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A reaction has $\Delta H_{rxn} = -107$ kJ and $\Delta S_{rxn} = 285$ J/K. At what temperature is the change in entropy for the reaction equal to the change in entropy for the surroundings?

My thoughts:

I wasn't really sure if there was an equation that related the entropy and the enthalpy and would like to know if there is such an equation.

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You're looking for the equation of Gibbs free energy:

$$\Delta G^\circ =\Delta H^\circ - T\Delta S^\circ$$

Per Wikipedia's Gibbs free energy page:

The equation can be also seen from the perspective of the system taken together with its surroundings (the rest of the universe). First assume that the given reaction at constant temperature and pressure is the only one that is occurring. Then the entropy released or absorbed by the system equals the entropy that the environment must absorb or release, respectively. The reaction will only be allowed if the total entropy change of the universe is zero or positive. This is reflected in a negative ΔG, and the reaction is called exergonic.

So effectively you're trying to deduce the $T$ that produces $\Delta G = 0$ which is the "tipping point" (then the reaction will occur for $\Delta G < 0$)

So you solve for $T$ in:

$0 = \Delta H - T \Delta S$

Thereafter as $T \rightarrow \infty$ you see that $\Delta G$ keeps decreasing, making the reaction more favorable.

Thus $S_{system} = S_{surroundings}$ is a fancy prerequisite for reaction occurrence, which we commonly associate with $\Delta G \le 0$.

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When we talk about heat of reaction and change in entropy of reaction, we are talking about going from pure reactants to pure products in some (unspecified reversible) process at constant temperature and pressure. Since the process is at constant pressure, the heat exchange with the surroundings, according to the first law, is equal to the change in enthalpy. So, $$Q_{sys}=\Delta H_{rxn}$$In this context, when we talk about heat exchange with the surroundings, all the heat transfer is taken to occur to an ideal constant temperature reservoir. So the change in entropy of the reservoir is$$\Delta S_{res}=\frac{-Q}{T}=-\frac{\Delta H_{rxn}}{T}$$If the process is carried out reversibly, the sum of the change in entropy of the system plus the change in entropy of the surroundings must be zero:$$\Delta S_{rxn}+\Delta S_{res}=\Delta S_{rxn}-\frac{\Delta H_{rxn}}{T}=0$$(The problem statement must only be referring to the magnitudes of the entropy changes for the system and the surroundings, since clearly they will be opposite in sign.)

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