Chapter 20 Practice Exam Questions

Spontaneity and Thermodynamics

True/False Statements

• Spontaneous processes are favored by negative $DH$ values.

• The second law of thermodynamics: the entropy of the universe always increases for a spontaneous process.

• Entropy: not the heat flow measured under 1 atm pressure (definition of enthalpy).

• When a solid melts, the entropy of the liquid is higher than the entropy of the solid.

• The Third Law of Thermodynamics: A perfect crystal at 0 K has zero entropy.

• Negative Gibbs Free Energy changes: indicate a spontaneous process.

• Elements in their free standard state: do not have absolute entropies of zero (only at 0 K).

Standard Entropy Change

• Formula: $∆S<em>{rxn} = ∑ nS</em>{products} - ∑ nS_{reactants}$

• Absolute entropies, $S^o$, only have a value of 0 J/mol K at 0K.

• Elements in their free standard state have non-zero entropies.

• Only enthalpy and Gibbs free energy changes have 0 values for elements in their free standard state.

Second Law of Thermodynamics

• States that for any spontaneous process, the entropy of the universe increases.

Entropy Increase Conditions

• Entropy usually increases when:

  • A molecule is broken into two or more smaller molecules.

  • A reaction results in an increase in the number of moles of gas.

  • A solid changes to a liquid.

  • A liquid changes to a gas.

Spontaneity and Temperature Range

• To determine the temperature range for spontaneity, calculate $∆H$ and $∆S$.

• Use $T = {\Delta H}{\Delta S}$ to find the temperature at which $T\Delta S = \Delta H$. $DG$ goes negative above this $T$.

• Check reasoning by plugging a temperature above calculated T into $DG = DH - TDS$ to confirm the sign.

• If $DH$ is positive and $DS$ is negative, the reaction is non-spontaneous at all temperatures.

Entropy Changes in Processes

• Evaporation (liquid to gas) increases entropy.

• Precipitation decreases entropy.

• Reactions with decreasing moles of gas decrease entropy.

• Organizing items (e.g., pennies) decreases entropy.

Gibbs Free Energy

• $DG = DH - TDS$

• Negative $DG$: spontaneous process.

• Positive $DG$: nonspontaneous process.

• If $DS$ is negative, $DH$ must be negative for a spontaneous process.

• When $DS$ and $DH$ have negative signs, the Gibbs free energy is negative below some temperature $T$.

• When $DS$ and $DH$ have positive signs, the Gibbs free energy is negative above some temperature $T$.

Spontaneity and Entropy of the Universe

• For a spontaneous process, the entropy of the universe ($\Delta S_{universe}$) must be greater than 0.

Calculating Entropy Change

• System: $\Delta S_{sys}$

• Surroundings: $\Delta S<em>{surr} = - \frac{\Delta H</em>{system}}{T}$

• Universe: $\Delta S<em>{universe} = \Delta S</em>{sys} + \Delta S_{surr}$

Spontaneity Based on $\Delta H$ and $\Delta S$

• If both ${\Delta H}{\Delta S}$ are positive, the reaction is spontaneous at high temperatures.

Processes Decreasing Entropy

• Changes of phase from liquid to solid, gas to liquid, and gas to solid result in decreases in entropy.

Second Law of Thermodynamics

• States that for a spontaneous process, the entropy of the universe is the sum of the entropy change of the system and the surroundings and must be greater than 0.

Vaporization of Tin(IV) Chloride

• To calculate the boiling point for $SnCl_4$ the enthalpy change must be calculated and plugged into $T = \frac{\Delta H}{\Delta S}$.

Gibbs Free Energy and Spontaneity

• For negative $DH$ and $DS$, the process is spontaneous below a certain temperature $T$.

Standard Gibbs Free Energy of Formation

• For elements in their free standard state, ${\Delta H<em>f^o = 0}{\Delta G</em>f^o = 0}$.

• Examples: Sodium metal and hydrogen gas.

Spontaneous Reactions

• Will proceed without outside intervention.

Reaction Between Lead(II) Sulfide and Oxygen

• The reaction is spontaneous above 3950 K or 3678 oC.

• Calculating the Gibbs free energy from the DGf is possible but will not give the temperature range over which the reaction is spontaneous. The DH and DS must be calculated and plugged into $T = \frac{\Delta H}{\Delta S}$.

Calculation of Entropy Change

• $\Delta S = (2 \,mol \,AlCl<em>3 \times 110.7 \,J/mol \,K) - [(2 \,mol \,Al \times 28.3 \,J/mol \,K) + (3 \,mol \,Cl</em>2 \times 223.0 \,J/mol \,K)] = 221.4 \,J/K - 725.6 \,J/K = -504.2 \,J/K$

Gibbs Free Energy Change

• $\Delta G = \Delta H - T\Delta S$

Reactions Producing a Decrease in Entropy

• $2 H<em>2(g) + O</em>2(g) \rightarrow 2 H_2O(l)$: Synthesis of liquid water from hydrogen and oxygen gas, decrease in entropy.

Standard Gibbs Free Energy Change Calculation

• $\Delta G<em>{rxn} = \Sigma \,moles \times G</em>f(products) - \Sigma \,moles \times G_f(reactants)$

Spontaneity Conditions

• A process cannot be spontaneous (product-favored) if it is endothermic, and there is a decrease in disorder.