# The sign of enthalpy of formation of magnesium oxide

I'm currently doing a lab to calculate the enthalpy of formation for $\ce{MgO}$. However at the moment me and my lab partner are having a disagreement. We've both calculated and agreed upon the same heat released in the reactions ($\pu{4.2\frac {kJ}{mol}}$ (reaction of $\ce{MgO +2HCl})$ and $\pu{7.5\frac {kJ}{mol}}$ (reaction of $\ce{Mg +2HCl}$)). Where we disagree is over how to calculate the enthalpy of reaction.

My argument:

Enthalpy is the amount of energy put into the reaction, what we measured is the heat released thus the equation should be: $$\Delta H=-\frac{Q}{n}$$ Also, this would make sense considering our reactions were blatantly exothermic, and using this equation provides negative values.

Her argument:

The equation is simple and the teacher agreed with her results: $$\Delta H=\frac{Q}{n}$$

My question: Who is correct?

• She is right, and the Q should be negative. – Chet Miller Apr 26 '18 at 1:08
• I'm pretty sure you are correct. Q is the change in heat for the surroundings. Thus, you have to put a negative sign in front of it to determine the change in heat for the system. If the system loses heat, the surroundings gain heat, vice versa. – Frank Apr 26 '18 at 1:10
• @Frank, That is not correct. Q represents the amount of heat you have to add to the system so that the final temperature of the products is the same as the initial temperature of the reactants. So Q represents heat added to the system, not heat added to the surroundings. – Chet Miller Apr 26 '18 at 1:37
• I think we're talking about the same thing in opposite ways. What I'm talking about is, when you measure the temperature change in the calorimeter, and you convert that to heat, you are calculating the heat added/removed from the surroundings. In a calorimetry experiment, you measure the temperature change of water. OP even noted that the calculated Q (heat) was positive, and since the reaction was exothermic, the Q he measured was obviously the heat added to the surroundings. I think what you are talking about is the inverse of this, although I'm not completely sure what you mean. – Frank Apr 26 '18 at 2:59
• In thermodynamics, the convention is that Q represents the heat transferred from the surroundings to the system. So, for an exothermic reaction, both Q and $Delta H$ are negative. There is no need to multiply by -1 if the convention is followed properly. – Chet Miller Apr 26 '18 at 10:36

## 1 Answer

I guess you've been answered in the comments already, but anyway. You are correct. In thermodynamics, enthalpy is the energy supplied to the system at constant pressure. If a reaction is accompanied by the release of heat, i.e. if it is exothermic, its $$\Delta H$$ is negative. If $$x$$ kJ is released into the calorimeter upon dissolving 1 mol of substance, resulting in the rise of calorimeter temperature, then $$\Delta H_{diss} = -x$$ kJ/mol.

As for the heat, depends on the type of the calorimeter, i.e. on the direction in which the heat flows. For conventional "coffee cup" calorimeter, where $$Q=mC\Delta T$$, $$Q$$ is the energy released, so $$Q = -\Delta H$$. For some power-compensated calorimeter, where you vary the heater power so as to maintain the temperature, $$Q$$ would be equal to the energy supplied. Since you have to supply excess energy ($$Q>0$$) to offset the endothermic reaction ($$\Delta H>0$$), $$Q = \Delta H$$.

I've heard that a long time ago, at least in some countries, there were two systems of signs for enthalpies, so-called "thermodynamic" and "thermochemical," with reversed signs in the latter. However, I haven't seen "thermochemical" one in real life, and it seems to be long since abandoned. It is not even mentioned in modern textbooks on thermodynamics. So only the thermodynamic sign system, where $$\Delta H>0$$ for endothermic processes, remains.