# How different concentration of HCl on magnesium affect the enthalpy change

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In fact all three reactions produced the same amount of heat energy and hence had the same enthalpy change. By changing the concentration of the HCl only effects the reaction rate and how much energy is produced in a certain time. However eventually, each reaction will produce the same amount of energy provided that all the Mg reacts which does happen in your experiment as HCl is in excess. The reason for this can be explained quick easily using the formula: $$\Delta H=\Delta U+pV$$ Then using the ideal gas equation, we can substitute $pV$ with $\Delta n(g)RT$. Note that the only variable that actually changes is n(g) which is the change in the number of moles of gas molecules. So the equation becomes: $$\Delta H= \Delta U + \Delta n(g)RT$$ So the only thing that effects the enthalpy change of the reaction in the change in internal energy (U) and the number of moles of gas molecules formed which are both constant for all three reactions.

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### Ethan Hunt

Updated on August 01, 2022

I had conducted an experiment where I reacted a $0.3\ \mathrm{g}$ strip of magnesium with $100\ \mathrm{mL}$ of varying concentrations of $\ce{HCl}$: $1\ \mathrm{mol/L}$, $2\ \mathrm{mol/L}$ and $3\ \mathrm{mol/L}$. From the data I collected, I noticed that the reaction with $1\ \mathrm{mol/L}$ $\ce{HCl}$ produced the most energy and the reaction with $3\ \mathrm{mol/L}$ $\ce{HCl}$ the least. I just wanted to know if my data is correct or not because I hypothesized that the $3\ \mathrm{mol/L}$ reaction would produce the most heat energy.