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 standard Gibbs free energy change for the reaction \(6C(s) + 3H_2(g) → C_6H_6(l)\) is \(9349.5 \; kJ\).

Step-by-step solution

Analyze the given reactions

We are given three reactions with their respective standard Gibbs free energy changes: 1. \(2C_6H_6(l) + 15O_2(g) \rightarrow 12CO_2(g) + 6H_2O(l) \quad \Delta G^{\circ} = -6399 \; kJ\) 2. \(C(s) + O_2(g) \rightarrow CO_2(g) \quad \Delta G^{\circ} = -394 \; kJ\) 3. \(H_2(g) + \frac{1}{2}O_2(g) \rightarrow H_2O(l) \quad \Delta G^{\circ} = -237 \; kJ\)

Manipulate equations to match desired reaction

Let's consider the given equations and manipulate them so that we can eventually arrive at our desired reaction. Firstly, multiply the first reaction by \(-\frac{1}{2}\), second reaction by -12, and third reaction by -6. This allows us to cancel out the unwanted terms in the reactions: 1. \(-\frac{1}{2} \; \rightarrow C_6H_6(l) - \frac{15}{2}O_2(g) + 6CO_2(g) + 3H_2O(l) \quad \Delta G^{\circ} = \frac{1}{2}(6399 \; kJ)\) 2. \(-12 \; \rightarrow 12C(s) → 12CO_2(g) + 12O_2(g) \quad \Delta G^{\circ} = 12 (394 \; kJ) \) 3. \(-6 \; \rightarrow 3H_2(g) → 6H_2O(l) + 3O_2(g) \quad \Delta G^{\circ} = 6(237 \; kJ)\) Now, sum up the manipulated reactions: \(6C(s) + 3H_2(g) + {\frac{15}{2}}O_2(g) → C_6H_6(l) + 6CO_2(g) + 3H_2O(l) + 12CO_2(g) + 12O_2(g) + 6H_2O(l) + 3O_2(g)\) Now, we can cancel out some terms on the left and right side of the equation: \({\frac{15}{2}}O_2(g) + 12O_2(g) + 3O_2(g) \) will cancel out, and \(6CO_2(g) + 12CO_2(g)\) will cancel out.

Calculate the standard Gibbs free energy change for the desired reaction

Now that we have obtained the desired reaction, we can sum up the ΔG° values of the modified reactions to find the ΔG° for the desired reaction: \(\Delta G^{\circ}_{desired} = \frac{1}{2}(6399 \; kJ) + 12(394 \; kJ) + 6(237 \; kJ) = 3199.5 + 4728 + 1422 = 9349.5 \; kJ\) The standard Gibbs free energy change for the reaction \(6C(s) + 3H_2(g) → C_6H_6(l)\) is \(9349.5 \; kJ\).

Key concepts

Thermodynamics

Thermodynamics is the branch of physics that deals with the relationships between heat, work, and energy. In the context of chemical reactions, it helps us understand how energy is transferred and transformed.
One key concept in thermodynamics is the Gibbs Free Energy (\( \Delta G^{\circ} \)), which tells us whether a reaction is spontaneous. A negative \( \Delta G^{\circ} \) indicates that a reaction can proceed without any input of energy, while a positive value means it is non-spontaneous and requires energy input.
In this exercise, we calculated \( \Delta G^{\circ} \) for the formation of benzene (\( C_6H_6 \)) from carbon and hydrogen using data from other reactions. This involved using Hess's law, which states that the total \( \Delta G^{\circ} \) is the same for a reaction regardless of the multiple stages it occurs in. We analyzed given reactions, manipulated them to fit the desired reaction, and summed up their \( \Delta G^{\circ} \) values.

Chemical Reactions

Chemical reactions occur when molecules interact to form new substances. These reactions involve rearranging atoms and breaking/forming chemical bonds. To predict the feasibility of these reactions under certain conditions, we can use thermodynamic principles.
The given exercise provides reactions where benzene combusts in oxygen, carbon forms carbon dioxide, and hydrogen forms water with oxygen. Each of these reactions has a known \( \Delta G^{\circ} \) value, which is crucial for understanding the energy changes involved.

• Combustion of benzene releases a large amount of energy, making it highly exergonic.
• Formation of carbon dioxide and water also involves significant energy changes, contributing to the overall energy profile of a process.

By manipulating these reactions, we derived the desired reaction and calculated the Gibbs free energy for producing benzene. This showcases how understanding chemical reactions and their energy profiles helps in applications ranging from industrial synthesis to environmental science.

Balancing Chemical Equations

Balancing chemical equations is a foundational skill in chemistry necessary for both qualitative and quantitative analysis of reactions. It ensures the law of conservation of mass is upheld, meaning the same number of each type of atom appears on both sides of a chemical equation.
In the problem, balancing was crucial when altering the given reactions to construct the desired reaction. When reactions are represented correctly, balancing ensures that stoichiometry matches with the chemical changes involved.

• The original combustion equation of benzene was balanced to show that two moles of benzene react with fifteen moles of oxygen.
• Secondary reactions involving carbon and hydrogen required scaling adjustments to match the desired number of carbon and hydrogen atoms involved in the formation of benzene.

Every step in the manipulation of equations was accompanied by careful balancing. This helped in not only properly cancelling out unwanted reactants and products but also achieving accurate calculations for computing \( \Delta G^{\circ} \). This exercise underscores the importance of balancing equations for both theoretical computations and practical laboratory chemistry.