Calculate ΔG for the following reactions two different ways:...

Calculate $\Delta G$ for the following reactions two different ways: (1) use Hess's law and the standard Gibbs free energies of formation, and ( 2 ) use $\Delta G=\Delta H-T \Delta S$. Compare the two values and judge whether you get the same $\Delta G_{\mathrm{rxn}}$ either way. Assume $T=298 \mathrm{~K}$. (a) $2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow 2 \mathrm{H}_{2} \mathrm{SO}_{4}(\ell)$ (b) $\mathrm{H}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\ell)$

Calculate $\Delta G$ for the following reactions two different ways: (1) use Hess's law and the standard Gibbs free energies of formation, and ( 2 ) use $\Delta G=\Delta H-T \Delta S$. Compare the two values and judge whether you get the same $\Delta G_{\mathrm{rxn}}$ either way. Assume $T=298 \mathrm{~K}$.
(a) $2 \mathrm{SO}_{2}(\mathrm{~g})+\mathrm{O}_{2}(\mathrm{~g})+2 \mathrm{H}_{2} \mathrm{O}(\ell) \rightarrow 2 \mathrm{H}_{2} \mathrm{SO}_{4}(\ell)$
(b) $\mathrm{H}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\mathrm{aq}) \rightarrow \mathrm{H}_{2} \mathrm{O}(\ell)$

Key Concepts

Hess's Law

Hess's Law is a fundamental principle in thermodynamics stating that the total enthalpy or free energy change of a chemical reaction is the same, regardless of the pathway between the initial and final states. This principle allows one to calculate the overall Gibbs free energy change (?G) of a reaction by summing the Gibbs free energy changes of individual steps, and it underpins methods using standard Gibbs free energies of formation to determine reaction spontaneity.

Standard Gibbs Free Energies of Formation

Standard Gibbs free energies of formation are tabulated values that represent the change in free energy when one mole of a compound is formed from its elements in their standard states under specified conditions (usually 298 K and 1 atm). These values are used to calculate the Gibbs free energy change (?G) for a reaction by taking the difference between the sum of the free energies of the products and that of the reactants.

Gibbs Free Energy and the Relation ?G = ?H - T?S

The thermodynamic relation ?G = ?H - T?S connects the Gibbs free energy change to the enthalpy change (?H) and entropy change (?S) of a reaction at a given temperature (T). This equation is used to assess the spontaneity of a reaction, as a negative ?G indicates a thermodynamically favorable process. It provides an alternative approach to calculating ?G compared to using Hess’s Law with formation values.

Consistency of Methods under Standard Conditions

When chemical reactions are evaluated under standard conditions (typically 298 K), both methods—using Hess's Law with standard Gibbs free energies of formation and calculating ?G from ?H and ?S—should yield the same result for the reaction’s Gibbs free energy change. This consistency reinforces the interconnectivity of thermodynamic properties and offers a robust way to verify calculations in physical chemistry.

Transcript

00:01 So in this problem, calculate delta g for the reactions using hess's law and delta g is equal to delta h minus t delta s.

00:10 Temperature is 298 kelvin.

00:13 Compare the values.

00:14 So for the first one, we're going to use hess's law.

00:17 So the change in standard gives free energy of the reaction is equal to the sum of the change in gives free energy of formation of the products multiplied by their stoichiometric coefficients, minus the change in gives free energy of formation of the reaction.

00:31 Reactants multiplied by their stoichi metric coefficients.

00:33 So we're going to look up the values in the appendix and plug them into the formula.

00:38 We get the answer in kilojoules and we multiply by 1 ,000 to get that in joules.

00:42 Then we're going to calculate the change in standard enthalpy of the reaction.

00:48 So we're going to take the change in standard enthalpy of formation of the products multiply by the stoichometric coefficients minus the change in standard enthalpy of formation of the reactants multiply by the historic metric coefficients.

01:03 Be sure you're looking up the correct phases and the change in standard enthalpy information, as well as standard gives free energy information of elements in their standard states are going to be zero.

01:15 So then we get that in kilojoules and then convert that to jewels.

01:18 Then we're going to do the same thing to get the change in molar entropy.

01:27 In this case, the change in molar entropy of an element in a standard state is not.

01:32 Not zero, so be sure that you're looking up the correct value...

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