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I conducted an experiment in which $2.0\ \mathrm g$ of $\ce{NaOH}$ was dissolved in $200\ \mathrm{mL}$ of distilled water. This took place in a well isolated calorimeter. A temperature probe was used to measure the temperature change of the solution, which had an initial temperature of $24.9\ \mathrm{^\circ C}$ and a final temperature of $39.6\ \mathrm{^\circ C}$. Then, the enthalpy change associated with the reaction is $$\begin{align} \Delta H &= -mc\Delta T \\ &= -(200\ \mathrm g)(4.184\ \mathrm{J/(g\ ^\circ C)})(39.7\ \mathrm{^\circ C} - 24.9\ \mathrm{^\circ C}) \\ &= -12.3\ \mathrm{kJ} \end{align}$$ However, according to the accepted enthalpy of solution, the actual enthalpy change should be $$\Delta H = (-44.2\ \mathrm{kJ/mol})(2.0\ \mathrm g/(40.0\ \mathrm{g/mol}))=-2.2\ \mathrm{kJ}$$

How is it possible that there should be more energy released experimentally than what is predicted theoretically? If anything, there should be less energy, as some will escape. The obvious factors can be ruled out, the temperature probe was only in contact with the solution, the calorimeter was very well insulated, exactly $2.0\ \mathrm g$ and $200\ \mathrm{mL}$ of sodium hydroxide and water was used. This experiment was also repeated, yielding similar results. I'm at my wit's end trying to figure this out.

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    $\begingroup$ Did you stir the solution in the calorimeter well? The NaOH would have sunk to the bottom. Convection would mix the solution some but not thoroughly. I'm also guessing that the temperature probe was near the bottom too. $\endgroup$ – MaxW Feb 21 '17 at 3:44
  • $\begingroup$ I'm pretty sure it was stirred well, but since I used the stirrer provided with the calorimeter it may not have reached all the way to the bottom. The probe was also near the bottom, so I think you're right. Would this really cause such a big difference though? $\endgroup$ – Nambiar M. Feb 21 '17 at 4:33
  • $\begingroup$ An error of +460% is pretty massive. Something was definitely wrong. Exactly what who knows... $\endgroup$ – MaxW Feb 21 '17 at 5:31
  • $\begingroup$ Yeah, I'll take another hard look at it. I think you're right about the stirring and temperature probe at the bottom however. Maybe add that as an answer. Thanks for your help. $\endgroup$ – Nambiar M. Feb 21 '17 at 11:41

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