increasing temperature in this exothermic reaction will lower \(K_{p}\), as heat acts as a product and the system shifts to absorb the excess heat.

Understanding the equilibrium constant helps predict how changes in conditions like temperature will affect the system's dynamics and product formation.