# Hess's Law and ΔH

Target equation: $$\ce{Mg(s) + 1/2 O2 -> MgO(s)}$$

\begin{align} \ce{MgO(s) + 2 HCl(aq) &-> MgCl2(aq) + H2O(l)} &\quad ΔH &= \pu{-1300 kJ} \tag{1}\\ \ce{Mg(s) + 2 HCl(aq) &-> MgCl2(aq) + H2(g)} &\quad ΔH &= \pu{-602 kJ} \tag{2}\\ \ce{H2(g) + 1/2 O2 &-> H2O(l)} &\quad ΔH &= \pu{-286 kJ} \tag{3} \end{align}

The target equation should have a negative $$ΔH$$ meaning that it's exothermic, but when I add it up, $$ΔH$$ ends up being positive. Is there anything I'm missing?

• If you are calculating $ΔΗ$ as $(2) + (3) - (1)$, then it looks correct, however, you are right, it should be negative. Is there a typo for $ΔΗ_1?$ It looks way to high: $\pu{-130 kJ mol-1}$ would make more sense than $\pu{-1300 kJ mol-1}$ IMO (I assume molar enthalpy based on the value for (3)). – andselisk Nov 16 '19 at 18:03
• I got 1300 kJ mol−1 by taking 3.34 kJ (the q value for energy absorbed by HCl) and the number of mols of MgO used, 0.00257 mol, and dividing 3.34 by 0.00257. One thing I forgot to mention is that this is a lab. Is there anything in the q = mcΔT calculation for HCl that I messed up on? – Justin Mathew Nov 16 '19 at 18:34
• It's hard to tell, but most likely yes, enthalpy of the first reaction has been determined incorrectly. You probably should edit in more details as to how it was obtained, as well as the value you expect and why (I already commented on that). – andselisk Nov 16 '19 at 19:01
• I agreed with @andselisk. I already give an answer, but I'm looking at where you have made mistake, as a part of following andselisk's suggestions. I assumed you made mistake on amount of $\ce{Mg}$ you have used. It should be $\pu{\approx 0.025 mol}$, if you have used $\pu{\approx 1 g}$ of $\ce{Mg}$. So please edit your post to include all these details. – Mathew Mahindaratne Nov 16 '19 at 19:11

Your target equation is: $$\ce{Mg(s) + 1/2O2 -> MgO(s)}$$. You have given following set of equations to get your target's enthalpy of formation: $$\ce{MgO(s) + 2HCl(aq) -> MgCl2(aq) + H2O(l)} \quad ΔH = \pu{-1300 kJ} \tag {1}$$ $$\ce{Mg(s) + 2HCl(aq) -> MgCl2(aq) + H2(g)} \quad ΔH = \pu{-602 kJ} \tag {2}$$ $$\ce{H2(g) + 1/2O2 -> H2O(l)} \quad ΔH = \pu{-286 kJ} \tag {3}$$

Yet, if you added equations $$(1)$$ through $$(3)$$, you won't get your target equation because you have set equation $$(1)$$ in opposite direction so that $$\ce{MgO}$$ would ends up in reactant side (LHS). What you should have done is reverse that reaction. The reverse of Equation $$(1)$$ has identical $$\Delta H$$ but with opposite sign:

$$\ce{MgCl2(aq) + H2O(l) -> MgO(s) + 2HCl(aq)} \quad ΔH = \pu{+1300 kJ} \tag {4}$$

Now you can add equations $$(2)$$ through $$(4)$$, and you should get a negative $$ΔH$$ for your target equation (meaning the reaction is exothermic, and given $$\Delta H^\circ_f$$ for $$\ce{MgO}$$ is $$\pu{−601.6 kJ/mol}$$ according to Wikipedia). However, your given $$\Delta H$$ values are seemingly incorrect and I leave you to find the correct values (Note: The given $$\Delta H^\circ_f$$ for water is acceptable).

Late edition: According to OP's late comment to andselisk's suggestion on first comment, it seems OP has made a mistake on amount of $$\ce{MgO}$$ calculation, of which OP has used in the calarimetric experiment. It should have been $$\pu{\approx 0.025 mol}$$, if OP has used $$\pu{\approx 1 g}$$ of $$\ce{MgO}$$. Thus, $$\Delta H$$ for the reaction $$\ce{MgO(s) + 2HCl(aq) -> MgCl2(aq) + H2O(l)}$$ should have been $$\approx \pu{3.34 kJ}/\pu{0.025 mol}= \approx \pu{134 kJ/mol}$$. Thus, $$\Delta H^\circ_f$$ of $$\ce{MgO}$$ from the experiment is: $$\pu{(+134-602-286) kJ/mol}=\pu{-754 kJ/mol}$$ (literature value is $$\pu{−601.6 kJ/mol}$$).

Until the time of this edition, OP has not edited the question. Thus, I assume OP has also calculated the data for the reaction $$\ce{Mg(s) + 2HCl(aq) -> MgCl2(aq) + H2(g)}$$, actual value of which has been around $$\pu{−460 kJ/mol}$$. Hence, approximate experimental value for $$\Delta H^\circ_f{_\ce{MgO}}$$ is: $$\pu{(+134-460-286) kJ/mol}=\pu{-612 kJ/mol}$$, which is better acceptable value for literature value, $$\pu{−601.6 kJ/mol}$$.

• As I already mentioned in the comment, the first value for neutralization looks way off. Since you already posted an attempt to answer, probably it would make sense to add that $ΔH_1$ should be approx. double the value of standard enthalpy of neutralization, $\pu{-57.62 kJ mol-1}.$ That's why I assumed $\pu{-130 kJ mol-1}$. Still it would be nicer if you dig up actual literature values; in the absence of those it's just speculations. – andselisk Nov 16 '19 at 18:33
• @ andselisk: Couldn't fond real literature values for them, but approximate values were calculated using standard enthalpies of formation given in en.wikipedia.org/wiki/Standard_enthalpy_of_formation. – Mathew Mahindaratne Nov 16 '19 at 20:45

Why do you add the three equations ?

By first adding (2) and (3) and then SUBSTRACTING (1), you find Mg + (1/2) O2 --> MgO.

On the other end, I think some of these numerical values are wrong, because one should obtain -602 kJ/mol, which is the heat of formation of MgO. And the combination of the given values do not yield to this result, by far.

• I got 1300 kJ/mol by calculating q of HCl divided by number of moles of MgO used, which ended being 3.34 kJ divided by 0.00257 mol. This was a lab by the way. – Justin Mathew Nov 16 '19 at 18:36