Question

Consider the following data at 298 K:

Compound	∆Hf° (kJ mol−1)
H2S (g)	-20.5
H2O (g)	-242

For the reaction   4 Ag(s) + 2 H₂S(g) + O₂(g) --> 2 Ag₂S(s) + 2 H₂O(g)   

at a temperature of 25 °C, ∆H° = −507 kJ

Calculate the ∆H_f° of Ag₂S (s) is (in kJ mol⁻¹):

		-285.5
		-32
		-64
		+ 475

Answer 1

4 Ag(s) + 2 H₂S(g) + O₂(g) --> 2 Ag₂S(s) + 2 H₂O(g)

∆H°rex   = ∆H°f products - ∆H°f reactants

              = [2*∆H°f Ag2S + 2*∆H°f H2O] -[4*∆H°fAg + 2*∆H°f H2S + ∆H°f O2]

-507      = 2*∆H°f Ag2S +2*-242 -(4*0+2*-20.5 +0)

-507     = 2*∆H°f Ag2S -484+41

2*∆H°f Ag2S = -507+484-41

2*∆H°f Ag2S    = -64

∆H°f Ag2S      = -32KJ/mole >>>>>answer