Chemistry problem

Question

Calculate ΔH^o for the process Zn(s) + S(s) + 2 O₂(g) → ZnSO₄(s)

from the following information

Zn(s) + S(s) → ZnS(s)	ΔHo = -118.1 kJ/mol
ZnS(s) + 2 O2(g) → ZnSO4(s)	ΔHo = -329.9 kJ/mol

 ^kJ/_mol

Answer 1

(1) Zn + S ==> ZnS ... ∆H = -118.1 kJ/mol

(2) ZnS + 2O2 ==> ZnSO4 ... ∆H = -329.9 kJ/mol

Target equation: Zn + S + 2O2 ==> ZnSO4

Use Hess' Law:

Copy eq. (1): Zn + S ==> ZnS ... ∆H = -118.1 kJ/mol

Copy eq (2) : ZnS + 2O2 ==> ZnSO4 ... ∆H = -329.9 kJ/mol

Add them to get...

Zn + S + ZnS + 2O2 ==> ZnS + ZnSO4 and cancel like terms to get ...

Zn + S + 2O2 ==> ZnSO4

Add ∆H values to get...

-118.1 + -329.9 = -448 kJ/mol

Answer 2

As for reaction one

-118.1= H of Zns

For reaction 2

H reaction = H znso4- H zns

-329.9= H znso4- (-118.1)

Hznso4= - 211.8 kj mol^-