Problem 106 - Chemistry The Central Science

Question

The hydrocarbons acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\) and benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) have the same empirical formula. Benzene is an "aromatic" hydrocarbon, one that is unusually stable because of its structure. (a) By using data in Appendix \(\mathrm{C}\), determine the standard enthalpy change for the reaction $3 \mathrm{C}_{2} \mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{6}(l)$.

Step-by-Step Solution

Answer

The standard enthalpy change for the reaction \(3 \mathrm{C}_{2} \mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{6}(l)\) is \(-631.1 \,\mathrm{kJ/mol}\).

Step 1: Identify the given standard enthalpies of formation

In Appendix C, we can find the standard enthalpies of formation for acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\) and benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\). The values are: - Standard enthalpy of formation of acetylene: \(\Delta H_f^{⦵}(\mathrm{C}_{2} \mathrm{H}_{2}) = 226.7 \,\mathrm{kJ/mol}\) - Standard enthalpy of formation of benzene: \(\Delta H_f^{⦵}(\mathrm{C}_{6} \mathrm{H}_{6}) = 49.0 \,\mathrm{kJ/mol}\)

Step 2: Calculate the enthalpy change for the reaction

To determine the enthalpy change for the given reaction, we will use the following equation: \(\Delta H_{rxn}^{⦵} = \sum n_p \Delta H_f^{⦵}(products) - \sum n_r \Delta H_f^{⦵}(reactants)\) where \(\Delta H_{rxn}^{⦵}\) is the standard enthalpy change for the reaction, \(\Delta H_f^{⦵}(products)\) is the standard enthalpy of formation of the products, \(\Delta H_f^{⦵}(reactants)\) is the standard enthalpy of formation of the reactants, \(n_p\) is the stoichiometric coefficient of product and \(n_r\) is the stoichiometric coefficient of reactant. For our reaction, \(3 \mathrm{C}_{2} \mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{6}(l)\): \(\Delta H_{rxn}^{⦵} = (1) \Delta H_f^{⦵}(\mathrm{C}_{6} \mathrm{H}_{6}) - (3) \Delta H_f^{⦵}(\mathrm{C}_{2} \mathrm{H}_{2})\) Substituting the values of standard enthalpies of formation from Step 1: \(\Delta H_{rxn}^{⦵} = (1)(49.0 \,\mathrm{kJ/mol}) - (3)(226.7 \,\mathrm{kJ/mol})\)

Step 3: Compute the standard enthalpy change for the reaction

Perform the calculation: \(\Delta H_{rxn}^{⦵} = 49.0 \,\mathrm{kJ/mol} - 680.1 \,\mathrm{kJ/mol} = -631.1 \,\mathrm{kJ/mol}\) The standard enthalpy change for the reaction \(3 \mathrm{C}_{2} \mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{6}(l)\) is \(-631.1 \,\mathrm{kJ/mol}\).