Use data in Appendix C to calculate ΔH^∘, ΔS^∘, and ΔG^∘ at 25^∘ C for each of the following rea

Use data in Appendix C to calculate ΔH^∘, ΔS^∘, and ΔG^∘ at...

Use data in Appendix C to calculate $\Delta H^{\circ}, \Delta S^{\circ}$, and $\Delta G^{\circ}$ at $25^{\circ} \mathrm{C}$ for each of the following reactions. In each case show that $\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ}$ (a) $2 \mathrm{Cr}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CrO}_{3}(s)$ (b) $\mathrm{BaCO}_{3}(s) \longrightarrow \mathrm{BaO}(s)+\mathrm{CO}_{2}(g)$ (c) $2 \mathrm{P}(s)+10 \mathrm{HF}(g) \longrightarrow 2 \mathrm{PF}_{5}(g)+5 \mathrm{H}_{2}(g)$ (d) $\mathrm{K}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{KO}_{2}(s)$

Use data in Appendix C to calculate $\Delta H^{\circ}, \Delta S^{\circ}$, and $\Delta G^{\circ}$ at $25^{\circ} \mathrm{C}$ for each of the following reactions. In each case show that $\Delta G^{\circ}=\Delta H^{\circ}-T \Delta S^{\circ}$
(a) $2 \mathrm{Cr}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{CrO}_{3}(s)$
(b) $\mathrm{BaCO}_{3}(s) \longrightarrow \mathrm{BaO}(s)+\mathrm{CO}_{2}(g)$
(c) $2 \mathrm{P}(s)+10 \mathrm{HF}(g) \longrightarrow 2 \mathrm{PF}_{5}(g)+5 \mathrm{H}_{2}(g)$
(d) $\mathrm{K}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{KO}_{2}(s)$

Key Concepts

Fundamental Thermodynamic Relationship

This key concept relates the standard Gibbs free energy change to the standard enthalpy and entropy changes through the equation ?G° = ?H° – T?S°. This equation provides a direct means of calculating one thermodynamic parameter if the other two are known, and reinforces the interplay between heat, disorder, and spontaneity in chemical reactions.

Standard Gibbs Free Energy (?G°)

Gibbs free energy combines the system's enthalpy and entropy to predict the spontaneity of a reaction at constant temperature and pressure. ?G° is calculated from the standard enthalpy and entropy changes, and its sign indicates whether a reaction is spontaneous under standard conditions. A negative ?G° implies a spontaneous process.

Using Tabulated Thermodynamic Data

Reactions are often analyzed using values from reference data such as standard enthalpies and entropies found in tables or appendices. This method involves summing the appropriate values for products and subtracting those for reactants, enabling the calculation of overall reaction parameters. It is a fundamental technique in chemical thermodynamics for system evaluation.

Standard Enthalpy (?H°)

This is the heat change associated with a reaction under standard conditions (1 bar pressure, 25°C). It represents the difference in enthalpy between products and reactants and is calculated using tabulated standard enthalpy values. ?H° is a state function, meaning its value depends only on the initial and final states, not on the pathway taken.

Standard Entropy (?S°)

Standard entropy is a measure of the disorder or randomness in a system at standard conditions. It is determined from tabulated absolute entropy values for substances. The overall entropy change of a reaction (?S°) is found by considering the entropy differences between the products and the reactants, using the principle that entropy is also a state function.

Transcript

00:01 So like question number 60, 61 is asking you to use data that is in the thermodynamic tables at the back of your book.

00:09 Except this time, we're just looking at the delta g, and then we need to decide whether this is spontaneous or non -spontaneous.

00:18 So our first reaction is so2 gas plus o2 gas yields to s .o3 gas.

00:34 You're always looking up products minus reactants.

00:39 So you're doing the value for so3 times 2, subtract out the s .o2 times 2 and the o2.

00:48 And remember you're multiplying because there are coefficients in front of some of these.

00:53 So we're going to calculate our delta g by doing 2 times negative 370 .4, which is the value for s .03, minus.

01:06 2 times negative 300 .4, which is the value for so2, and then zero because oxygen is zero in the charts, it's in its standard state.

01:20 When i do this math, i'm going to get a negative 140 .0.

01:32 That is kilojoules.

01:36 This is a negative value, and because that is negative, this is non -spontane.

01:41 Sorry, this is spontaneous.

01:45 If you're taking ap cam, the new term for that is thermodynamically favorable.

01:54 And that's what you'll be looking for if you're using this book for an ap course.

02:00 All right.

02:01 The next portion of this question is no2.

02:07 That's a gas plus n2o, also a gas, yields 3 no, gas.

02:18 Again, we're just looking them up, products minus reactants.

02:22 So three times 86 .71 minus 51 .84 minus 103 .59 equals 104.

02:38 0 .4 .7 kilojoules.

02:42 Now, this one is positive.

02:44 And because this is positive, this is non -sponantaneous.

02:53 If you're in ap chem, again, that is thermodynamically unfavorable.

03:03 All right...

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