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Calculate ΔH⁰298 (in kJ) for the process P(s) + 5/2 Cl2(g) → PCl5(g) from the following information.
P(s) + 3/2 Cl2(g) → PCl3(g) ΔH⁰298 = −287 kJ
PCl3(g) + Cl2(g) → PCl5(g) ΔH⁰298 = −88 kJ

Respuesta :

Answer:

[tex]\Delta H_{298}^{0}[/tex] for the process is -375 kJ

Explanation:

  • Given reaction is a combination of the two given elementary steps.
  • Summation of change in standard enthalpy ([tex]\Delta H_{298}^{0}[/tex])of the two elementary reactions give [tex]\Delta H_{298}^{0}[/tex] of the reaction [tex]P(s)+\frac{5}{2}Cl_{2}(g)\rightarrow PCl_{5}(g)[/tex].

[tex]P(s)+\frac{3}{2}Cl_{2}(g)\rightarrow PCl_{3}(g)[/tex]......[tex]\Delta H_{1}=-287kJ[/tex]

[tex]PCl_{3}(g)+Cl_{2}(g)\rightarrow PCl_{5}(g)[/tex]......[tex]\Delta H_{2}=-88kJ[/tex]

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[tex]P(s)+\frac{5}{2}Cl_{2}(g)\rightarrow PCl_{5}(g)[/tex]

[tex]\Delta H_{298}^{0}[/tex] = [tex]\Delta H_{1}+\Delta H_{2}=-287kJ-88kJ=-375kJ[/tex]

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