Sn (s) + 2Cl2(g) --> SnCl4(l).Sn(s) + Cl2(g) --> Sn Cl2(l) DH = -186.2 kJSnCl2(s) + Cl2(g) --> SnCl4(l) DH = -325.1 kJWhat is the change in enthalpy?
Sn (s) + 2Cl2(g) --> SnCl4(l)
Change in enthalpy = DH1 + DH2 + DH3
Sn(s) + Cl2(g) --> Sn Cl2(l)
DH1 = -186.2 kJ
Sn Cl2(l) --> SnCl2(s)
DH2 = ? <-- find this either in a table in your book or online
SnCl2(s) + Cl2(g) --> SnCl4(l)
DH3 = -325.1 kJ
To find the change in enthalpy (ΔH) for a reaction, you need to sum up the enthalpy changes for the individual steps involved in the reaction. Let's break down the given reaction and calculate the overall enthalpy change.
Given reactions:
1. Sn(s) + 2Cl2(g) → SnCl4(l) (ΔH = ?)
2. Sn(s) + Cl2(g) → SnCl2(l) (ΔH = -186.2 kJ)
3. SnCl2(s) + Cl2(g) → SnCl4(l) (ΔH = -325.1 kJ)
To obtain the overall reaction, we can subtract reaction 2 from reaction 1 and add reaction 3:
[Reaction 1] - [Reaction 2] + [Reaction 3]:
(Sn(s) + 2Cl2(g) → SnCl4(l)) - (Sn(s) + Cl2(g) → SnCl2(l)) + (SnCl2(s) + Cl2(g) → SnCl4(l))
Now, we cancel out the common species:
2Cl2(g) - Cl2(g) + SnCl2(s) - SnCl2(s) + SnCl4(l) - SnCl4(l)
Left with:
Cl2(g) + Sn(s) → Sn(s) + SnCl4(l)
Since Sn(s) appears on both sides of the reaction, it cancels out, leaving us with:
Cl2(g) → SnCl4(l)
The overall enthalpy change (ΔH) can be calculated by summing the individual enthalpy changes:
ΔH = ΔH1 - ΔH2 + ΔH3
= 0 kJ - (-186.2 kJ) + (-325.1 kJ)
= 186.2 kJ - 325.1 kJ
= -138.9 kJ (rounded to one decimal place)
Therefore, the change in enthalpy for the given reaction is approximately -138.9 kJ.
To find the change in enthalpy, you need to calculate the overall change in enthalpy for the reaction. This can be done by summing up the enthalpy changes for each step.
Step 1: Sn (s) + 2Cl2 (g) → SnCl4 (l)
∆H₁ = -186.2 kJ
Step 2: Sn (s) + Cl2 (g) → SnCl2 (l)
∆H₂ = -325.1 kJ
Step 3: SnCl2 (s) + Cl2 (g) → SnCl4 (l)
∆H₃ = unknown (let's call it ∆H)
Overall reaction: Sn (s) + 3Cl2 (g) → SnCl4 (l)
To calculate ∆H₃, you can simply rearrange the equations and signs of Steps 1 and 2:
SnCl2 (l) → Sn (s) + Cl2 (g)
∆H₂ = +325.1 kJ
Add this to Step 3:
SnCl2 (s) + Cl2 (g) → SnCl4 (l)
∆H₃ + ∆H₂ = -325.1 kJ
Since the overall reaction has the same stoichiometry as Step 3, the change in enthalpy for the overall reaction can be expressed as:
∆H overall = ∆H₃
Substituting the values:
∆H overall + 325.1 kJ = -325.1 kJ
Simplifying, we get:
∆H overall = -325.1 kJ - 325.1 kJ
∆H overall = -650.2 kJ
Therefore, the change in enthalpy (∆H) for the overall reaction is -650.2 kJ.