Question

Given the following data: $$2 \mathrm{C}_{6} \mathrm{H}_{6}(l)+15 \mathrm{O}_{2}(g) \longrightarrow 12 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l)\Delta G^{\circ}=-6399 \mathrm{kJ}$$ $$\mathrm{C}(s)+\mathrm{o}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) \quad \Delta G^{\circ}=-394 \mathrm{kJ}$$ $$\mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) \quad \Delta G^{\circ}=-237 \mathrm{kJ}$$ calculate \(\Delta G^{\circ}\) for the reaction $$6 \mathrm{C}(s)+3 \mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{6}(l)$$

Short answer

The Gibbs free energy change, \(\Delta G^\circ\), for the desired reaction \(6 \mathrm{C}(s)+3 \mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{6}(l)\) is +124.5 kJ.

Step-by-step solution

Analyze the given reactions and the desired reaction

The goal is to arrive at the desired reaction using the given reactions. Our given reactions are: 1. 2 C₆H₆(l) + 15 O₂(g) → 12 CO₂(g) + 6 H₂O(l) \(ΔG^\circ = -6399 kJ\) 2. C(s) + O₂(g) → CO₂(g) \(ΔG^\circ = -394 kJ\) 3. H₂(g) + 1/2 O₂(g) → H₂O(l) \(ΔG^\circ = -237 kJ\)

Reverse given reaction 1

In order to isolate C₆H₆(l) in the desired reaction, we will reverse reaction 1: 1. 12 CO₂(g) + 6 H₂O(l) → 2 C₆H₆(l) + 15 O₂(g) \(ΔG_r^\circ = +6399 kJ\) Now, the equation looks like this: 6 C(s) + 3 H₂(g) → 2 C₆H₆ (l)

Divide reaction 1 by 2

In order to have only one molecule of C₆H₆ on the right side of the equation, we will divide the reversed 1st equation by 2: 1. 6 CO₂(g) + 3 H₂O(l) → C₆H₆(l) + 7.5 O₂(g) \(ΔG_r^\circ = +3199.5 kJ\) Now, the equation looks like this: 6 C(s) + 3 H₂(g) → C₆H₆ (l)

Convert CO₂ and H₂O to their constituents

Now, we would like to rewrite CO₂ and H₂O in terms of C and H₂. We will do that using reactions 2 (for CO₂) and 3 (for H₂O), because they provide \(\Delta G^\circ\) for obtaining C and H₂ from CO₂ and H₂O.

Multiply given reaction 2 by 6

Since we have 6 moles of CO₂ in the first reaction, we will multiply reaction 2 by 6 to obtain 2. 6 C(s) + 6 O₂(g) → 6 CO₂(g) \(ΔG^\circ = -394 \times 6 = -2364 kJ\)

Multiply given reaction 3 by 3

Since we have 3 moles of H₂O in the first reaction, we will multiply reaction 3 by 3 to obtain 3. 3 H₂(g) + 1.5 O₂(g) → 3 H₂O(l) \(ΔG^\circ = -237 \times 3 = -711 kJ\)

Combine the reactions

Now adding reactions 2 and 3 to the reversed and divided 1st reaction, we will cancel out CO₂ and H₂O: (+3199.5 kJ): 6 CO₂(g) + 3 H₂O(l) → C₆H₆(l) + 7.5 O₂(g) (+(-2364 kJ)): 6 C(s) + 6 O₂(g) → 6 CO₂(g) (+(-711 kJ)): 3 H₂(g) + 1.5 O₂(g) → 3 H₂O(l) Sum reaction: 6 C(s) + 3 H₂(g) → C₆H₆(l)

Calculate the Gibbs free energy change for the desired reaction

With the sum of the reactions being the desired reaction, we can now calculate ΔG⁰ by taking the sum of all energy changes : ΔG⁰ = (+3199.5) + (-2364) + (-711) = +124.5 kJ Thus, the Gibbs free energy change, ΔG⁰, for the desired reaction is +124.5 kJ.

Key concepts

Thermodynamics

Thermodynamics is the study of energy, heat, and work, and how they affect matter. It's crucial in understanding chemical reactions, especially in calculating energy changes. In thermodynamics, we often talk about the system, the part of the universe we are studying, and the surroundings, which is everything else. When chemical reactions occur, they can release or absorb energy, often in the form of heat. This is where concepts like enthalpy and entropy come in. Enthalpy, denoted by \(H\), represents the total energy of a system, while entropy, represented by \(S\), measures disorder. Gibbs Free Energy, \(G\), is a vital concept connecting enthalpy and entropy. It tells us whether a reaction can occur spontaneously. For any process at constant temperature and pressure, Gibbs Free Energy is defined as: \[ \Delta G = \Delta H - T \Delta S \] where:

• \(\Delta G\) is the change in Gibbs Free Energy.
• \(\Delta H\) is the change in enthalpy.
• \(T\) is the temperature in Kelvin.
• \(\Delta S\) is the change in entropy.

A negative \(\Delta G\) indicates a spontaneous reaction.

Chemical Reactions

Chemical reactions involve breaking bonds in reactants and forming new ones in products. This rearrangement of atoms results in energy changes that can be calculated and predicted. Three main types of reactions include synthesis, decomposition, and combustion. In a synthesis reaction, multiple reactants combine to form a single product, like the formation of water from hydrogen and oxygen. Decomposition reactions reverse this process, where a single compound breaks down into simpler substances. Combustion reactions involve reactants combining rapidly with oxygen, releasing heat and light.Understanding these reactions involves recognizing reactants and products and writing balanced chemical equations. For the exercise given, we are looking at a synthesis reaction where carbon and hydrogen gases are forming benzene, \(C_6H_6(l)\). Each of these reactions comes with its own Gibbs Free Energy change \(\Delta G\), which helps in predicting the feasibility and spontaneity of the reaction.

Hess's Law

Hess's Law is a principle that states the total enthalpy change for a reaction is the same, no matter how it is carried out in multiple steps. This law allows us to calculate \(\Delta G\) for complex reactions by breaking them into simpler steps. It works because enthalpy is a state function, meaning it depends only on the initial and final states, not the path taken. Using Hess's Law, we can sum the \(\Delta G\) values of known reactions to find the \(\Delta G\) for a target reaction.In the given problem, Hess’s Law is applied by manipulating the given reactions and adding their \(\Delta G\) values. By reversing and dividing the benzene combustion reaction, and then appropriately scaling the given \(CO_2\) and \(H_2O\) formation reactions, we find that the desired \(\Delta G\) can be calculated as +124.5 kJ, showing it is a non-spontaneous process under standard conditions.

Free Energy Calculations

Free energy calculations involve determining changes in Gibbs Free Energy to predict if a reaction occurs spontaneously. These calculations are essential in chemical thermodynamics, particularly for industrial and biological processes.To calculate \(\Delta G\) for a reaction, we follow these steps:

• Identify the target reaction and the given reactions.
• Use Hess's Law to manipulate and combine given reactions to match the target reaction.
• Add the \(\Delta G\) values of the modified reactions.

In this exercise, we calculated the Gibbs Free Energy for the formation of benzene, \(C_6H_6(l)\), from carbon and hydrogen. We carefully adjusted the given reactions by reversing and scaling them, then summed their \(\Delta G\) values. The final \(\Delta G\) of +124.5 kJ indicates that the reaction is not spontaneous under standard conditions, as free energy is positive.