0
$\begingroup$

As we know volume change in bomb calorimeter is zero. So if we use $\Delta H = \Delta U + p\Delta V$ then we get $\Delta H = \Delta U$. But in some reaction there is change in number of moles like $$\ce{NH2CN(g) + 3/2O2(g) -> N2(g) + CO2(g)+ H2O(l)}$$ so here $\Delta H = \Delta U + (-0.5)RT$.

$\endgroup$
2
$\begingroup$

First and foremost your $-0.5$ should have units of moles (i.e. should be $-0.5~\mathrm{mol}$). Don't leave out units. They're important.

$$\Delta H = \Delta U + p\Delta V$$

If you were not taught what conditions this equation holds true, shame on your teacher. If you were taught it, you should probably try to remember that this holds for constant pressure.

In general

$$\begin{align} H &= U + pV \\ H_2 - H_1 &= (U_2 + p_2V_2) - (U_1 + p_1V_1) \\ &= (U_2 - U_1) + (p_2V_2 - p_1V_1) \\ \Delta H &= \Delta U + \Delta (pV) \end{align}$$

Only under constant pressure $p_2 = p_1 = p$ does this simplify to

$$\begin{align} \Delta H &= (U_2 - U_1) + (pV_2 - pV_1) \\ &= \Delta U + p(V_2 - V_1) \\ &= \Delta U + p\Delta V \\ \end{align}$$

If this is not true (and think about it, if your volume is constant, then what do you think is happening to the pressure inside the calorimeter?), then you may use the ideal gas approximation

$$p_2V_2 = n_2RT_2;\quad p_1V_1 = n_1RT_1$$

and if you further assume constant temperature $T_2 = T_1 = T$ (which is probably not true, but it is close enough to the truth since the heat capacity of a calorimeter is usually large leading to a very small difference between $T_2$ and $T_1$), then

$$\begin{align} \Delta H &= (U_2 - U_1) + (n_2RT - n_1RT) \\ &= \Delta U + (n_2 - n_1)RT \\ &= \Delta U + RT\Delta n \end{align}$$

The further approximation is that the non-gaseous species do not contribute significantly to the $\Delta(pV)$ term, so $\Delta n$ here can be identified with $\Delta n_g$, the difference in the number of moles of gas.

TL;DR

$$\color{red}{\textbf{Learn when the equations you are using are appropriate to use.}} \\ \large\color{red}{\textbf{Don't just blindly use them!}}$$

$\endgroup$
  • 1
    $\begingroup$ I wish I could frame the lines in red and put them up on a banner on the top of this site... $\endgroup$ – Zhe Jan 4 '17 at 13:26
  • $\begingroup$ While I agree with your 'red' comment we do not help ourselves when teaching and often use a variety of notation, e.g. such as G, $\Delta G$ and dG or create equations out of 'thin air' as $dU=(\partial U/\partial T)_VdT+(\partial U/\partial V)_TdV$. As an example, consider a change at constant volume then $\Delta V =0$ and so by the first law $q=\Delta U$. But $\Delta U$ is a function of state and q a path variable so how can this be? The answer is that the path is fixed so q is now a path function and should be changed, say, to $q_V$ . So the possibilities for confusion are many. $\endgroup$ – porphyrin Jan 13 '17 at 14:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.