Question: What is the enthalpy change for the unknown reaction? Pb(s) + Cl2(g) ( PbCl2(s)H = 359 kJ PbCl2(s) + Cl2(g) ( PbCl4()H = ? Pb(s)

What is the enthalpy change for the unknown reaction?
Pb(s) + Cl2(g) ( PbCl2(s)ΔH = −359 kJ
PbCl2(s) + Cl2(g) ( PbCl4(ℓ)ΔH = ?
Pb(s) + 2Cl2(g) ( PbCl4(ℓ) ΔH = −329 kJ

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The unknown reaction has PbCl 2 as a reactant and PbCl 4 as the product So if we reverse the ... View full answer

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