Determine Δr H^∘(298 K) for each of the following reactions....

Determine $\Delta_{\mathrm{r}} H^{\circ}(298 \mathrm{K})$ for each of the following reactions. For data, see Appendix $11 .$ (a) $\mathrm{SO}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g})\longrightarrow\mathrm{SO}_{3}(\mathrm{s})$ (b) $\mathrm{PCl}_{5}(\mathrm{s}) \longrightarrow \mathrm{PCl}_{3}(\mathrm{l})+\mathrm{Cl}_{2}(\mathrm{g})$ (c) $4 \mathrm{FeS}_{2}(\mathrm{s})+11 \mathrm{O}_{2}(\mathrm{g})\longrightarrow2 \mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{s})+8 \mathrm{SO}_{2}(\mathrm{g})$ (d) $\mathrm{SF}_{6}(\mathrm{g})+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\longrightarrow\mathrm{SO}_{3}(\mathrm{g})+6 \mathrm{HF}(\mathrm{g})$

Determine $\Delta_{\mathrm{r}} H^{\circ}(298 \mathrm{K})$ for each of the following reactions. For data, see Appendix $11 .$
(a) $\mathrm{SO}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g})\longrightarrow\mathrm{SO}_{3}(\mathrm{s})$
(b) $\mathrm{PCl}_{5}(\mathrm{s}) \longrightarrow \mathrm{PCl}_{3}(\mathrm{l})+\mathrm{Cl}_{2}(\mathrm{g})$
(c) $4 \mathrm{FeS}_{2}(\mathrm{s})+11 \mathrm{O}_{2}(\mathrm{g})\longrightarrow2 \mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{s})+8 \mathrm{SO}_{2}(\mathrm{g})$
(d) $\mathrm{SF}_{6}(\mathrm{g})+3 \mathrm{H}_{2} \mathrm{O}(\mathrm{g})\longrightarrow\mathrm{SO}_{3}(\mathrm{g})+6 \mathrm{HF}(\mathrm{g})$

Key Concepts

Stoichiometry in Thermochemistry

Understanding reaction stoichiometry is essential in thermochemistry since it dictates the proportion in which reactants combine and products form. This proportionality is crucial when applying enthalpy values to a reaction because the enthalpy change must be scaled according to the number of moles of substances involved in the reaction.

Standard Enthalpy Change of Reaction Calculation

This key concept involves the calculation of the overall heat change of a reaction under standard conditions (298 K, 1 atm) using tabulated standard enthalpies of formation. The reaction enthalpy is calculated by subtracting the sum of the standard enthalpies of formation of the reactants, each multiplied by their stoichiometric coefficients, from that of the products.

Standard Enthalpy of Formation

This concept refers to the heat change when one mole of a compound is formed from its elemental constituents in their standard states. It provides the fundamental building blocks for calculating reaction enthalpies by offering a way to quantify the energy associated with forming each compound from its pure elements.

Hess's Law

Hess's Law states that the total enthalpy change for a chemical reaction is the same, regardless of the pathway taken from reactants to products. This principle allows for the calculation of reaction enthalpies by combining standard enthalpies of formation for reactants and products, even when the direct measurement of the reaction enthalpy is not feasible.

Transcript

00:01 Okay, so here we'll be calculating the standard entropy change for a few different reactions.

00:10 The way that you calculate delta h for a specific reaction is to take the sum of the delta h formation for the products and minus the sum of the delta h of formation for the reactants.

00:37 And you want to make sure to use your coefficients that are in your reaction as well.

00:41 So let's go through these examples.

00:46 If we have s .o2 gas and we add one half o2, that goes to s .o3 solid.

01:04 Okay, so to calculate delta h here, let's start by doing the delta h formation of the products.

01:12 So we're going to sum that all together.

01:13 In this case, it's easy because we only have one product.

01:17 So what we're going to do is we're going to say the product that we have, s .o3 solid.

01:20 We have one of those, so make sure to include that coefficient.

01:23 And then we'll multiply that by the delta h of formation for s o3 solid which is negative 454 kilojoules per mole so that's it for the products part and now we'll subtract the sum of the delta h of formation for the reactants so in this case we have one times the delta h of formation of s02 because we only have one of those that's negative 297 kilojoules per mole and then we'll add so within this whole bracket we'll add the other reactant which is 02 gas times one half so one half times the delta h formation of o2 which in this case is zero and delta h for this reaction when we propagate this math so delta h equals all of this this will be negative 157 kilojoules per mole okay so let's do b in part b we have pcl 5 solid and that's going to go to pcl3 liquid and co2 gas.

02:54 So we'll do the same thing here.

02:56 The sum of the delta h of formation of the products minus the sum of the delta h of formation of the reactors.

03:03 So delta h will equal do the products first.

03:07 So 1 times the delta h of formation of pcl3 liquid, which is negative 320 kilojoules per mole.

03:18 And then we'll add on 1 times the delta h of of cl2 gas, which is zero.

03:24 So that's the products part.

03:25 And now for the reactants, we'll minus one times the delta h of formation of pcl5 solid, which is negative 443 .5 kilojoules per mole.

03:42 And that will give us a delta h for the reaction of positive 123 .5 kilojoules per mole.

04:00 Part c, now we're starting to get higher coefficient here four fes2 and this is a solid and we're going to add 1102 gas so that's 1102 gas and that looks kind of weird sorry and then the product side will have two to f .e .203 and this is going to be a solid and the other product is 8 s .o2 gas.

04:40 Okay so here we're going to do is the same procedure as before but we have a few more coefficients than in the others.

04:47 So here we'll start with the products...

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