$0.500\,\mathrm g$ of magnesium chips are placed in a coffee-cup calorimeter and $100.0\,\mathrm{mL}$ of $1.00\,\mathrm M$ $\ce{HCl}$ is added to it. The reaction is:

$$\ce{Mg(s) + 2HCl(aq) -> H2(g) + MgCl2(aq)}$$

The temperature of the solution increases from $22.2\,^{\circ}\mathrm{C}$ to $44.8\,^{\circ}\mathrm{C}$. What’s the enthalpy change for the reaction, per mole of $\ce{Mg}$? Assume specific heat capacity of solution is $4.20\,\mathrm {J\,g^{-1}K^{-1}}$ and density of the $\ce{HCl}$ solution is $1.00\,\mathrm {g/mL}$.

I don't understand how to solve this; does this have to do with heat capacity equation? (Or is that just a completely different thing).

$$q = mc\Delta T$$

(Also note this is not a homework problem. The answer should be $\Delta H = -4.64 \cdot 10^{5}\,\mathrm{J\,mol^{-1}}$ $\ce{Mg}$, but I don't know how to get there and would like to understand.)

  • 1
    $\begingroup$ You're on the right track. The enthalpy change is reflected in the heat released the reaction (you immediately know it should be negative). This heat is absorbed by the water, raising its temperature. You need to back compute the amount of heat using the temperature change and then determine how much heat per mole of magnesium. $\endgroup$
    – Zhe
    Commented Nov 29, 2016 at 4:08

1 Answer 1

  • Step 1 Calculate $q_\text{solution}$. To do so, you have to have in mind that the mass of the solution is aproximatly equals to $100\ \mathrm{ml}$ of $\ce{HCl}$ plus the mass of $\ce{Mg}$.

    \begin{align*} q_\text{solution}&=100.5\ \mathrm g\cdot4.20\ \frac{\mathrm J}{\mathrm{mol\cdot K}}\cdot\left(318.0\ \mathrm K-295.4\ \mathrm K\right)\\ &=9.54\cdot 10^3\ \mathrm J \end{align*}

  • Step 2 Calculate $q_\mathrm{rxn}$

    \begin{align*} q_\mathrm{rxn} + q_\text{solution} &=0\\ q_\mathrm{rxn} + 9.54\cdot 10^3\ \mathrm J &= 0 \end{align*}

  • Step 3 Calculate the value of $\Delta H$ per mole.

    \begin{align*} \Delta H&= \left(\frac{-9.54\cdot 10^3\ \mathrm J}{0.500\ \mathrm g\ \text{of}\ \ce{Mg}}\right)\left(\frac{24.31\ \mathrm g\ \text{of}\ \ce{Mg}}{1\ \mathrm{mol}\ \text{of}\ \ce{Mg}}\right)\\ &= -4.64 \cdot 10^5\ \frac{\mathrm J}{\mathrm{mol}}\ \text{of}\ \ce{Mg} \end{align*}


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