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Relation between Enthalpy change for formation and enthalpy change for the reaction?

I need DESPERATE help, I am doing past papers, exam for Unit 1 is after tomorrow, and I don't understand what the relationship between the stated enthalpy change of formation can be used to solve for the enthalpy change. I don't want a straight forward answer to a question, I actually want an explanation.

Here is the question:

The standard enthalpy change for the formation of ethene, C2H4 is +52.2 kJ mol-1 and that of ethane is -84.7 kj mol-1

Calculate the standard enthalpy change for the reaction below in kJ mol-1

C2H4 + H2 --> C2H6

A) -32.5
B) -136.9
C) 136.9
D) This cannot be calculated using only the data above.

*I Know the answer, but I still don't get it, please someone explain qq*
(edited 8 years ago)
Original post by AhmedSiddiqui
I need DESPERATE help, I am doing past papers, exam for Unit 1 is after tomorrow, and I don't understand what the relationship between the stated enthalpy change of formation can be used to solve for the enthalpy change. I don't want a straight forward answer to a question, I actually want an explanation.

Here is the question:

The standard enthalpy change for the formation of ethene, C2H4 is +52.2 kJ mol-1 and that of ethane is -84.7 kj mol-1

Calculate the standard enthalpy change for the reaction below in kJ mol-1

A) -32.5
B) -136.9
C) 136.9
D) This cannot be calculated using only the data above.

*I Know the answer, but I still don't get it, please someone explain qq*


D, because there is no equation
Original post by charco
D, because there is no equation


I forgot to include the equation, it's

C2H4 + H2 --> C2H6
Original post by AhmedSiddiqui
I forgot to include the equation, it's

C2H4 + H2 --> C2H6


To get enthalpy of reaction you can 'pretend' to go via the elements in their standard states:

reactants --> elements in standard states ---> products

So you add the NEGATIVE of the enthalpy of formation of the reactants to the actual enthalpy of formation of the products.
Original post by charco
To get enthalpy of reaction you can 'pretend' to go via the elements in their standard states:

reactants --> elements in standard states ---> products

So you add the NEGATIVE of the enthalpy of formation of the reactants to the actual enthalpy of formation of the products.


Ahhh, you're a beast man, thank you so much. I think I understand it. So it's basically like doing hess law in your mind but you include elements in it?
Original post by AhmedSiddiqui
Ahhh, you're a beast man, thank you so much. I think I understand it. So it's basically like doing hess law in your mind but you include elements in it?


All energetics questions use Hess's law. Check out this interactive on Hess

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