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$$\begin{alignat}{2}\ce{X(s) + 1/2O2(g) &-> XO}&&\qquad\Delta H=-895.5\ \mathrm{kJ} \\ \ce{XCO3(s) &-> XO(s) + CO2(g)}&&\qquad\Delta H= +484.3\ \mathrm{kJ}\\ \ce{X(s) + 1/2O2(g) + CO2 &-> XCO3(s)}&&\qquad\Delta H = \;?\ \mathrm{kJ} \end{alignat}$$

I wish to know how to do it. I have seen some questions where there's 3+ equations and at the end a final equation that I would need to determine the $\Delta H$ for the final equation. I have looked online on how to solve this but the tutorials are very confusing and I can't understand them. Can someone explain how to do this and other similar problems like this?

One of the tutorials talks about making sure they cancel out and flipping the equations? Can someone please explain this?

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  • $\begingroup$ Well, that's pretty much about it: you can cancel things out and flip the equations, just like you would do in algebra, say, with $x=x^2+x-1$. $\endgroup$ – Ivan Neretin Oct 15 '15 at 8:37
  • $\begingroup$ @IvanNeretin Ok but why would you need to flip it? Like can you explain step by step how to solve these types of problems? I'm totally lost here. $\endgroup$ – user3882522 Oct 15 '15 at 8:44
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    $\begingroup$ @surelyourejoking I'm assuming that's not proper formatting, but that's how I think of chemical equations in my head, as equations. But can you please explain how to solve these types of problems step by step? $\endgroup$ – user3882522 Oct 15 '15 at 8:47
  • $\begingroup$ Flip the second equation, add it to the first, cancel out whatever you can. See where this gets you. $\endgroup$ – Ivan Neretin Oct 15 '15 at 8:48
  • $\begingroup$ @IvanNeretin Ok so under what conditions would I need to flip an equation? Is there anyway to tell when I will need to flip an equation? I mean thank you for telling me what to do to solve it but I want to understand the topic so I can apply it to other problems $\endgroup$ – user3882522 Oct 15 '15 at 8:52
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As enthalpy is a state function, so regardless the multiple stages of a reaction, the total enthalpy change for the reaction is the sum of all changes.This is Hess's law. So, you can see clearly that the third reaction results from adding the first reaction to the reverse of the second reaction, and so is the enthalpy change.

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Ok so for future reference if anyone needs help with this you look at the last equation and then at the other ones, if the compounds(ex: CO2) of the other equations don't appear to be on the same side of the final equation you flip that equation and as a result the sign. After you have done this add everything up and it will give you the answer

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  • $\begingroup$ While the method used to obtain a result is definitely useful, the reasoning behind that method is much more important. Please don't rote learn methods like this, the answer given by Yomen is better as it also captivates the reasoning. $\endgroup$ – Gaurang Tandon Feb 23 '18 at 1:45

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