Question

Using the equations 2 Fe (s) + 3 Cl₂ (g) → 2 FeCl₃ (s) ΔH° = -800.0 kJ/mol Si(s) + 2 Cl₂ (g) → SiCl₄ (s) ΔH° = -640.1 kJ/mol Determine the molar enthalpy (in kJ/mol) for the reaction 3 SiCl₄ (s) + 4 Fe (s) → 4 FeCl₃ (s) + 3 Si (s).

A) -1159.9
B) 1159.9
C) -459.9
D) 459.9

Answer 1

The molar enthalpy for the reaction 3 SiCl₄ (s) + 4 Fe (s) → 4 FeCl₃ (s) + 3 Si (s) is -251.45 kJ/mol.

Step-by-step explanation:

To determine the molar enthalpy for the reaction 3 SiCl₄ (s) + 4 Fe (s) → 4 FeCl₃ (s) + 3 Si (s), we need to sum up the enthalpies of the individual reactions and cancel out any common species. First, we double the equation 2 Fe (s) + 3 Cl₂ (g) → 2 FeCl₃ (s) and multiply its enthalpy by 2. Next, we multiply the equation Si(s) + 2 Cl₂ (g) → SiCl₄ (s) by 3 and its enthalpy by 3.

Now we can combine the equations and enthalpies to get the overall reaction: 6 SiCl₄ (s) + 8 Fe (s) → 8 FeCl₃ (s) + 6 Si (s). Adding the enthalpies of the individual reactions, we get: (2 * -800.0 kJ/mol) + (3 * -640.1 kJ/mol) = -1600.0 kJ/mol + -1920.3 kJ/mol = -3520.3 kJ/mol. Finally, we divide the enthalpy by the sum of the stoichiometric coefficients to get the molar enthalpy: -3520.3 kJ/mol ÷ (6 + 8) = -3520.3 kJ/mol ÷ 14 = -251.45 kJ/mol.