x Turn on thread page Beta
 You are Here: Home

# Enthalpy cycles / Hess's law (A level) watch

1. Calculate the enthalpy change of formation of bromoethane C2H5Br from the following data:

- the enthalpy change for 2C + 2H2 ---------> C2H4 is +52.2 kJ mol-1

- the enthalpy change for 1/2H2 + 1/2Br2 -----------> HBr is - 36.4 kJ mol-1

- the enthalpy change for C2H4 + HBr -----------> C2H5Br is -106.3 kJ mol-1

First of all, why do we need to calculate the enthalpy change of formation of bromoethane when it already tells us it's -106.3 kJ mol-1 ?

I've been told that the actual answer is -90.5 How do I draw the enthalpy cycle for this?
2. Enthalpy change of formation: The enthalpy change when 1 mole of a compound is formed from it's elements which are in their standard states under standard conditions

The reactants are not in their standard states you see. Let me try it out now...

EDIT:
[DRAW ARROWS FROM DOWN TO UP]
Ok to draw it put the elements in standard states the bottom: C(s) + H2(g) + Br2(g
Then put: C2H4 + HBr -----> C2H5Br at the top
Draw the arrow from the elements to the left side and to the right side

The enthalpy change to form C2H4 is +52.2, while the enthalpy change to form HBr is -36.4
The total enthalpy change on the left arrow is 15.8 then

The enthalpy change for the right arrow is -106.3 as you know

So enthalpy change is:
(-106.3)-(15.8) = -122.1

It kind of makes sense. Breaking a molecule requires energy (so it is endothermic) and making bonds which are are doing gives out energy (so it is exothermic)
Our value is -122.1, which is exothermic
3. (Original post by depptastic)
Calculate the enthalpy change of formation of bromoethane C2H5Br from the following data:

- the enthalpy change for 2C + 2H2 ---------> C2H4 is +52.2 kJ mol-1

- the enthalpy change for 1/2H2 + 1/2Br2 -----------> HBr is - 36.4 kJ mol-1

- the enthalpy change for C2H4 + HBr -----------> C2H5Br is -106.3 kJ mol-1

First of all, why do we need to calculate the enthalpy change of formation of bromoethane when it already tells us it's -106.3 kJ mol-1 ?

I've been told that the actual answer is -90.5 How do I draw the enthalpy cycle for this?
The main reason you are having problem with this question is because you haven't completely understood most of the terms used in thermodynamics(a level), ie standard enthalpy of formation, standard enthalpy of combustion, what are them, can you write an eqn for them?

One more thing when dealing with these questions are to label the state. The initial and final states are very important for you to label the correct enthalpy values.
4. Thank you guys!

See I've spent hours on this question and I also kept getting -122.1 as the final answer. So why does the book say it's -90.5?
I don't normally have problems with these questions, but this one just confuses me...
5. I'm sorry having to tell you, but 90.5 is the correct answer. I have no idea how you guys do these calculations, but I have always found it easiest to treat them just as equations and add the sum up the enthalpies of formation... anyway, you get the result by adding -212.6 (twice enthalpy of formation for C2H5Br) + 104.4 (C2H4) minus 72.8 (HBr). Then divide by two as the unit is 'per mol'- as your chemical equation for the formation of C2H5Br has 2 as a coefficient before C2H5Br.

Hope that helps...

TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Updated: February 12, 2010
Today on TSR

### Four things top students are doing

Over the Easter break

### Her parents made her dump me

Poll

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE