Question

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

Name: using the equations 2 fe s 3 cl2 g 2 fecl3 s h 8000 kjmol sis 2 cl2 g sicl4 s h 6401 kjmol determine the enthalpy in kjmol for the reaction 3 sicl4 s 4 fe s 4 fecl3 s 3 si s 51903
Uploaded: 2023-10-01T09:58:53-08:00
Duration: 2 min 55 s
Channel: Narayan Hari
Description: using the equations 2 fe s 3 cl2 g 2 fecl3 s h 8000 kjmol sis 2 cl2 g sicl4 s h 6401 kjmol determine the enthalpy in kjmol for the reaction 3 sicl4 s 4 fe s 4 fecl3 s 3 si s 51903

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

Added by Donna B.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Narayan Hari

10/01/2023

Step 1

1 kJ/mol Now, we need to multiply the first equation by 2 and the second equation by 3 to get the desired coefficients: 4 Fe (s) + 6 Cl₂ (g) → 4 FeCl₃ (s) ∆H° = -1600.0 kJ/mol 3 SiCl₄ (s) → 3 Si(s) + 6 Cl₂ (g) ∆H° = 1920.3 kJ/mol Now, we can add the two Show more…

Show all steps

Thanks for your feedback!

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