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I need help with this one homework question:

I am combining $100\,\text{mL}$ of $0.5~\text{M}~\ce{HCl}$ and $100~\text{mL}$ of $0.5~\text{M}~\ce{NaOH}$ to obtain $200~\text{mL}$ of $0.25~\text{M}~\ce{NaCl}$. The change in temperature of this reaction was $\Delta \vartheta =3.38~\text{°C}$. I have to calculate the heat of reaction for 1 mole of the reactant. I also need to calculate the $c_\text{total}$ of this reaction. How do I go about doing this?

So the equation is $$\ce{HCl + NaOH -> H2O + NaCl}$$

Suppose the heats of formation of $\ce{HCl}$, $\ce{NaOH}$, $\ce{H2O}$, and $\ce{NaCl}$ are $-166.19$, $-469.07$, $-285.85$, and $-406.77$, respectively, all in $\text{kJ/mol}$.

I went about doing this calculating the heat $$q=\Delta H=[-(-166.19-469.07)+(-285.85-406.77)]~\text{kJ/mol} = -57.36~\text{kJ/mol}$$

I then found the mass of the liquids used for the reaction:

$$\begin{aligned}m &= 0.1~\text{L}\times 0.500 ~\text{M}\times 36.4611~\mathrm{g\,mol^{-1}} \\ &\;\;+0.1~\text{L} \times 0.500~\text{M} \times 39.99715 ~\mathrm{g\,mol^{-1}}\\ &=3.822~ \mathrm{g} \end{aligned}$$

I then solved for $c$ using $q=mc\Delta T$.

$0.05\times 57360~\mathrm{J} = 3.822\ \mathrm{g}\times c_\text{total}\times 3.38~\mathrm{K} \Rightarrow c_\text{total}=222.01 ~\mathrm{J \, g^{-1}\, K^{-1}}$

By the way, I multiplied the $q$ by $.05$ because that's how many moles of reactant I have (since we have the molarity and volume of the reactant).

I am pretty sure I am doing something wrong, but I am not sure what.

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  • $\begingroup$ This question is nicely showing your work, but could you elaborate why you think something is wrong? What conceptual problem you have with the question? As it is, this is a 'check my calculation'-type of question that is off-topic on this site. $\endgroup$ – Michiel Mar 6 '14 at 6:20
  • $\begingroup$ You correctly calculated the heat per mole, $-57.36kJ/mol$...Consider that you do not have a mole of acid and base and then you're done ! $\endgroup$ – mannaia Mar 8 '14 at 7:58
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Two things are wrong in your calculation. Well, it's really just one thing, but the other one bugs me more.

  1. The total mass of the solution.
  2. Units.

1. Mass of the Solution

To calculate the mass of the total solution, I think it is fair to make the following assumption: $$ \rho_\ce{HCl} \approx \rho_\ce{NaOH} \approx \rho_\ce{NaCl} \approx \rho_\mathrm{H_2O} \approx 1000~\mathrm{kg\, m^{-3}}$$ $$ V_\text{tot} = V_1 + V_2$$

It says that we assume the densities of all the solutions to be the same, and that the total volume is the same as the sum of the two initial volumes.

Using the above assumptions, we can calculate the total mass of the reaction mixture: $$ m_\text{tot} = \rho V_\text{tot} = \rho (V_1 + V_2) = 1~\mathrm{g\, cm^{-3}} \times (200 ~\mathrm{cm^{3}}) = 200~\mathrm{g}$$

You have calculated the reaction enthalpy to be $\Delta H = -57.36~\mathrm{kJ\,mol^{-1}}$.

Now we have to calculate the total heat that was released during this reaction, which leads me to

2. Units

Using unit analysis (we have $\mathrm{kJ\,mol^{-1}}$ and need $\mathrm{J}$) we know that we need to multiply the reaction enthalpy with some amount of substance (unit: $\text{mol}$). Now that we know what we have to plug in, we need to know how much of it. Since the reaction enthalpy that you calculated is per $\text{mol}$ of reactant, and we have exactly $0.25~\mathrm{M} \times 0.2~\mathrm{L} = 0.05~\mathrm{mol}$ of reactants, we calculate the total heat as follows: $$ q_\text{tot} = n_\text{tot} \times -\Delta H_\text{reaction} = 0.05~\mathrm{mol} \times -57.36~\mathrm{kJ\,mol^{-1}} = 2.868~\mathrm{kJ}$$ Why the $-\Delta H$? Well, the heat that goes out from the reaction goes into the mixture. That is why the sign has to change here.

We also need the temperature difference, which was measured to be $\Delta T = 3.38~\mathrm{K}$.

The final result (the heat capacity) is calculated via $$ q_\text{tot} = m_\text{tot}c_\text{tot}\Delta T \Rightarrow c_\text{tot} = \frac{q_\text{tot}}{m_\text{tot} \Delta T} = \frac{2.868~\mathrm{kJ}}{200~\mathrm{g} \times 3.38~\mathrm{K}} = 4.243~\mathrm{J\, g^{-1}\, K^{-1} }$$ This is a little bit more than the specific heat capacity of water, so we're good!

I hope my walk-through has demonstrated all crucial points of where you went wrong, and that it clarified things for you.

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