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In most, if not all, general chemistry books, you will find that in from constant volume calorimeters, $\Delta H = \Delta U + RT \Delta n_\mathrm{g}$, which is, of course derived from $H = U + PV$.

My question is, given that the measured quantity is the change of temperature of the calorimeter, how can we assume that $T$ is constant to get to $\Delta H = \Delta U + RT\Delta n_\mathrm{g}$?

Edit:

As said in the comments, I found this link where it says that this approximation is valid for a small $\Delta T$ value. But if we have the values, why not just use them to get a more accurate number? Furthermore, for example, let's assume an experiment we measure a $\pu{10 kJ mol-1}$ heat from a $T_\mathrm{i} = \pu{298 K}$ to $T_\mathrm{f} = \pu{303 K}$, and a $n_\mathrm{i} = 0$, $n_\mathrm{f} = 2$. Then the approximation gives:

$$\Delta H = \pu{-10 kJ mol-1} + (\pu{8.314 J mol-1 K-1})(2)(\pu{298 K}) = \pu{-5.04 kJ mol-1}$$

$$\Delta H = -\pu{10 kJ mol-1} + (\pu{8.314 J mol-1 K-1})\left[(2)(\pu{303 K}) - (0)(\pu 298 K)\right] = \pu{-4.96 kJ mol-1}$$

And the error is, indeed, small ($\pu{80 J}$) (side question, isn't $\Delta n$ supposed to be in moles? my dimensional analysis doesn't match if it does). However, if the reaction was from $n_\mathrm{i} = 2$ to $n_\mathrm{f} = 4$, and we keep the other values, we get:

$$\Delta H = \pu{-10 kJ mol-1} + (\pu{8.314 J mol-1 K-1})(2)(\pu{298 K}) = \pu{-5.04 kJ mol-1}$$

$$\Delta H = -\pu{10 kJ mol-1} + (\pu{8.314 J mol-1 K-1})\left[(4)(\pu{303 K}) - (2)(\pu 298 K)\right] = \pu{-4.88 kJ mol-1}$$

And the error is, yes, still small but it doubled (of course). So I'm guessing the $\Delta T \approx 0$ to be valid only for small values of $\Delta T$, but also dependent on the actual values for $n_\mathrm{i}$ and $n_\mathrm{f}$. Is there a rule I'm missing?

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  • $\begingroup$ On the contrary, I have a book that writes $\Delta U$ is the enthalpy change at constant volume, thus indirectly skipping the very problem posed in your question. But it does seem right given that $\Delta U=nC_\mathrm{v}\Delta T$. I am not sure though. $\endgroup$ – Gaurang Tandon Mar 19 '18 at 6:28
  • $\begingroup$ How can $\Delta U$ be equal to $\Delta H$ at constant volume, if there is necessarily a $V\Delta P$ term in the definition of $\Delta H$? I agree $\Delta U$ is the heat at constant volume, but the definition of enthalpy is at constant pressure. $\endgroup$ – ralk912 Mar 19 '18 at 6:44
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    $\begingroup$ Indeed, I do not know :( I am a sad victim of the frequent misuse of the terms "heat" and "ethalpy" in my textbook. $\endgroup$ – Gaurang Tandon Mar 19 '18 at 6:47
  • $\begingroup$ Here says that as long that $\Delta T$ is small, this is a decent approximation. I believe that, but why not just use the $\Delta T$ value if it's known? $\endgroup$ – ralk912 Mar 19 '18 at 6:54
  • $\begingroup$ But then what's the right one? And goes back to, if we have the data, why don't we calculate the more precise one? $\endgroup$ – ralk912 Mar 19 '18 at 7:54
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You have more or less answered your own question. Enthalpy is a function of state, so you can write an expression of the form $\Delta {H} = {H_f}-{H_i}$, and we know that this depends only on the initial and final states, and not the path taken between them.

You also have the relation, $H = U + PV$, using this and the previous equation, and the fact that $U$ is also a state function (so we can write $\Delta {U} = {U_f}-{U_i}$)

$\Delta {H} = \Delta {U} + P_fV_f - P_iV_i \tag{1}$

One comment, in the link that you posted equation (6) namely,

$\Delta {H} = \Delta {U} + \Delta(PV) = \Delta {U} + P\Delta V + V\Delta P $ seems absurd, for finite, non-infinitesimal differences.

For infinitesimal changes, i.e if we write the differential

$\mathrm{d}H = \mathrm{d}U + \mathrm{d}(PV) = \mathrm{d}U + P\mathrm{d}V + V\mathrm{d}P $ this holds true; (chain rule from mathematics)

Anyway, for solids in liquids, the $PV$ term can be neglected on account of their small molar volume. I guess this would be one approximation/assumption you are interested in.

Second, assumption/approximation would be that a gas behaves ideally, and thus follows the equation of state: $PV = nRT$

Plugging it into (1);

$\Delta {H} = \Delta {U} + P_fV_f - P_iV_i = \Delta {U} + R(n_fT_f-n_iT_i) $

This expression is true for finite differences. Now, the finally approximation is indeed $T_f \approx T_i \equiv T$, i.e $\delta T$ is "sufficiently" small, then indeed we can write

$\Delta {H} = \Delta {U}+ RT\Delta n $

Note however, that we don't necessarily require $\Delta T \to 0$ in a strict, mathematical sense. What can be considered "sufficiently" small depends on your system, your problem, and the kind of solution you need. What I mean to say is in some cases/applications $-4.96\ \mathrm{kJ}$ is just as good as $-4.88\ \mathrm{kJ}$

There are many other simplifying assumptions in calorimetry when it is taught such as heat-capacities are independent of temperature for example; this only, really true over narrow ranges of temperature and in general, heat capacity is in fact a function of temperature.

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With the second edit, you're basically wavering towards answering your own question i.e. the approximation is wrong and may not always yield the right result. $\Delta T$ is not always small, and when the above formula if used for large $\Delta T$, the answer may end up being way off.

"Why don't we calculate the more precise one?"

It doesn't matter. You would also have encountered a similar situation in ionic equilibrium calculations (the most popular neglection is taking $K_\mathrm{a}=C\alpha^2$ instead of the actual $K_\mathrm{a}=\frac{C\alpha^2}{1−\alpha}$) Ultimately, it boils down to how many decimal places you want in your calculations. In competitive exams, time is more important than accuracy, so a few digits off doesn't matter for a reasonable estimate. That is probably the reason why the books use the neglection in this formula

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