Learning Objectives

To acquire an knowledge of the relationship in between spontaneity, complimentary energy, and temperature.To have the ability to calculate the temperature in ~ which a procedure is at equilibrium under conventional conditions.

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In the Gibbs totally free energy adjust equation, the only component we as scientists deserve to control is the temperature. We have seen just how we can calculate the standard change in Gibbs free energy, ΔG⁰, yet not every reactions we room interested in take place at precisely 298 K. The temperature plays critical role in identify the Gibbs complimentary energy and also spontaneity the a reaction.

ΔG = ΔHTΔS

If we research the Gibbs free energy readjust equation, we have the right to cluster the contents to create two basic terms, an enthalpy term, ΔH, and also an entropy term, –TΔS. Relying on the sign and magnitude of each, the amount of these terms determines the authorize of ΔG and therefore the spontaneity (Table 18.2 “Spontaneity and also the Signs of Enthalpy and Entropy Terms”).

Table 18.2. Spontaneity and the signs of Enthalpy and Entropy Terms

Since all temperature worths are optimistic in the Kelvin scale, the temperature influence the size of the entropy term. As presented in Table 18.2 “Spontaneity and also the Signs of Enthalpy and Entropy Terms,” the temperature have the right to be the deciding element in spontaneity as soon as the enthalpy and also entropy terms have actually opposite signs. If ΔH is negative, and –TΔS positive, the reaction will be voluntary at low temperatures (decreasing the magnitude of the entropy term). If ΔH is positive, and also –TΔS negative, the reaction will certainly be spontaneous at high temperatures (increasing the magnitude of the entropy term).

Sometimes it deserve to be beneficial to recognize the temperature when ΔG⁰ = 0 and also the procedure is in ~ equilibrium. Understanding this value, we can readjust the temperature to drive the process to spontaneity or additionally to stop the process from occurring spontaneously. Remember that, at equilibrium:

ΔG⁰ = 0 = ΔH⁰- TΔS

We can rearrange and solve for the temperature T:

TΔS⁰ =ΔH

T =frac m Delta m H^o m Delta m S^o


Example 6

Using the postposition table of conventional thermodynamic quantities, identify the temperature in ~ which the following process is at equilibrium:

CHCl3(ℓ) ⇌ CHCl3(g)

How go the value you calculated compare to the boiling point of chloroform offered in the literature?

Solution

At equilibrium: ΔG⁰ = 0 = ΔH⁰- TΔS

We have to estimate ΔH⁰ and S⁰ from your enthalpies that formation and also standard molar entropies, respectively.

ΔH⁰ = ∑nΔHf(products) – ∑mΔHf(reactants)

ΔH⁰=102.7 kJ/mol – (–134.1 kJ/mol)

ΔH⁰ = + 31.4 kJ/mol

ΔS⁰ = ∑nΔS⁰ (products) – ∑mΔS⁰ (reactants)

ΔS⁰ = 295.7 J/mol K – (201.7 J/mol K)

ΔS⁰ = 94.0 J/mol K (or 94.0 x 10-3 kJ/mol K)

Now we can use these values to deal with for the temperature:

T =frac m Delta m H^o m Delta m S^o

T =   extit frac31.4 kJ/mol94.0 x 10^-3 kJ/mol K extit = 334 K = 60.9⁰C

The literature boiling suggest of chloroform is 61.2⁰C. The worth we have calculated is very close but slightly lower because of the presumption that ΔH⁰ and S⁰ carry out not readjust with temperature as soon as we estimate the ΔH⁰ and S⁰ from your enthalpies of formation and also standard molar entropies.

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Key Takeaways

The temperature have the right to be the deciding variable in spontaneity when the enthalpy and also entropy terms have opposite signs:

– If ΔH is negative, and also –TΔpositive, the reaction will certainly be voluntary at low temperatures (decreasing the magnitude of the entropy term).– If ΔH is positive, and also –TΔnegative, the reaction will certainly be spontaneous at high temperature (increasing the size of the entropy term).