Final answer:
The enthalpy change (∆Hrxn) for the reaction CCl4(g) + H2(g) → CHCl3(g) + HCl(g) is calculated by using the bond energies given for the broken and formed bonds. The calculation reveals that ∆Hrxn is -80 kJ/mol, indicating the reaction is exothermic. The correct option is E. -80 kJ/mol.
Step-by-step explanation:
To estimate the ∆Hrxn for the given reaction CCl4(g) + H2(g) → CHCl3(g) + HCl(g), we need to calculate the total bond energy for the bonds broken and formed during the reaction. The energy required to break the bonds is calculated by adding the bond energies of all the bonds that are broken, and the energy released is calculated by adding the bond energies of all the bonds that are formed.
For the reactants, 1 mole of CCl4 will involve breaking 4 C-Cl bonds and 1 H-H bond. That is 4 × 328 kJ/mol (C-Cl) + 436 kJ/mol (H-H) = 1756 kJ/mol. For the products, 1 mole of CHCl3 will involve forming 3 C-Cl bonds and 1 C-H bond, and 1 mole of HCl will involve forming 1 H-Cl bond. That is 3 × 328 kJ/mol (C-Cl) + 413 kJ/mol (C-H) + 431 kJ/mol (H-Cl) = 1828 kJ/mol.
Thus, the total bond energy of products is lower than that of reactants, indicating that the reaction is exothermic. The enthalpy change (∆Hrxn) can be estimated by subtracting the energy of the bonds formed from the energy of the bonds broken:
∆Hrxn = Total energy of bonds formed – Total energy of bonds broken
∆Hrxn = 1828 kJ/mol – 1756 kJ/mol
∆Hrxn = -80 kJ/mol
Therefore, the enthalpy change for the given reaction is -80 kJ/mol, which corresponds to option E.