Answer:
ΔH°C2H2Cl4(l) = -333,36 kJ/mol
ΔH°r₂ = -35,14 kJ/mol
Explanation:
The ΔH°r of the first reaction is:
ΔH°r = -385,76 kJ/mol = (ΔH°C2H2Cl4(l) + ΔH°H2(g)) - (ΔH°C2H4(g) + 2ΔHCl2 (g))
ΔH°H2(g) = 0 kJ/mol
ΔH°C2H4(g) = 52,4 kJ/mol
Δ°HCl2 (g) = 0 kJ/mol
Replacing:
ΔH°C2H2Cl4(l) = -385,76 kJ/mol + 52,4 kJ/mol = <em>-333,36 kJ/mol</em>
The standard heat of the second reaction is:
ΔH°r₂ = ΔH°C2HCl3(l) + ΔH°HCl(g) - ΔH°C2H2Cl4(l)
Where:
ΔH°C2HCl3(l) = -276,2 kJ/mol; ΔH°HCl(g) = -92,3 kJ/mol; ΔH°C2H2Cl4(l) = -333,36 kJ/mol
Replacing:
ΔH°r₂ = -276,2 kJ/mol -92,3 kJ/mol + 333,36 kJ/mol
<em>ΔH°r₂ = -35,14 kJ/mol</em>
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I hope it helps!
