Using values of ΔHj and S^∘, calculate the standard molar free energy of formation, ΔGj^∘, for e

Using values of ΔHj and S^∘, calculate the standard molar...

Using values of $\Delta H_{j}$ and $S^{\circ},$ calculate the standard molar free energy of formation, $\Delta G_{j}^{\circ},$ for each of the following: (a) $\mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{s})$ (b) $\mathrm{Cl}(\mathrm{g})$ (c) $\mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{s})$ Compare your calculated values of $\Delta G_{1}^{\circ}$ with those listed in Appendix L. Which compounds are predicted to be formed spontancously?

Using values of $\Delta H_{j}$ and $S^{\circ},$ calculate the standard molar free energy of formation, $\Delta G_{j}^{\circ},$ for each of the following:
(a) $\mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{s})$
(b) $\mathrm{Cl}(\mathrm{g})$
(c) $\mathrm{Na}_{2} \mathrm{CO}_{3}(\mathrm{s})$
Compare your calculated values of $\Delta G_{1}^{\circ}$ with those listed in Appendix L. Which compounds are predicted to be formed spontancously?

Key Concepts

Standard Molar 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. It is calculated using the relationship ?G = ?H - T?S, which balances the energetic contributions from the enthalpy and the disorder contribution from the entropy, thereby providing information on the thermodynamic favorability of a reaction.

Standard Enthalpy

Standard enthalpy (?H°) measures the heat absorbed or released during the formation of one mole of a compound under standard conditions. It provides insight into the energy changes involved in the reaction and is a crucial factor in determining the overall energy balance of the formation process.

Standard Entropy

Standard entropy (S°) quantifies the degree of randomness or disorder in a system at standard conditions. In the context of formation reactions, the entropy change reflects how the arrangement of molecules shifts during the reaction, influencing the free energy through the temperature-dependent term T?S.

Temperature Dependence of Free Energy

The relation ?G = ?H - T?S illustrates how the Gibbs free energy change depends on temperature. As temperature increases, the entropy term (T?S) may contribute more significantly, potentially altering the balance between enthalpy and entropy and thus affecting the spontaneity of the reaction.

Spontaneity Criterion

A reaction or formation process is considered spontaneous if the standard free energy change (?G°) is negative. Calculating ?G° from the known ?H and S° values and comparing these results with established tabulated values helps determine whether a compound is thermodynamically favored to form under standard conditions.

Transcript

00:01 All right, so here we have three compounds, and we need to find the delta g, the free energy of each.

00:08 So this is the gibbs free energy equation that we'll use.

00:13 Delta g is equivalent to delta h, enthalpy minus t, delta s, as being entropy.

00:20 So we will use these thermodynamic information from our appendix l in the textbook.

00:29 So let's go to black.

00:35 So first off, we have calcium hydroxide.

00:41 So it's delta g.

00:45 To calculate it, we will find the, it's delta h and it's delta s and the appendix l.

00:54 So once we found those, we know that it's 986.

01:05 And we have to keep units because there's a little bit.

01:09 Of a difference, joules per kelvin or joules per mole, subtracted by the temperature, which is the 298 .15 kelvin.

01:39 And this temperature is multiplied by the entropy, which is 83 .39 joules per mole kelvin.

02:04 Okay, and we need consistent units here.

02:15 We can't have kilojoules and joules.

02:17 So we will convert the joules to kilojoules by multiplying by a factor of one to keep the same value.

02:28 So we know that one kilojoule is equivalent to a thousand joules.

02:40 So that division is equal to one.

02:43 So we are multiplying by one.

02:46 And we are able to cancel out jewels on top and we're left with kilojoules.

02:55 So if we run this all through a calculator, we get negative 1 ,010 kilojoules per mole.

03:17 Okay, so what does this tell us about spontaneity? so first we look at the sign...

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