Answer to Question #192168 in General Chemistry for Jade

Question

2) Calculate the Δ

Δ

H value heat for the following reaction: Ag2S(s)  + 2 HCl(g) -->  2 AgCl(s) + H2S(g) Δ

Δ

H = ? kJ/mol Solve by writing formation equations for each reactant and product and using the Δ

Δ

H values for each.

Answer 1

The given reaction is-

Ag₂S_(S) +2HCl_(g) →2AgCl_(S) +H₂S_(g)

Now, independent reactions are:

2Ag⁺+S^2-→Ag₂S(s);  ∆H_f=−31.8 kJ/mol

H⁺+Cl^-→HCl(g);  ∆H_f=−92.3 kJ/mol

Ag⁺+Cl^-→AgCl(s);  ∆H_f=−127.0 kJ/mol

2H⁺+S^2-→H₂S(g);  ∆H_f=−20.6 kJ/mol

Therefore, the enthalpy change of reaction can be given as follows:

∆H_rxn=Σn∆H_f(Products)−Σm∆H_f(Reactants)

=2∆H_f(AgCl(s))+1∆H_f(H₂S(g))]

−[1∆H_f(Ag₂S(s))+2∆H_f(HCl(g))]

=[2×(−127.0 kJ/mol)+1×(−20.6 kJ/mol)]

−[1×(−31.8 kJ/mol)+2×(−92.3 kJ/mol)]

=[-254 kJ/mol−20.6 kJ/mol]−[−31.8 kJ/mol−184.6 kJ/mol]

=[-274.6 kJ/mol−(−216.4 kJ/mol)]

=[−274.6 kJ/mol+216.4 kJ/mol]

=−58.2 kJ/mol

Answer: ΔH value heat for the following reaction:

Ag₂S_(S) +2HCl_(g) →2AgCl_(S) +H2S_(g)

 is -58.2 kJ/mol.