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Which of the following is/are correct for spontaneous isothermal chemical reaction?

(A) $\Delta H = 0$, because $\Delta T = 0$

(B) $\Delta S = 0$

(C) $\Delta U = 0$, because $\Delta T = 0$

(D) $\Delta G < 0$

My answer: $\mathrm{A, C, D}$


$U$ is a state function and only depends on the initial and final states and their temperature. Since $\Delta T = 0$, $\Delta U = 0$

$\Delta H = \Delta U + \Delta(PV)$, but $PV = \text{constant}$, so $\Delta H = 0$

$\Delta G <0$ - the requirement for any reaction to be spontaneous.

But answer given is only option D.

I think the flaw in my $\Delta H = 0$ argument is that reactants can be solids and liquids as well and we cannot apply $PV = \text{constant}$ to them. Am I right?

But I still can't figure out why $\mathrm C$ is wrong.


marked as duplicate by Mithoron, Melanie Shebel, Tyberius, A.K., Todd Minehardt Feb 28 at 1:25

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  • 3
    $\begingroup$ For ideal gases U is a function of T only, but not generally. Even for ideal gases $\Delta H = \Delta U + RT \Delta n$ and you are considering a reaction where $\Delta n$ may not be 0. $\endgroup$ – Night Writer Dec 18 '18 at 11:31
  • $\begingroup$ I want to say this is a duplicate of chemistry.stackexchange.com/q/39645/16683... $\endgroup$ – orthocresol Dec 19 '18 at 1:00

The internal energy and enthalpy of a pure ideal gas depends only on temperature. But, the internal energy and enthalpy of an ideal gas mixture depends both on temperature and on the amount of each chemical species that is present. To be more precise, it is equal to the internal energy and enthalpy per mole of each component as if it were a pure species times the number of moles of that species present, summed over all the species in the mixture: $$U=\sum{n_iu_i}$$and$$H=\sum{n_ih_i}$$So, even at constant temperature, $$\Delta U=\sum{(\Delta n_i)u_i}$$and$$\Delta H=\sum{(\Delta n_i)h_i}$$You should have been taught this when they were covering the thermodynamics of mixtures and chemical reactions.

  • $\begingroup$ I'm confused why your answer is about ideal gases when the question doesn't say anything about the nature of the reactants and products. Does the question assume they are gases, much less ideal ones? $\endgroup$ – Curt F. Dec 19 '18 at 5:32
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    $\begingroup$ No. My answer just uses ideal gases as an example. If one of the given choices is not true for an ideal gas (mixture), it certainly isn't true for a chemical reaction in a mixture of real gases (which exhibits ideal gas behavior in the limit of low pressures) or for a liquid phase reaction. Incidentally, OPs analysis was strictly valid also only for ideal gases. In real gases, the internal energy and enthalpy are functions of pressure, and PV is not constant even if temperature is constant. What I didn't want to do in my discussion was to start introducing partial molar properties. $\endgroup$ – Chet Miller Dec 19 '18 at 12:25
  • $\begingroup$ That makes complete sense. (If you add the third sentence of your comment to the answer, you'll get a +1 from me, if such things matter to you.) $\endgroup$ – Curt F. Dec 19 '18 at 17:22

The correct answer is (D).

Systems can be held at constant temperature by being placed in a heat bath. This lets heat flow into and out of the system as needed to ensure that the temperature stays constant.

A. $\Delta H=0$ is not necessarily true, because an exothermic (or endothermic) reaction releases enthalpy. For example, consider the reaction $\ce{C6H12O6 + 6 O2 -> 6 H2O + 6 CO2}$. This is the reaction that your body uses to get energy from eating sugar, and it happens at nearly isothermal temperatures (37 °C). But enthalpy is released, and used to power your body's metabolism.

B. $\Delta S=0$ is not necessarily true, because systems held in a heat bath can exchange heat with their surroundings, leading to entropy changes. The glucose oxidation reaction above releases enthalpy, and without heat flow to the heat bath, it would result in a temperature increase. But with heat flow, the change in entropy cannot be zero.

C. $\Delta U=0$ is not necessarily true, because (among other reasons) $U = H - PV$, and we have already established that $\Delta H$ is not necessarily zero.

D. $\Delta G < 0$ is true because that's essentially the definition of a spontaneous reaction. If it weren't true your body couldn't get energy from metabolizing sugar.


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