You are Here: Home >< A-levels

# Hess law calculation Watch

1. CO + 2H2 > CH3OH

Enthalpy change of formation for CO is -110 and for CH3OH it's -204
I'm a little lost with this as idk how to work it out without the value for H? Can anyone shed some light on this for me?
2. Anyone?
3. The H2 because hydrogen is an element, therefore enthalpy of formation for an element is zero. you can calculate enthalpy change by -204-(-110) = -94
4. (Original post by AqsaMx)
Anyone?
Hey!
So just as explained above, the answer will be -94 KJmol^-1.
To work out the enthalpy of formation, the rule is: ''sum of the enthalpies of the product(s)- sum of the enthalpies of the reactants''.
H2 is in its elemental form and so by definition will have a enthalpy of formation equal to zero- there is nothing to form as H2 exists exactly in this way.
So, the enthalpy of formation calculation should look like this:
= Enthalpy of formation of CH3OH - Enthalpy of formation of CO
= -204-(-110)
= -204+110
= -94 KJmol^-1.
Hope this helps
5. (Original post by AqsaMx)
CO + 2H2 > CH3OH

Enthalpy change of formation for CO is -110 and for CH3OH it's -204
I'm a little lost with this as idk how to work it out without the value for H? Can anyone shed some light on this for me?
Hey, a lot of people are explaining the short cut method, but i think you need to understand why this works. First of all, let us consider some definitions. Hess law tells us that the enthalpy change for a reaction is independent of the route taken, which means you can take any route you like as long as you start and finish in the same place. Also the standard molar enthalpy of formation means, the enthalpy change when one mole of a substance is formed from its constituent elements under standard conditions, all products and reactants being in their standard state. The enthalpy change of formation of any element is zero by definition. WHY? well the element remains in its standard state already, we don't need to add enthalpy change to form it, its there already made (hehe)! However if we had (H instead of H2) then yes we would need enthalpy change of formation for this because hydrogen exist as h2 in standard conditions(298k and 100kpa). Right, to find the enthalpy change for this reaction we have to consider a different route, from products to reactants. As we have been given to use enthalpy changes of formation, we will essentially think of breaking all reactants to their constituent elements and then forming the product from the constituent elements! Now we want to break carbon monoxide up, into its constituent elements! isn't this just the opposite of formation? yes it is! Now we are reversing the reaction of formation of carbon monoxide so we use the negative of the enthalpy of formation, this is a consequence of hess law, we are going in the opposite direction from product to reactant, so we reverse the sign of the formation value. So to convert carbon monoxide to its elements we need (-)(-110) kj mol-1 of enthalpy change = +110. This is the enthalpy change to break all the bonds (Hint: breaking bonds = endothermic, so we get positive enthalpy as you can see). Now the reactants we have just broken up into elements in standard state, we want to form the new product methanol from these. So we use the enthalpy of formation simply for this, why? well we are going form constituent elements and forming one mole of this, so we simply need -204kj mol-1 of energy. Now remember we could have gone directly from carbon monoxide to methanol, but we picked a different direction however enthalpy change will be same as we went from same reactants to same product, just different route. So to break up carbon monoxide we needed 110 and to make methanol we needed -204, so enthalpy change of the reaction is 110 + -204 =-94 kj mol-1. As you can see it's a negative value therefore this reaction was an exothermic, more bonds were formed than broken.
6. Thank you everyone

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: April 3, 2016
Today on TSR

What should I do?

### How to go from a 4 to a 9 in GCSE Maths?

Discussions on TSR

• Latest
• ## See more of what you like on The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

• Poll

## All the essentials

### Student life: what to expect

What it's really like going to uni

### Essay expert

Learn to write like a pro with our ultimate essay guide.

### Create a study plan

Get your head around what you need to do and when with the study planner tool.

### Resources by subject

Everything from mind maps to class notes.

### Study tips from A* students

Students who got top grades in their A-levels share their secrets

## Groups associated with this forum:

View associated groups
Discussions on TSR

• Latest
• ## See more of what you like on The Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

• 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