# 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. – Chester Miller Apr 26 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 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. – Chester Miller Apr 26 at 1:37
• @ChesterMiller I'm pretty sure this experiment is carried out in a calorimeter, and Q is calculated using Q = mcΔT. I feel like that would be the most reasonable method to carry out such an experiment. Also, why would you want to add heat to make the products the initial temperature instead of just measuring the temperature change? – Frank Apr 26 at 1:58
• 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 at 2:59