Calculate ΔH^∘ and ΔE^∘ for the following reactions at 25^∘ C. (If necessary, refer to the data

Calculate ΔH^∘ and ΔE^∘ for the following reactions at 25^∘...

Calculate $\Delta H^{\circ}$ and $\Delta E^{\circ}$ for the following reactions at $25^{\circ} \mathrm{C}$. (If necessary, refer to the data in Table C.2 in Appendix C.) (a) $3 \mathrm{PbO}(s)+2 \mathrm{NH}_{3}(g) \longrightarrow 3 \mathrm{~Pb}(s)+\mathrm{N}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(g)$ (b) $\mathrm{NaOH}(s)+\mathrm{HCl}(g) \longrightarrow \mathrm{NaCl}(s)+\mathrm{H}_{2} \mathrm{O}(l)$ (c) $\mathrm{Al}_{2} \mathrm{O}_{3}(s)+2 \mathrm{Fe}(s) \longrightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+2 \mathrm{Al}(s)$ (d) $2 \mathrm{CH}_{4}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)+\mathrm{H}_{2}(g)$

Calculate $\Delta H^{\circ}$ and $\Delta E^{\circ}$ for the following reactions at $25^{\circ} \mathrm{C}$. (If necessary, refer to the data in Table C.2 in Appendix C.)
(a) $3 \mathrm{PbO}(s)+2 \mathrm{NH}_{3}(g) \longrightarrow 3 \mathrm{~Pb}(s)+\mathrm{N}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(g)$
(b) $\mathrm{NaOH}(s)+\mathrm{HCl}(g) \longrightarrow \mathrm{NaCl}(s)+\mathrm{H}_{2} \mathrm{O}(l)$
(c) $\mathrm{Al}_{2} \mathrm{O}_{3}(s)+2 \mathrm{Fe}(s) \longrightarrow \mathrm{Fe}_{2} \mathrm{O}_{3}(s)+2 \mathrm{Al}(s)$
(d) $2 \mathrm{CH}_{4}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g)+\mathrm{H}_{2}(g)$

Key Concepts

Standard Enthalpy of Formation

This concept refers to the enthalpy change when one mole of a substance is formed from its constituent elements in their standard states. It is a fundamental building block in thermodynamics used to calculate the overall enthalpy change (?H°) of a reaction by summing the enthalpies of formation of products and subtracting the sum for the reactants via Hess's law.

Standard Reaction Enthalpy (?H°)

The standard reaction enthalpy is the total heat change during a reaction carried out under standard conditions, typically 25°C and 1 atm pressure. It is calculated using the sum of the standard enthalpies of formation of the products minus those of the reactants, reflecting the net energy change that occurs during the reaction.

Standard Internal Energy Change (?E°)

The standard internal energy change of a reaction represents the change in the internal energy of the system under standard conditions. In chemical reactions, this is related to the enthalpy change by accounting for work done due to pressure-volume changes, particularly in the case of gases.

Relationship between ?H° and ?E°

This key concept in thermodynamics connects enthalpy change and internal energy change by the equation ?E° = ?H° - ?n_g(RT), where ?n_g is the difference in moles of gaseous products and reactants, R is the gas constant, and T is the absolute temperature. This relation highlights that for reactions involving gases, the work done during volume change affects the difference between enthalpy and internal energy.

Hess's Law

Hess's Law states that the total enthalpy change for a chemical reaction is independent of the pathway taken from reactants to products, meaning that it is a state function. This principle enables the calculation of reaction enthalpies by summing the enthalpies of multiple steps or using standard enthalpies of formation, regardless of the complexity of the reaction mechanism.

Transcript

00:01 So in this problem, we're given four chemical equations and we're asked to find delta h and the delta e values for them.

00:10 For that, we're actually just going to use the thermodynamic data for the selected elements in the back of your book, and in particular, the standard enthothel formation, or delta h .f.

00:25 I'm just going to write it down.

00:27 Okay, so a few things that we need to cover before we go into.

00:32 Solving these problems is that we actually find the delta h value by subtracting the delta h f values of the products of the reactions from those of the products and we find a delta e value by subtracting delta and gas times the r constant times the temperature from delta h or the thing that we actually found prior to that and delta and gas is basically to change in most of gase molecules in your reactions.

01:13 And of course, our our constant, we're actually going to use the one with juice in this example, and that's a .314 juice per mole times k.

01:30 And if delta in gas is actually zero our delta e value is going to equal delta h okay so here we have delta h equals well you have three moles of lead on the side of the product and for that we have zero kilojews per mole which is entopia formation to add, well, for your nitrogen gas, which is right here, you also have zero kilojews, entopia formation, or zero kilojuse per mole.

02:30 And then for the three water molecules, you have three moles times negative 241.

02:46 0 .82 kilojouze per mole.

02:52 And now from all of this, you're actually expected to subtract the sun of the antopies of formations of the reactions.

03:09 Well, again, using the thermodynamic data for the selected elements, table c2, your appendix c in your book.

03:22 You multiply 3 times negative 218 .99 kilojews per mole and that's a standard entropy of formation for the first reactant and to that you add 2mose of the second reactant, nh3, and the entopia for major, that is, 45 .9 kilojews per mo.

04:08 The malls cancelled out, and here you're left wet, negative 725 .46 kilojew juice, minus 748 .77 kilos, and you have 23 .31.

04:29 Kilo juice now for the delta e first let's find the delta in gas well you have three mows of gas and the side of the reactants of the products plus one mole that's four for the side of the products minus two most and that's two most and two most and total.

05:09 So our delta e actually equals 23 .31 kilojews minus 2mo times 8 .314 juice per mo times k times the temperature which is 298 .15 k.

05:45 Now let's just cancel and we're left what delta but e equals 23 .31 kilojews minus 4 ,957 .664 jews.

06:03 Now we need to convert a juice into kilojews by dividing by 1 ,000.

06:13 And that's just going to leave you with 23 .31 kilojews minus 4.

06:29 4 .96 kilo juice which gives you well 18 .35 kilojews and that's the final answer for the delta e.

06:52 Now for this one well again we have delta h equals one mole times negative 92 .307 kj per mole.

07:13 Moles can sew out to that we add again one mole times a negative for 111 .2 kilo juice per mole and again this is the entropy formation for water before that we had then to be formation for sodium chloride.

07:46 Now we can sew out the moles again and then from that we subtract one mole times negative 92 .307 kilo juice per mole, deentopia formation for one of the reactants for the first one.

08:17 Then to that we add one mole times 426...

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