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Question :

Draw a Hess`s law cycle and use it, together with the data given, to calculate the enthalpy change for the reaction :

P(s) + 1,5 Cl2(g) ---> PCl3(l)

Data:

P(s)+ 2,5 Cl2(g) ---> PCl5 (s) H= -463 kj mol-1

PCl3 (l) + Cl2 (g) ---> PCl5 (s) H= -124 kj mol-1

Please could anybody let me know what to write on the bottom of Hess`s cycle, as I got confused by the data given...what do they talk about PCl5 ???

thanks a lot.

Draw a Hess`s law cycle and use it, together with the data given, to calculate the enthalpy change for the reaction :

P(s) + 1,5 Cl2(g) ---> PCl3(l)

Data:

P(s)+ 2,5 Cl2(g) ---> PCl5 (s) H= -463 kj mol-1

PCl3 (l) + Cl2 (g) ---> PCl5 (s) H= -124 kj mol-1

Please could anybody let me know what to write on the bottom of Hess`s cycle, as I got confused by the data given...what do they talk about PCl5 ???

thanks a lot.

The question wants you to find P(s) + 1,5 Cl2(g) ---> PCl3(l) .

You can't do this under standard conditions so you have to by a different route ( the other two equations)

You need to manipulate/change the two equations to find a value for the first equation.

I would inverse the second equation to give PCl5 ---> PCl3 + Cl2 (so the enthalpy reverses too to give +124kj mol-1)

Then if you add this equation and the equation P(s)+ 2,5 Cl2(g) ---> PCl5 (s) H= -463 kj mol-1

You get PCl5 + PCl3 + Cl2 --> PCl5 + P + Cl2

Anything that is the same on both sides, cancels so you get rid of it :

PCl3--> PCl5 + P

I know that doesnt end up with the answer but maybe it will help? You can also times and divide the equations by what you want, so long as you do the same to both sides. Im sorry I couldnt be of more use! from emzaz

You can't do this under standard conditions so you have to by a different route ( the other two equations)

You need to manipulate/change the two equations to find a value for the first equation.

I would inverse the second equation to give PCl5 ---> PCl3 + Cl2 (so the enthalpy reverses too to give +124kj mol-1)

Then if you add this equation and the equation P(s)+ 2,5 Cl2(g) ---> PCl5 (s) H= -463 kj mol-1

You get PCl5 + PCl3 + Cl2 --> PCl5 + P + Cl2

Anything that is the same on both sides, cancels so you get rid of it :

PCl3--> PCl5 + P

I know that doesnt end up with the answer but maybe it will help? You can also times and divide the equations by what you want, so long as you do the same to both sides. Im sorry I couldnt be of more use! from emzaz

Janka3112

Question :

Draw a Hess`s law cycle and use it, together with the data given, to calculate the enthalpy change for the reaction :

P(s) + 1,5 Cl2(g) ---> PCl3(l)

Data:

P(s)+ 2,5 Cl2(g) ---> PCl5 (s) H= -463 kj mol-1

PCl3 (l) + Cl2 (g) ---> PCl5 (s) H= -124 kj mol-1

Please could anybody let me know what to write on the bottom of Hess`s cycle, as I got confused by the data given...what do they talk about PCl5 ???

thanks a lot.

Draw a Hess`s law cycle and use it, together with the data given, to calculate the enthalpy change for the reaction :

P(s) + 1,5 Cl2(g) ---> PCl3(l)

Data:

P(s)+ 2,5 Cl2(g) ---> PCl5 (s) H= -463 kj mol-1

PCl3 (l) + Cl2 (g) ---> PCl5 (s) H= -124 kj mol-1

Please could anybody let me know what to write on the bottom of Hess`s cycle, as I got confused by the data given...what do they talk about PCl5 ???

thanks a lot.

If you directly subtract equation 2 from equation 1 you get:

P(s)+ 2,5 Cl2(g) ---> PCl5 (s) H= -463 kj mol-1

PCl3 (l) + Cl2 (g) ---> PCl5 (s) H= -124 kj mol-1

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

P(s) + 1,5 Cl2(g) - PCl3(l) ---> nothing ΔH= -339 kj mol-1

Now rearrange

P(s) + 1,5 Cl2(g) ---> PCl3(l) ΔH= -339 kj mol-1

charco

If you directly subtract equation 2 from equation 1 you get:

P(s)+ 2,5 Cl2(g) ---> PCl5 (s) H= -463 kj mol-1

PCl3 (l) + Cl2 (g) ---> PCl5 (s) H= -124 kj mol-1

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

P(s) + 1,5 Cl2(g) - PCl3(l) ---> nothing ΔH= -339 kj mol-1

Now rearrange

P(s) + 1,5 Cl2(g) ---> PCl3(l) ΔH= -339 kj mol-1

P(s)+ 2,5 Cl2(g) ---> PCl5 (s) H= -463 kj mol-1

PCl3 (l) + Cl2 (g) ---> PCl5 (s) H= -124 kj mol-1

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

P(s) + 1,5 Cl2(g) - PCl3(l) ---> nothing ΔH= -339 kj mol-1

Now rearrange

P(s) + 1,5 Cl2(g) ---> PCl3(l) ΔH= -339 kj mol-1

ok...from ur explanation i understand how u worked out the value for the enthalpy change of the reaction.

The only thing I still don`t get is what to write on the bottom part of Hess`s cycle?

Janka3112

ok...from ur explanation i understand how u worked out the value for the enthalpy change of the reaction.

The only thing I still don`t get is what to write on the bottom part of Hess`s cycle?

The only thing I still don`t get is what to write on the bottom part of Hess`s cycle?

P(s)+ 2,5 Cl2(g) ---> PCl5 (s) <--- PCl3 (l) + Cl2 (g)

Then the bottom is just

P(s)+ 2,5 Cl2(g) --------------------> PCl3 (l) + Cl2 (g)

which cancelling the chlorines is the same as writing:

P(s)+ 1,5 Cl2(g) --------------------> PCl3 (l)

charco

P(s)+ 2,5 Cl2(g) ---> PCl5 (s) <--- PCl3 (l) + Cl2 (g)

Then the bottom is just

P(s)+ 2,5 Cl2(g) --------------------> PCl3 (l) + Cl2 (g)

which cancelling the chlorines is the same as writing:

P(s)+ 1,5 Cl2(g) --------------------> PCl3 (l)

Then the bottom is just

P(s)+ 2,5 Cl2(g) --------------------> PCl3 (l) + Cl2 (g)

which cancelling the chlorines is the same as writing:

P(s)+ 1,5 Cl2(g) --------------------> PCl3 (l)

...to be honest, im totally confused and don`t know what to do anymore

Janka3112

...to be honest, im totally confused and don`t know what to do anymore

Have a cup of tea and come back to it later on...

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