From data in Appendix 4 , calculate ΔH^∘, ΔS^∘, and ΔG^∘ for each of the following reactions at

From data in Appendix 4 , calculate ΔH^∘, ΔS^∘, and ΔG^∘ for...

From data in Appendix 4 , calculate $\Delta H^{\circ}, \Delta S^{\circ}$, and $\Delta G^{\circ}$ for each of the following reactions at $25^{\circ} \mathrm{C}$. a. $\mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)$ b. $6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g)$ Glucose c. $\mathrm{P}_{4} \mathrm{O}_{10}(s)+6 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 4 \mathrm{H}_{3} \mathrm{PO}_{4}(s)$ d. $\mathrm{HCl}(g)+\mathrm{NH}_{3}(g) \longrightarrow \mathrm{NH}_{4} \mathrm{Cl}(s)$

From data in Appendix 4 , calculate $\Delta H^{\circ}, \Delta S^{\circ}$, and $\Delta G^{\circ}$ for each of the following reactions at $25^{\circ} \mathrm{C}$.
a. $\mathrm{CH}_{4}(g)+2 \mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)$
b. $6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g)$
Glucose
c. $\mathrm{P}_{4} \mathrm{O}_{10}(s)+6 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow 4 \mathrm{H}_{3} \mathrm{PO}_{4}(s)$
d. $\mathrm{HCl}(g)+\mathrm{NH}_{3}(g) \longrightarrow \mathrm{NH}_{4} \mathrm{Cl}(s)$

Key Concepts

Standard Entropy Change (?S°)

Standard entropy change refers to the difference in entropy between products and reactants under standard conditions. It is calculated by taking the sum of the standard molar entropies of the products, each multiplied by its stoichiometric coefficient, and subtracting the corresponding sum for the reactants. This calculation gauges the dispersal of energy and matter during the reaction.

Hess’s Law and Formation Data

Hess’s Law states that the total enthalpy change for a reaction is the same regardless of the path by which the reaction takes place. By using standard formation data (enthalpies of formation, standard molar entropies, etc.), one can apply Hess’s Law to determine net enthalpy changes and thereby calculate the thermodynamic parameters of complex reactions from tabulated values.

Standard States and Temperature (25°C)

Thermodynamic calculations are typically performed under standard conditions, which include a defined temperature (298 K or 25°C), pressure (1 atm), and standard states for substances. Using these conditions ensures consistency when comparing thermodynamic data such as enthalpy, entropy, and free energy values, which are provided at 25°C in standard references.

Standard Enthalpy Change (?H°)

This concept involves calculating the heat evolved or absorbed during a reaction under standard conditions. It is determined using the standard enthalpies of formation for products and reactants, where ?H° for the reaction is computed as the sum of the enthalpy contributions of the products minus that of the reactants. This approach, often based on Hess’s Law, allows the determination of heat changes even when the direct measurement is not feasible.

Standard Gibbs Free Energy Change (?G°)

Gibbs free energy, ?G°, is a thermodynamic quantity used to predict the spontaneity of a reaction at constant temperature and pressure. It is calculated using the relationship ?G° = ?H° - T?S°, linking the enthalpy and entropy changes of the reaction. A negative ?G° indicates a spontaneous reaction under standard conditions, while a positive value suggests non-spontaneity.

Transcript

00:01 In this problem, our first order of business will be to calculate these three entities for the following reactions at standard conditions.

00:23 So let's begin.

00:24 We'll do equation one and equation two.

00:27 For equation one, we will use our delta h will be equal to, are we ready? this is negative 484 and this is 160.

01:00 I should leave you a little bit more room here, but you'll be able to tell where they all are.

01:14 Ok, so let's start.

01:15 Negative 484 times one minus methane is negative 75 times one plus one times negative 393 .5.

01:37 So this will equal negative 16 kilojoules.

01:44 For my delta s, this will equal 1 times 160 minus 1 times 186 plus 1 times 214.

02:10 And this will equal negative 240 joules per kelvin.

02:19 And my delta g will equal one times, oh, let's do it this way.

02:31 This will equal my delta h minus my delta h times my delta s, t times delta s.

02:52 So this will equal negative 16 kilojoules minus 298 times negative 240 joules per kelvin times 1000 joules per kilojoule.

03:12 And this will equal positive 56 kilojoules.

03:19 So this will be spontaneous below t equals negative h over, excuse me, and that will be 16 divided by 0 .240.

03:55 That'll equal 67 kelvin.

04:03 That's too cold to be practical.

04:12 Oh, my pen just fell apart.

04:14 Hang on here.

04:15 I captured it.

04:18 Don't want it to fall on the floor.

04:21 That was so weird.

04:28 Okay, practical.

04:30 Okay, it's still working...

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