Problem 57 - Chemistry An Atoms First Approach

Question

Given the following data: $$2 \mathrm{H}_{2}(g)+\mathrm{C}(s) \longrightarrow \mathrm{CH}_{4}(g) \quad \Delta G^{\circ}=-51 \mathrm{kJ}$$ $$2 \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{H}_{2} \mathrm{O}(l) \quad \Delta G^{\circ}=-474 \mathrm{kJ}$$ $$\mathrm{C}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) \quad \Delta G^{\circ}=-394 \mathrm{kJ}$$ Calculate \(\Delta G^{\circ}\) for \(\mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\)

Step-by-Step Solution

Answer

The standard Gibbs free energy change for the target reaction, \(\mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\), is \(\Delta G^{\circ}=-1291 \mathrm{kJ}\).

Step 1: Write down the target reaction

Write down the target reaction: $$\mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$$

Step 2: Reverse Reaction 1

Reverse Reaction 1 to get \(\mathrm{CH}_{4}(g)\) on the reactant side: $$\mathrm{CH}_{4}(g) \longrightarrow 2 \mathrm{H}_{2}(g)+\mathrm{C}(s) \quad \Delta G_{1}^{\circ}=51 \mathrm{kJ}$$

Step 3: Double Reaction 2

Double Reaction 2 to get \(2 \mathrm{O}_{2}(g)\) and \(2 \mathrm{H}_{2}\mathrm{O}(l)\) on the product side: $$4 \mathrm{H}_{2}(g)+2\mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{H}_{2}\mathrm{O}(l) \quad \Delta G_{2}^{\circ}=-2 \times 474 \mathrm{kJ}=-948 \mathrm{kJ}$$

Step 4: Add the manipulated reactions

Add the manipulated Reaction 1 and the double of Reaction 2, with Reaction 3. This will give the target reaction: $$\mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$$ Now, sum the corresponding Gibbs free energy changes: $$\Delta G_{total}^{\circ} = \Delta G_{1}^{\circ} + \Delta G_{2}^{\circ} + \Delta G^{\circ}_3 = 51 \mathrm{kJ} + (-948 \mathrm{kJ}) + (-394 \mathrm{kJ}) = -1291 \mathrm{kJ}$$

Step 5: Report the result

The standard Gibbs free energy change for the target reaction is: $$\Delta G^{\circ}(\mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l))=-1291 \mathrm{kJ}$$

Key Concepts

Chemical Reactions

When looking at the chemical reaction given in the exercise, it's essential to understand what happens at the molecular level. The target reaction is - \(\mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)\).Chemical reactions involve breaking bonds in reactants and forming new bonds in products. During this process, bonds in methane \(\mathrm{CH}_{4}\) and oxygen \(\mathrm{O}_{2}\) are broken and new bonds are formed to create carbon dioxide \(\mathrm{CO}_{2}\) and water \(\mathrm{H}_{2}\mathrm{O}\). Each of these steps requires energy and involves rearranging atoms to make the products. - Reactants: Substances that undergo change (\(\mathrm{CH}_{4}\) and \(\mathrm{O}_{2}\)).- Products: New substances formed (\(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2}\mathrm{O}\)).Chemical reactions are a vital part of thermodynamics, helping us understand the balance of energy in reactions. Here, we focus on Gibbs Free Energy, which tells us if a reaction is spontaneous or if it requires external intervention.

Enthalpy

Enthalpy is a concept that captures the total heat content of a system during a chemical reaction. While it is related to Gibbs Free Energy, it is not the same. Enthalpy is concerned with the heat absorbed or released at constant pressure.- Endothermic reactions absorb heat.- Exothermic reactions release heat.In the context of the reaction \(\mathrm{CH}_{4}+2\mathrm{O}_{2}\rightarrow\mathrm{CO}_{2}+2\mathrm{H}_{2}\mathrm{O}\), the reaction is known to be exothermic. It releases energy in the form of heat. This aligns with the calculation showing a negative Gibbs Free Energy (\(-1291 \mathrm{kJ}\)) confirming the spontaneity of the reaction. Enthalpy changes, along with entropy changes, allow us to calculate Gibbs Free Energy and decide if a reaction will proceed on its own. Understanding enthalpy helps predict reaction behavior and energy changes when substances react.

Thermodynamics

Thermodynamics is the study of energy and its transformations. In a chemical reaction, it helps us predict if a reaction will take place based on energy changes. Three main components contribute to this:1. **Enthalpy (\(H\)):** - Deals with heat changes at constant pressure.2. **Entropy (\(S\)):** - A measure of disorder or randomness in a system. - Higher entropy means more disorder.3. **Gibbs Free Energy (\(G\)):** - Combines enthalpy and entropy to determine reaction spontaneity. - Calculated by the formula: \[\Delta G = \Delta H - T\Delta S\]- **Negative \(\Delta G\):** Spontaneous reaction.- **Positive \(\Delta G\):** Non-spontaneous reaction.In the reaction analyzed, the negative \(-1291 \mathrm{kJ}\) indicates that it is spontaneous under standard conditions. Thermodynamics provides a framework to understand how energy changes in reactions, particularly through Gibbs Free Energy, assisting in predicting whether reactions need energy input or occur naturally.