Question

Given the following data: $$ \begin{aligned} 2 \mathrm{C}_{6} \mathrm{H}_{6}(l)+15 \mathrm{O}_{2}(g) \longrightarrow & 12 \mathrm{CO}_{2}(g)+& 6 \mathrm{H}_{2} \mathrm{O}(l) \\ & \Delta G^{\circ}=-6399 \mathrm{~kJ} \\ \mathrm{C}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) & \Delta G^{\circ}=-394 \mathrm{~kJ} \\ \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) & \Delta G^{\circ}=-237 \mathrm{~kJ} \end{aligned} $$ 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 \(6 \mathrm{C}(s) + 3 \mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{6}(l)\) is \( \Delta G^{\circ}_{final} = 3324 \mathrm{~kJ} \).

Step-by-step solution

Reverse the first reaction and calculate its reversed standard Gibbs free energy change

Reverse the first given reaction: $$ 12 \mathrm{CO}_{2}(g) + 6 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 2 \mathrm{C}_{6} \mathrm{H}_{6}(l) + 15 \mathrm{O}_{2}(g) $$ The standard Gibbs free energy change for a reversed reaction is the opposite to the original reaction. Therefore, the reversed reaction's \(\Delta G^{\circ}\) is: $$ \Delta G_{1}^{\circ} = -(-6399 \mathrm{~kJ}) = 6399 \mathrm{~kJ} $$

Scale the second and third reactions

We need 6 moles of C(s) in our final reaction, which requires scaling up the second reaction 6 times: $$ 6 (\mathrm{C}(s) + \mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g)) $$ This gives us: $$ 6 \mathrm{C}(s) + 6 \mathrm{O}_{2}(g) \rightarrow 6 \mathrm{CO}_{2}(g) $$ Scaling the standard Gibbs free energy change of a reaction is also directly proportional, giving us: $$ \Delta G_{2}^{\circ} = 6 \times (-394 \mathrm{~kJ}) = -2364 \mathrm{~kJ} $$ Similarly, we need 3 moles of \(\mathrm{H}_{2}(g)\) in our final reaction, which requires scaling up the third reaction 3 times: $$ 3 (\mathrm{H}_{2}(g) + \frac{1}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{H}_{2} \mathrm{O}(l)) $$ This gives us: $$ 3 \mathrm{H}_{2}(g) + \frac{3}{2} \mathrm{O}_{2}(g) \rightarrow 3 \mathrm{H}_{2} \mathrm{O}(l) $$ And the scaled Gibbs free energy change becomes: $$ \Delta G_{3}^{\circ} = 3 \times (-237 \mathrm{~kJ}) = -711 \mathrm{~kJ} $$

Add the adjusted reactions together

Now we can add the adjusted reactions together to find the final reaction: $$ \begin{aligned} (12 \mathrm{CO}_{2}(g) + 6 \mathrm{H}_{2} \mathrm{O}(l) &\rightarrow 2 \mathrm{C}_{6} \mathrm{H}_{6}(l) + 15 \mathrm{O}_{2}(g)) \\ +(6 \mathrm{C}(s) + 6 \mathrm{O}_{2}(g) &\rightarrow 6 \mathrm{CO}_{2}(g)) \\ +(3 \mathrm{H}_{2}(g) + \frac{3}{2} \mathrm{O}_{2}(g) &\rightarrow 3 \mathrm{H}_{2} \mathrm{O}(l)) \\ \hline 6 \mathrm{C}(s) + 3 \mathrm{H}_{2}(g) &\rightarrow \mathrm{C}_{6} \mathrm{H}_{6}(l) \end{aligned} $$

Calculate the standard Gibbs free energy change for the final reaction

Add the standard Gibbs free energy changes of the adjusted reactions together to find the desired \(\Delta G^{\circ}\): $$ \Delta G^{\circ}_{final} = \Delta G_{1}^{\circ} + \Delta G_{2}^{\circ} + \Delta G_{3}^{\circ} = 6399 \mathrm{~kJ} - 2364 \mathrm{~kJ} - 711 \mathrm{~kJ} = 3324 \mathrm{~kJ} $$ So, the standard Gibbs free energy change for the reaction \(6 \mathrm{C}(s) + 3 \mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{6}(l)\) is \( \Delta G^{\circ}_{final} = 3324 \mathrm{~kJ} \).

Key concepts

Thermochemistry

Thermochemistry is a sub-discipline of thermodynamics that focuses on the energy and heat associated with chemical reactions. One of the fundamental concepts in thermochemistry is the understanding of how energy is absorbed or released during a reaction. Energy can be stored in chemical bonds, and when these bonds are broken or formed, energy is either consumed or released, often in the form of heat. This energy change is quantified as enthalpy (( \( \Delta H \) )).

An exothermic reaction releases heat (( \( \Delta H < 0 \) )), making the surroundings warmer, while an endothermic reaction absorbs heat (( \( \Delta H > 0 \) )), resulting in a temperature drop of the surroundings. However, to truly understand the spontaneity of a reaction under constant pressure and temperature (which is most common in everyday scenarios), we look at another thermodynamic function: Gibbs free energy.

Chemical Reactions

Chemical reactions involve the transformation of reactants into products through the breaking and forming of chemical bonds. The way we represent these changes is through a chemical equation, which details the chemical species involved, their phase (solid, liquid, gas), and the stoichiometry, which shows the relative amounts of reactants and products.

A balanced chemical equation ensures that the law of conservation of mass is met, meaning the quantity of each element is the same in the reactants and products. In the exercise provided, for instance, balancing the reaction required manipulating the coefficients of the reactants and products so that the number of atoms for each element is consistent on both sides of the equation. This process is essential as it lays the foundation for further calculations in thermochemistry, including the determination of \( \Delta G^{\text{\textcircled{a}}} \) for a chemical reaction.

Standard Gibbs Free Energy Change

The standard Gibbs free energy change (( \( \Delta G^{\text{\textcircled{a}}} \) )) is a thermodynamic property that indicates the spontaneity of a chemical reaction at constant pressure and temperature. It combines the concepts of enthalpy (( \( \Delta H \) )), entropy (( \( \Delta S \) )), and temperature (( T \) )) into a single value. When \( \Delta G^{\text{\textcircled{a}}} < 0 \) ), the reaction tends to occur spontaneously, and when \( \Delta G^{\text{\textcircled{a}}} > 0 \) ), the reaction is considered nonspontaneous. A reaction with \( \Delta G^{\text{\textcircled{a}}} = 0 \) is at equilibrium.

The \( \Delta G^{\text{\textcircled{a}}} \) can be calculated from standard enthalpies and entropies of the reactants and products or determined experimentally. In our exercise, it was computed by reversing, scaling, and combining the \( \Delta G^{\text{\textcircled{a}}} \) values of related reactions to furnish the \( \Delta G^{\text{\textcircled{a}}} \) for a target reaction (( 6 \( \mathrm{C}(s) \) + 3 \( \mathrm{H}_{2}(g) \) \longrightarrow \( \mathrm{C}_{6} \( \mathrm{H}_{6}(l) \) )), indicative of its spontaneity under standard conditions.