Question

2Fe(s)+3 Cl 2 (g) 2 FeCl 3 (s) Delta H^ =-800,0 kJ/mol Si(s)+2 Cl 2 (g) SiCl 4 (s) Delta H^ =-640.1 KJ/mol Determine the enthalpy ( in kJ/mol) for the reaction 3 SiCl 4 (s)+4 Fe(s) 4 FeCl 3 (s)+3 Si(s)

Answer 1

The enthalpy change (ΔH) for the reaction 3 SiCl4(s) + 4 Fe(s) → 4 FeCl3(s) + 3 Si(s) is calculated using Hess's Law, resulting in a ΔH of 320.3 kJ/mol.

Step-by-step explanation:

The student is asking to determine the enthalpy change (ΔH, in kJ/mol) for the reaction 3 SiCl4(s) + 4 Fe(s) → 4 FeCl3(s) + 3 Si(s) using the given enthalpies of other reactions. To solve this, we'll use Hess's Law, which states that if a reaction is the sum of two or more other reactions, then the enthalpy change for the overall process is the sum of the enthalpy changes for the individual steps.

We start with the provided enthalpies:

• 2Fe(s) + 3 Cl2(g) → 2 FeCl3(s), ΔH = -800.0 kJ/mol
• Si(s) + 2 Cl2(g) → SiCl4(s), ΔH = -640.1 kJ/mol

For the reaction of interest, we reverse the second reaction and multiply it by 3, and multiply the first reaction by 2:

• 3 SiCl4(s) → 3 Si(s) + 6 Cl2(g), ΔH = 3 * 640.1 kJ/mol = 1920.3 kJ/mol
• 4 Fe(s) + 6 Cl2(g) → 4 FeCl3(s), ΔH = 2 * -800.0 kJ/mol = -1600.0 kJ/mol

Now add these reactions to get the target reaction:

3 SiCl4(s) + 4 Fe(s) → 4 FeCl3(s) + 3 Si(s), ΔH = 1920.3 kJ/mol + (-1600.0 kJ/mol) = 320.3 kJ/mol

Therefore, the enthalpy change for the reaction is 320.3 kJ/mol.