Question

Given the standard enthalpy changes for the following two reactions:

Answer 1

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

Answer 2

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.