Given the following data: 2 C6H6(l)+15 O2(g) ⟶ 12 CO2(g)+6 H2O(l) Δ G^∘=-6399 kJ

step 1 All right, problem 58, we're looking for the delta G for this reaction and we're given all these

Given the following data: 2 C6H6(l)+15 O2(g) ⟶ 12...

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 $$6 \mathrm{C}(s)+3 \mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{6}(l)$$

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
$$6 \mathrm{C}(s)+3 \mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{6}(l)$$

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Key Concepts

Hess's Law

Hess's Law states that the total change in a thermodynamic quantity, such as Gibbs free energy, during a chemical reaction is the same regardless of the steps taken to complete the reaction. This principle allows one to add, subtract, or reverse given reactions to obtain the overall reaction’s free energy, making it a powerful tool for determining unknown thermodynamic values.

Gibbs Free Energy

Gibbs free energy is a thermodynamic potential that combines enthalpy and entropy to predict the spontaneity of a reaction under constant temperature and pressure. A negative value indicates a spontaneous process, and standard Gibbs free energy values (?G°) provide a benchmark for comparing the favorability of reactions.

Reaction Manipulation and Stoichiometry

Adjusting chemical reactions by reversing or scaling them requires corresponding changes in their associated Gibbs free energies: reversing a reaction inverts the sign of ?G° and multiplying a reaction by a coefficient scales ?G° by that same factor. Mastery of these manipulation techniques, along with stoichiometric balancing, is essential for constructing thermochemical cycles and calculating unknown reaction energies.

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