Use standard free energies of formation to calculate ΔG^∘ at 25^∘ C for each reaction in Problem

Use standard free energies of formation to calculate ΔG^∘ at...

Use standard free energies of formation to calculate $\Delta G^{\circ}$ at $25^{\circ} \mathrm{C}$ for each reaction in Problem 61. How do the values of $\Delta G^{\circ}$ calculated this way compare to those calculated from $\Delta H^{\circ}$ and $\Delta S^{\circ} ?$ Which of the two methods could be used to determine how $\Delta G^{\circ}$ changes with temperature?

Use standard free energies of formation to calculate $\Delta G^{\circ}$ at $25^{\circ} \mathrm{C}$
for each reaction in Problem 61. How do the values of $\Delta G^{\circ}$ calculated this way compare to those calculated from $\Delta H^{\circ}$ and $\Delta S^{\circ} ?$ Which of the two methods could be used to determine how $\Delta G^{\circ}$ changes with temperature?

Key Concepts

Temperature Dependence of Gibbs Free Energy

The equation ?G° = ?H° - T?S° explicitly demonstrates the dependence of Gibbs free energy on temperature. Among the methods available for calculating ?G°, using standard enthalpy and entropy values is the one that can account for how ?G° changes with temperature. By analyzing the temperature component T?S°, one can predict the behavior of spontaneous processes under different thermal conditions.

Thermodynamic Relationships Involving Enthalpy and Entropy

?G° for a reaction can also be calculated using the relationship ?G° = ?H° - T?S°, where ?H° is the standard enthalpy change and ?S° is the standard entropy change. This formulation provides insight into how enthalpy and entropy contribute to the spontaneity of a reaction, linking the energy and disorder aspects of the process.

Standard Free Energy of Formation

This concept refers to the change in Gibbs free energy when one mole of a compound is formed from its elements in their standard states under specified conditions (typically 25°C and 1 bar pressure). It is a fundamental thermodynamic quantity that is used to calculate reaction free energies by summing the contributions from products and subtracting those from reactants.

Standard Gibbs Free Energy Change

This is the overall change in Gibbs free energy for a reaction carried out under standard conditions. It is determined by combining the standard free energies of formation of the reactants and products. A negative value indicates that the reaction is spontaneous under standard conditions, while a positive value indicates non-spontaneity.

Transcript

00:02 Here we're going to calculate the delta g values using the standard free energy values.

00:07 I've already listed the delta g values of each of these elements from the appendix.

00:14 So let's start.

00:19 So to calculate the delta g value using standard free energy values, we are going to use the formula.

00:26 Delta g is equal to summation of n delta g of the products minus summation of the summation of and delta g of the reactants.

00:38 So starting with equation a, we're just going to input these numbers in here for each of these equations.

00:49 So first we have a product to be no2, but we have two moles of it, so it's going to be two moles into the delta g value of no2 is 51 .3 kilojoules per mole.

01:08 And our reactant is into and there's only one mole of it, so it'll be one mole into 99 .8 kilojoules per mole.

01:22 So in both the cases, the moles and the mold get cancelled.

01:26 So you get a final value of 2 .8 kilojolz.

01:36 Next, we're going to do the same thing with b, but here we have two products.

01:40 So starting with hcl, we have minus 95 .3.

01:47 Kilojoules only one mole similar to one mole into o four here would be one mole into 95 .3 kilojoules per mole and the mole and mole will get cancelled so i'm just not going to put anything here i'll just do a kilojol we have another product here which is in h3 so according to the equation we're going to do summation of the products so plus in h3 is minus 16 .4 kilojoules again per mole and one more gets cancelled next our reactant is nh4 cals so this will be minus two o two point nine kilojoules so that gives us a value of 91 .2 kilojoules next for c we have two moles of f e which is zero plus three moles of h2 o which is minus 2 to 8 .6 to jou per mole minus we have three moles of h2 which is 0.

03:13 Now we have two reactants and from the equation we can see that it's summation of the reactants so i'm just going to put here plus f .e .203 which is minus 742 .2 .2 kilojoules...

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