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Not sure how to start- i have the enthalpies of formation though
Enthalpy of formation for H2 and Mg should be zero because they are both naturally occurring
Yes i have the enthalpies of formation, but not sure how to actually construct the diagram
Mg- 0 H2- 0 H2O- -241.8 MgO- -601.8
Subtract [Products - reactants] and if you get a number that's greater than zero it's endothermic if it's less than zero it's exptiermic. Then we will focus on constructing the diagram
Alright i did -601.8+241.8= -387
The zeros cancel out and the double negative becomes a positive
So its a - number= exothermic
So it's exothermic . That means that the products are at a lower energy than the reactants.' M
Yep- so do i start with the reactants?
So from here you need to construct your graph Y a is is energy X axis is reaction progress
Yeah so you would then put the enthLpy of formation of the reactants draw this as a horizontal line and do the same for the products
@Photon336 we also have to do the decomposition into elements and the recombination of elements into products- not sure how that factors in
Right now I'm using the phone app I can't use the draw feature. Will show you in a bit
Alright, thanks for your help so far
did they also give you the activation energy?
no they didn't
So the full directions are: show the enthalpy changes assoc with the decamp of the reactants into elements, recomb of the elements into products, overall enthalpy change for the reaction
http://s3.amazonaws.com/answer-board-image/5c34163e-6243-4ed4-b420-6b9ccb6fd2a7.jpeg similar to this
Oh okay I see
so you've already found the enthalpy for the reaction and that this is an exothermic reaction. Mg(s) + H2O(l) --> MgO + H2
yes, delta H of -387
Mg(s) is naturally occurring so it wont have an enthalpy of formation. same for hydrogen gas
I listed the enthalpies of formation above
I saw that, that's just to provide a reason why those numbers are zero in case you were wondering. other than that well I think it says decomposition of reactants into elements H2O--> H2 and O2 it's kind of the same I think. the enthalpy of formation you had for water was -241.8 so I think it would be +248 because you would have to put in energy to break the O-H bonds.
Other than that, I don't see how there could be more to this question. Well, I apologize because I thought you had to make an enthalpy diagram but I guess you cant do that unless you know the activation energy too.
thanks for your help!