Answer to Question #113187 in General Chemistry for Chan

Question

For the following reactions:
Ca(s) + 2HCl(aq) → CaCl2(aq) + H2(g) ΔHrxn = -543.0 kJ/mol
CaS(s) + 2HCl(aq) → CaCl2(aq) + H2S(g) ΔHrxn = -81.1 kJ/mol
CaS(s) → Ca(s) + 1/8 S8(s) ΔHrxn = +482.4 kJ/mol
What is the enthalpy of the reaction
H2(g) + 1/8 S8(s) → H2S(g) ?

Answer 1

$\Delta$ H_reaction = $\sum$ $\nu$ * Δ H_products - ν * Δ H_{starting materials}

ΔH (Ca) = ΔH (H₂) = 0

For firs reaction find ΔH(CaCl₂):

ΔH(CaCl₂) - 2ΔH(HCl) = -543 kJ/mol

ΔH(CaCl₂) = 2ΔH(HCl) - 543 kJ/mol

For second reaction find ΔH(H₂S):

ΔH(H₂S) + ΔH(CaCl₂) - ΔH(CaS) - 2ΔH(HCl) = -81.1 kJ/mol

Insert a value of ΔH(CaCl₂):

ΔH(H₂S) + 2ΔH(HCl) - 543 kJ/mol - ΔH(CaS) - 2ΔH(HCl) = -81.1 kJ/mol

ΔH(H₂S) - 543 kJ/mol - ΔH(CaS) = -81.1 kJ/mol

ΔH(H₂S) = ΔH(CaS) + 461.9 kJ/mol

For third reaction find ΔH(S₈):

1/8ΔH(S₈) - ΔH(CaS) = 482.4 kJ/mol

1/8ΔH(S₈) = ΔH(CaS) + 482.4 kJ/mol

Enthalpy of the reaction H2(g) + 1/8 S8(s) → H2S(g) is:

Δ H_reaction = ΔH(H₂S) - 1/8ΔH(S₈) = ΔH(CaS) + 461.9 kJ/mol - ΔH(CaS) - 482.4 kJ/mol = -20.5 kJ/mol