Answer:
ΔH° = -186.2 kJ
Explanation:
Hello,
This case in which the Hess method is applied to compute the required chemical reaction. Thus, we should arrange the given first two reactions as:
(1) it is changed as:
SnCl2(s) --> Sn(s) + Cl2(g)...... ΔH° = 325.1 kJ
That is why the enthalpy of reaction sign is inverted.
(2) remains the same:
Sn(s) + 2Cl2(g) --> SnCl4(l)......ΔH° = -511.3 kJ
Therefore, by adding them, we obtain the requested chemical reaction:
(3) SnCl2(s) + Cl2(g) --> SnCl4(l)
For which the enthalpy change is:
ΔH° = 325.1 kJ - 511.3 kJ
ΔH° = -186.2 kJ
Best regards.
Answer:
ΔH° = -186.2 kJ
Explanation:
Step 1: Data given
(1) Sn(s) + Cl2(g) ⇆ SnCl2(s) ΔH° = -325.1 kJ
(2) Sn(s) + 2Cl2(g) ⇆ SnCl4(l) ΔH° = -511.3 kJ
Step 2: The balanced equation
SnCl2(s) + Cl2(g) ⇆ SnCl4(l)
Step 3: Calculate the standard enthalpy change for the reaction
We have to take the reverse equation of the first reaction ( Because we need SnCl2 as reactant)
SnCl2(s) ⇆ Sn(s) + Cl2(g) ΔH° = 325.1 kJ
Then we add the second equation to this new one
SnCl2(s) ⇆ Sn(s) + Cl2(g) ΔH° = 325.1 kJ
Sn(s) + 2Cl2(g) ⇆ SnCl4(l) ΔH° = -511.3 kJ
SnCl2(s) + Sn(s) + 2Cl2(g) ⇆ Sn(s) + Cl2(g) + SnCl4(l)
SnCl2(s) + Cl2(g) ⇆ SnCl4(l)
ΔH° = 325.1 kJ + (-511.3kJ)
ΔH° = 325.1 kJ - 511.3 kJ
ΔH° = -186.2 kJ
