Question

Given the enthalpies of reaction: 2P(g) + 3Cl2(g) → 2PCl3(g) DH = –574 kJ 2P(g) + 5Cl2(g) → 2PCl5(g) DH = –887 kJ What is the enthalpy change of the following reaction: PCl3(g) + Cl2(g) → PCl5(g

Answer 1

Answer:
`PCl_3(g)+Cl_2(g)\rightarrow PCl_5(g)`
`\Delta H=-156.5kJ`

Step-by-step explanation: It's based on Hess's law. We will rearrange the two given equations and make the equation to which it asks to calculate the enthalpy change.

First equation needs to be reversed and divided by 2 so that we could get one
`PCl_3` on reactant side. When we reversed an equation then the sign of enthalpy change is also changed.

`PCl_3(g)\rightarrow P(g)+(3)/(2)Cl_2(g)`
`\Delta H=287kJ`

Second equation also needs to be divided by 2 and its not reversed.

`P(g)+(5)/(2)Cl_2(g)\rightarrow PCl_5(g)`
`\Delta H=-443.5kJ`

Both the equations are added now:

`PCl_3(g)+Cl_2(g)\rightarrow PCl_5(g)`
`\Delta H=-156.5kJ`

So, the enthalpy change for the equation is -156.5 kJ.