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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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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.

The enthalpy of formation of [tex]\rm MCl_3[/tex] will be -5849 kJ.

The given reactions represent:

Enthalpy of reaction of M with HCl = [tex]\rm \Delta H_1[/tex] = -556 kJ

Enthalpy of sublimation of HCl = [tex]\rm \Delta H_2[/tex] = -74.8 kJ

Enthalpy of HCl formation = [tex]\rm \Delta H_3[/tex] = -1845 kJ

Enthalpy of sublimation of [tex]\rm MCl_3\;=\;\Delta H_4[/tex] = -342.0 kJ

The enthalpy of formation of [tex]\rm MCl_3[/tex] will be: [tex]\rm \Delta H_5[/tex]

[tex]\rm \Delta H_5[/tex] = [tex]\rm \Delta H_1\;+\;6\;\times\;\Delta H_2\;+\;3\;\times\;\Delta H_3\;-\;2\;\times\;\Delta H_4[/tex]

= -550 + 6 (-74.8) + 3 (-1845) - 2 (342)

= -5849 kJ.

The enthalpy of formation of [tex]\rm MCl_3[/tex] will be -5849 kJ.

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