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Say I (perfectly) mix $1\ \mathrm{m^3}$ of 1 M $\ce{HCl}$ with $1\ \mathrm{m^3}$ of 1 M $\ce{NaOH}$ at standard air pressure and room temperature within a metal vessel. Heat of neutralization for strong acids and bases is $-55.8\ \mathrm{kJ/mol}$ (is that actually just the bond enthalpy of an $\ce{H-OH}$ bond?).

So the $2\ \mathrm{m^3}$ of $\ce{H2O}$ amount to around $2 \cdot 10^6\ \mathrm{mol}$ of water ($M(\ce{H2O}) \overset{\wedge}{=} 20\ \mathrm{g\ mol^{-1}}$). Therefore, $\Delta_\mathrm n H$ is around $2 \cdot 10^6 \times 55.8\ \mathrm{kJ/mol}$, around $110 \cdot 10^6\ \mathrm{kJ}$.

I would now like to calculate the temperature rise.

The heat released is $Q = m c_p \Delta T$, where $m$ is the mass of the solution ($2000\ \mathrm{kg}$), $c_p$ is the specific heat capacity of the solution and $\Delta T$ ist the change in temperature. But that's exactly what I'm trying to find. So unless I know $Q$, I can't find $\Delta T$. Therfore, how do I find $Q$?

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  • $\begingroup$ Please don't use MathJax in titles due to searchability issues and URL slug issue. (I don't think the compound $\ce{CeHCl}$ exists) $\endgroup$ – M.A.R. Aug 27 '15 at 19:55
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You have 1 mole/liter of each reactant. How many liters of each solution do you have? That gives you the number of moles of reactants. They will create way less than $2\cdot 10^6$ moles of water-you were counting the water that was already there. Once you have the right number of moles of reactant, that is the number of moles of water produced. Multiplying by your heat per mole gives $Q$. Then divide by $mc_p$ to get $\Delta T$

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  • $\begingroup$ Which theory tells me that here in this case $H$ is just another symbol for $Q$? $\endgroup$ – TMOTTM Aug 28 '15 at 4:28
  • $\begingroup$ They are defined differently, but in this problem you are supposed to recognize that they are the same. $H$ is the energy released by one class of reactions. You have claimed that this reaction is typical of the class and it is the only reaction going on. In that case it is the equal to $Q$. $\endgroup$ – Ross Millikan Aug 28 '15 at 4:39
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As Ross has said, $\Delta H$ is the same as $q$. This is true only under three conditions:

  1. Pressure is constant;
  2. Only p-V work is performed;
  3. The process occurs in a closed system.

The First Law of Thermodynamics states that $\mathrm{d}U = \mathrm{d}q + \mathrm{d}w$. The expression for p-V work is $\mathrm{d}w = -p\,\mathrm{d}V$, giving $\mathrm{d}U = dq - p\,\mathrm{d}V$.

Enthalpy, $H$, is defined as $H = U + pV$. At constant pressure, this means that $$\begin{align} \mathrm{d}H &= \mathrm{d}U + p\,\mathrm{d}V + V\,\mathrm{d}p \\ &= \mathrm{d}U + p\,\mathrm{d}V \quad\quad \mathrm{(since}\,\mathrm{d}p = 0) \\ &= \mathrm{d}q - p\,\mathrm{d}V + p\,\mathrm{d}V \\ &= \mathrm{d}q \end{align} $$

Integrating over the path of interest, you would find that $\Delta H = q$.


Apart from that, I think you would also find that the number of moles of water produced is wrong. You will have to calculate it stoichiometrically, using the equation: $$\ce{HCl + NaOH -> H2O + NaCl}$$

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2 cubic meters of water aren't produced in the reaction, but they come from the acid and base solutions, so I do not see why you are multiplying 2 cubic meters of water with the enthalpy of neutralization. By multiplying the amount of water actually produced during neutralisation with the enthalpy of neutralisation, you get Q, the amount of heat released, so the temperature is the only unknown.

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