Question

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

Answer 1

`\Delta H_(298)^(0)` for the process is -375 kJ

Step-by-step explanation:

• Given reaction is a combination of the two given elementary steps.

• Summation of change in standard enthalpy (
  `\Delta H_(298)^(0)`)of the two elementary reactions give
  `\Delta H_(298)^(0)` of the reaction
  `P(s)+(5)/(2)Cl_(2)(g)\rightarrow PCl_(5)(g)`.

`P(s)+(3)/(2)Cl_(2)(g)\rightarrow PCl_(3)(g)`......
`\Delta H_(1)=-287kJ`

`PCl_(3)(g)+Cl_(2)(g)\rightarrow PCl_(5)(g)`......
`\Delta H_(2)=-88kJ`

-----------------------------------------------------------------------------------------------------

`P(s)+(5)/(2)Cl_(2)(g)\rightarrow PCl_(5)(g)`

`\Delta H_(298)^(0)` =
`\Delta H_(1)+\Delta H_(2)=-287kJ-88kJ=-375kJ`