There is a problem in my textbook that asks to calculate the entropy change in surroundings for the Haber process, assuming that the enthalpy change is -92.4 kJ/mol. The solution to this problem is to use the formula ΔS_(surr) = - ΔH / T, for T = 25 degrees celsius. I understand the usage of this formula in general, but I am confused as to why this formula can be used if the problem never specified that the reaction was reversible? From my understanding, the problem should have specified this to be reversible, because only then does ΔS_(surr) = -Q_rev / T, and under a constant pressure situation, the formula becomes ΔS_(surr) = - ΔH / T.
2 Answers
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Whenever entropy changes are calculated we must apply the condition of 'reversibility' giving $dS=dq_{rev}/T$ at constant temperature $T$ and where $q_{rev}$ is the thermal energy. There is, however, no reason for the actual process to be reversible, because entropy is a function of state and $\Delta S$ is the same whatever the route taken between the starting and ending points. We usually need to devise, on paper, a reversible route merely to perform the calculation.
(btw. By 'reversible' we mean that the process occurs at an infinitesimal rate so that the thermodynamic system is always effectively in equilibrium. A minute change in one direction can be reversed by a similar minute change in the opposite direction. Under these conditions the maximum amount of work is obtainable. As an example think of a tug-of-war contest between two well matched teams. The amount of work done depends on the opposing teams strength not their own inherent strength, and the maximum amount of work will be done when the teams are very evenly matched and so are working flat-out. The performance will then take a long time.)
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$\begingroup$ so does that mean we can use ΔS_(surr) = - ΔH / T because entropy is calculated along a reversible route (irregardless of the actual process), and the heat along the reversible route will equal ΔH if it is conducted at constant pressure (which is true for many reactions), so - ΔH / T is essentially always correct to calculate the entropy change at constant pressure for the surroundings. I assume that would mean, by similar reasoning, we can use ΔS_(sys) = ΔH / T also then? $\endgroup$– MVVCommented May 4, 2023 at 17:52
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You are correct that $\mathrm{d}S = \dfrac{đq}{T}$ only holds for a reversible process. However, the usual convention is to assume that the “surroundings” are so big that they never change temperature no matter how much heat you pump into (or out of) them. If the temperature is constant, we can take it outside the integral:
$$\Delta S_\text{surr.} = \int\mathrm{d}S_\text{surr.} = \int\dfrac{đq_\text{surr., rev.}}{T} = \frac{1}{T}\intđq_\text{surr., rev.} = \frac{\Delta H_\text{surr.}}{T} = - \frac{\Delta H}{T} $$
You can see that we choose to take the integral over a reversible process (so that we know the right formula to use), but actually, because $S$ is a state function, the answer is the same regardless of whether or not the process followed in real life is reversible.
In practice it is very rare for chemical reactions to happen in a thermodynamically reversible way.
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$\begingroup$ Hello Aant. We agree that this is corret. However, you seem to affirm that $dq_\pu{surr,rev} = dH$ in one of the equalities. This is true for a mechanically reversible process. But if the path where reactants are converted into products is not reversible, then, how do you prove that that equality still holds? At the end you recognize that no chemical reaction traverses along a reversible path. $\endgroup$ Commented May 4, 2023 at 11:07
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$\begingroup$ Doesn't that simply follow from the definition of enthalpy and the assumption (admittedly tacet here) that the reaction proceeds at constant pressure? There's no requirement for reversibility here. $\endgroup$– AantCommented May 8, 2023 at 16:03