Consider these reactions, where M represents a generic metal. 2M (s) + 6HCl (aq) ⟶2MCl3 (aq) + 3H2 (g) ΔH1=-556.0 kJ HCl (g) ⟶HCl (aq) ΔH2=-74.8 kJ H2 (g) + Cl2 (g) ⟶2HCl (g) ΔH3=-1845.0 kJ MCl3 (s) ⟶MCl3 (aq) ΔH4=-342.0 kJ Use the given information to determine the enthalpy of the reaction 2M (s) + 3Cl2 (g) ⟶2MCl3 (s)

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Athena Drake

Chemistry

18 April, 14:36

Consider these reactions, where M represents a generic metal. 2M (s) + 6HCl (aq) ⟶2MCl3 (aq) + 3H2 (g) ΔH1=-556.0 kJ HCl (g) ⟶HCl (aq) ΔH2=-74.8 kJ H2 (g) + Cl2 (g) ⟶2HCl (g) ΔH3=-1845.0 kJ MCl3 (s) ⟶MCl3 (aq) ΔH4=-342.0 kJ Use the given information to determine the enthalpy of the reaction 2M (s) + 3Cl2 (g) ⟶2MCl3 (s)

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Kurly · Answer 1

Chemical reaction 1: 2M (s) + 6HCl (aq) ⟶ 2MCl3 (aq) + 3H2 (g); ΔH1=-556.0 kJ. Chemical reaction 2: HCl (g) ⟶ HCl (aq); ΔH2=-74.8 kJ.

Chemical reaction 3: H2 (g) + Cl2 (g) ⟶ 2HCl (g); ΔH3=-1845.0 kJ

Chemical reaction 4: MCl3 (s) ⟶ MCl3 (aq); ΔH4=-342.0 kJ.

Chemical reaction 5: 2M (s) + 3Cl2 (g) ⟶ 2MCl3 (s); ΔH5 = ?.ΔH5 = ΔH1 + 6·ΔH2 + 3·ΔH3 - 2·ΔH4.

ΔH5 = - 550 kJ + 6· (-74,8 kJ) + 3· (-1845 kJ) - 2· (-342 kJ).

ΔH5 = - 550 kJ - 448,8 kJ - 5535 kJ + 684 kJ.

ΔH5 = - 5849,8 kJ.