Question

Given the following data: $$\begin{array}{lr}2 \mathrm{C}_{6} \mathrm{H}_{6}(l)+15 \mathrm{O}_{2}(g) \longrightarrow 12 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \\ \Delta G^{\circ}=-6399 \mathrm{~kJ} \\\\\mathrm{C}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) & \Delta G^{\circ}=-394 \mathrm{~kJ} \\ \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) & \Delta G^{\circ}=-237 \mathrm{~kJ} \end{array}$$ 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 formation of benzene, \(C_6H_6(l)\), from its elements, carbon and hydrogen, is \(\Delta G^{\circ}_{target} = 124.5 \ \text{kJ}\).

Step-by-step solution

Analyze the reactions and manipulate them to cancel out unwanted terms

Start by looking at the three given reactions. The target reaction is: $$6 C(s) + 3 H_2(g) \longrightarrow C_6H_6(l)$$ We need to manipulate the given reactions so that we can add them together to get the target reaction. This may involve reversing and/or scaling the reactions.

Reverse the first reaction, and scale it by 1/2

Reverse the first reaction in order to obtain \(\Delta G^{\circ}\) for: $$12 CO_2(g) + 6 H_2O(l) \longrightarrow 2 C_6H_6(l) + 15 O_2(g)$$ The new free energy change associated with this reaction is: \(\Delta G^{\circ}_{1\_reversed} = 6399 \ \text{kJ}\) Now, divide both sides of the reversed reaction by 2: $$6 CO_2(g) + 3 H_2O(l) \longrightarrow C_6H_6(l) + 7.5 O_2(g)$$ The new free energy change associated with this reaction is: \(\Delta G^{\circ}_{1\_scaled} = \frac{1}{2} * 6399 \ \text{kJ} = 3199.5 \ \text{kJ}\)

Scale the second reaction by 6

Multiply the second reaction by 6 to have the same number of carbon atoms as in the target reaction: $$6 C(s) + 6 O_2(g) \longrightarrow 6 CO_2(g)$$ The new free energy change associated with this reaction is: \(\Delta G^{\circ}_{2\_scaled} = 6 * (-394) \ \text{kJ} = -2364 \ \text{kJ}\)

Scale the third reaction by 3

Multiply the third reaction by 3 to have the same number of hydrogen atoms as in the target reaction: $$3 H_2(g) + 1.5 O_2(g) \longrightarrow 3 H_2O(l)$$ The new free energy change associated with this reaction is: \(\Delta G^{\circ}_{3\_scaled} = 3 * (-237) \ \text{kJ} = -711 \ \text{kJ}\)

Add the manipulated reactions to obtain the target reaction

Now, add the reactions from Steps 2, 3, and 4. This will cancel out unwanted terms and result in the target reaction. \( (6 CO_2(g) + 3 H_2O(l) \longrightarrow C_6H_6(l) + 7.5 O_2(g)) \newline(+ 6 C(s) + 6 O_2(g) \longrightarrow 6 CO_2(g)) \newline(+ 3 H_2(g) + 1.5 O_2(g) \longrightarrow 3 H_2O(l))\) The sum of these reactions gives us the target reaction: $$6 C(s) + 3 H_2(g) \longrightarrow C_6H_6(l)$$ Now, sum up the associated free energy changes for each reaction: \(\Delta G^{\circ}_{target} = \Delta G^{\circ}_{1\_scaled} + \Delta G^{\circ}_{2\_scaled} + \Delta G^{\circ}_{3\_scaled} = 3199.5 + (-2364) + (-711) = 124.5 \ \text{kJ}\) Thus, the standard Gibbs free energy change for the formation of benzene, \(C_6H_6(l)\), from its elements is \(\Delta G^{\circ}_{target} = 124.5 \ \text{kJ}\).

Key concepts

Chemical Reactions

Chemical reactions are processes where substances interact to form new products. In our exercise, the reaction involves the formation of benzene (labeled as \( C_6H_6(l) \)) from carbon \( C(s) \) and hydrogen \( H_2(g) \). Understanding these reactions requires knowledge of breaking and forming chemical bonds, as energy is transformed. Within this context:

• The conversion of carbon and hydrogen into benzene requires very specific conditions.
• This particular reaction involves manipulating initial reactions to achieve a desired one.

Each compound reacts differently, so calculations must account for different energetic properties. This is where we apply concepts like Gibbs Free Energy, which helps determine if the reaction occurs spontaneously.

Thermodynamics

Thermodynamics deals with the principles governing energy transformations, especially as heat and work. In chemical reactions, it allows us to predict if and how a reaction will occur.
For the reaction of forming benzene, thermodynamic principles guide:

• Energy balance: Ensuring energy entering the system equals energy leaving the system.
• Entropy and enthalpy: These concepts come into play to predict reaction spontaneity.

The Gibbs Free Energy is a key component. By calculating the changes, we determine if the process energetically favors the formation of benzene. Negative \( \Delta G \) means spontaneous reactions under set conditions.

Enthalpy

Enthalpy is the measure of total heat content in a thermodynamic system, crucial in determining reaction feasibility. It represents the bond energy required or released when substances change.
For benzene formation:

• Reactions release heat when forming \( CO_2 \) and \( H_2O \).
• These processes require calculating changes in enthalpy to predict how energy is absorbed or released.

By evaluating enthalpy and combining it with entropy, the overall Gibbs Free Energy is found. This provides insight into the energy landscape of chemical reactions and helps predict the direction and feasibility of reactions.