Because entropy is a state function, we can calculate the entropy change of a reaction if we know the standard entropies, *S*^{o} of the reactants and products. Remember, a state function depends only on the initial and final values and not on the path that leads to these states. A classic example of a state function is attitude. Regardless of the path the hikers have taken, they both are at the same altitude:

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The standard entropy change of the reaction is calculated by **subtracting the standard entropies of the reactants ***multiplied by their stoichiometric coefficients* from the **standard entropies of the products** *multiplied by their stoichiometric coefficients*:

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The standard entropies of reactants and products are given in textbooks under the appendix for thermodynamics data.

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This is similar to how we calculate the enthalpy of a reaction from the standard enthalpies of formation.

Notice, however, that** S^{o}**, unlike Î”

*H*

^{o}f and Î”

*G*

^{o}

_{f}for elements is

**not zero at 25 Â°C**

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**For example**, using the data in an appendix for homonymic data, calculate the standard entropy changes, **Î”****S**** for the following reactions** at 25Â°C:

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2H_{2}S(*g*) + 3O_{2}(*g*) Â â†’ 2SO_{2}(*g*) + 2H_{2}O(*l*)

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The standard entropies of formation, *S***Â° **(J/K) for the reactants and products are: H_{2}S(*g*) = 205.8, for O_{2}(*g*) = 205.2, for SO_{2}(*g*) = 248.2, for H_{2}O(*l*) = 70

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Therefore, the standard entropy change for the reaction is:

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Î”*SÂ°*_{rxn} = [2 x *S*Â° (H_{2}O) + 2 x *S*Â° (SO_{2})] â€“ [2 x *S*Â° (H_{2}S) + 3 x *S*Â°(O_{2})]

Î”*S*Â°_{rxn} = [2 x 70 J/K + 2 x 248.2 J/K] â€“ [2 x 205.8 J/K + 3 x 205.2 J/K] = **-390.8 J/K**

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**Check Also**

- The Gibbs Free Energy
- The Effect of đťš«H, đťš«S, and T on đťš«G â€“ Spontaneity
- Entropy and State Change
- Entropy Changes in the Surroundings
- đťš«
*G*^{o}_{rxn}Â from the Free Energies of Formation - Gibbs Free Energy and Hessâ€™s Law
- Gibbs Free Energy Under Nonstandard Conditions
- Gibbs Free Energy and Equilibrium Constant
**Entropy, Enthalpy, and Gibbs Free Energy Practice Problems**

#### Practice

Using the data in the attached Appendix, calculate the standard entropy changes, **Î” S for the following reactions** at 25Â°C:

**a) **2KO_{2}(*aq*)+ 2H_{2}O(*l*) â†’ 2KOH(*aq*) + O_{2}(*g*)+ H_{2}O_{2}(*l*)

**b) **C_{2}H_{5}OH(*l*) + 3O_{2}(*g*) â†’ 2CO_{2}(*g*) + 3H_{2}O(*g*)

**c) **C_{2}H_{4}(*g*) + H_{2}(*g*) â†’ C_{2}H_{6}(*g*)

**d) **C(*s*)_{graphite} + H_{2}O(*g*) â†’ CO(*g*) + H_{2}(*g*)

**e) **3NO_{2}(*g*) + H_{2}O(*l*) â†’ 2HNO_{3}(*aq*) + NO(*g*)

**f) **2H_{2}S(*g*) + 3O_{2}(*g*) â†’ 2H_{2}O(*l*) + 2SO_{2}(*g*)

**g) **N_{2}O(*g*) + 3H_{2}(*g*) â†’ N_{2}H_{4}(*l*) + H_{2}O(*l*)

**h) **2C_{3}H_{7}OH(*l*) + 9O_{2}(*g*) â†’ 6CO_{2}(*g*) + 8H_{2}O(*l*)