Question

Using the supplemental data, calculate ΔH⁰ (in kJ/mol) for the following reaction:

C₆H₅Cl(l) + H₂O(g) → C₆H₅OH(s)+HCl(g)

Answer 1

To calculate ΔH⁰ (in kJ/mol) for the reaction C₆H₅Cl(l) + H₂O(g) → C₆H₅OH(s) + HCl(g), we need to consider the bond energies of the bonds broken and formed in the reaction. The ΔH⁰ for the reaction is -14 kJ/mol.

Step-by-step explanation:

To calculate ΔH⁰ (in kJ/mol) for the reaction C₆H₅Cl(l) + H₂O(g) → C₆H₅OH(s) + HCl(g), we need to consider the bond energies of the bonds broken and formed in the reaction.

First, we break the C-Cl bond in C₆H₅Cl(l), which requires 339 kJ/mol. Then, we break the O-H bond in H₂O(g), which requires 464 kJ/mol.

Next, we form the C-O bond in C₆H₅OH(s), which releases 360 kJ/mol. Finally, we form the H-Cl bond in HCl(g), which releases 431 kJ/mol.

To calculate ΔH⁰, we sum the energies required to break the bonds (339 kJ/mol + 464 kJ/mol) and subtract the energies released when the new bonds are formed (360 kJ/mol + 431 kJ/mol). This gives us a ΔH⁰ of -14 kJ/mol for the reaction.