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) $3 \mathrm{PbO}(s)+2 \mathrm{NH}_{3}(g) \longrightarrow$ (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) $3 \mathrm{PbO}(s)+2 \mathrm{NH}_{3}(g) \longrightarrow$
(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

Stoichiometry and Change in Moles of Gas (?n)

Stoichiometry involves the quantitative relationships between reactants and products in a chemical reaction. The change in the number of moles of gas (?n) is crucial for converting between enthalpy and internal energy changes, as it accounts for the work done by or against the pressure of the system during a reaction.

Standard States and Conditions

Standard states provide the reference conditions (such as 1 atm pressure for gases and pure substances in their standard states) used when reporting thermodynamic data like ?H° and ?E°. These conditions ensure that measurements and calculations of thermodynamic properties are consistent and comparable across different reactions.

Hess's Law

Hess's Law states that the overall enthalpy change of a chemical reaction is independent of the reaction pathway. This principle allows one to calculate the reaction’s enthalpy change by adding together the enthalpy changes for multiple steps that lead to the overall reaction, making use of standard enthalpies of formation.

Standard Internal Energy Change (?E°)

Standard internal energy change reflects the total change in energy of a system at constant volume. For reactions involving gases, it is closely related to ?H° by accounting for the work done due to volume change. The relationship is given by ?E° = ?H° - ?nRT, where ?n represents the change in the number of moles of gas and R and T are the gas constant and temperature, respectively.

Standard Enthalpy Change (?H°)

Standard enthalpy change is the heat absorbed or released during a chemical reaction at constant pressure under standard conditions (typically 25°C and 1 atm). It is usually determined using standard enthalpies of formation and Hess's Law, where the sum of the products’ enthalpies is subtracted from the sum of the reactants’ enthalpies, each multiplied by their stoichiometric coefficients.

Standard Enthalpy of Formation

The standard enthalpy of formation is the enthalpy change when one mole of a compound is formed from its elements in their most stable states under standard conditions. These values serve as the fundamental data for calculating reaction enthalpies using Hess’s Law, by summing the enthalpies for products and subtracting the sum for reactants.

Transcript

00:01 To answer all parts of this question, we first need to calculate the standard delta h for each of these reactions using the standard delta h of formation for all the reactants and the products in the back of the book.

00:20 So for this first reaction, we have three moles of lead to oxide reacting with two moles of ammonia gas to produce three moles of lead, two or one mole of nitrogen, and three moles of oil.

00:41 Water as a gas.

00:46 So delta h will be the delta h of formation of gaseous water multiplied by three plus the delta h of formation of lead which is zero and the delta h of formation of nitrogen gas which is zero will then subtract off the delta h is a formation of the reactants two times the delta h of formation of ammonia plus three times the delta h of formation of the of lead to oxide and we get 24 .58 kilojoules.

01:25 Now to solve for delta e standard we need our delta h standard which we just determined.

01:31 We need to know the change in moles of gas.

01:34 So it's going to be the final moles which will be 1 plus 3 minus the initial moles 2.

01:43 So 1 plus 3 is 4 minus 2 is 2.

01:47 So n is 2.

01:49 So we'll take our delta value and then do minus 2 times the r value of 8 .314 so we get units of joules multiplied by our temperature which is provided at 25 degrees celsius for 298 kelvin we'll then convert the jewels we get here to kilojoules so that we can sum it with the kilojoule value we determined up here and we get 19 .6 kilojoules.

02:18 For the next reaction, we have sodium hydroxide solid reacting with one mole of gaseous hcl, producing nacl solid and h2o liquid.

02:31 So the delta h for the reaction is going to be the delta h of formation of solid sodium chloride, plus the delta h of formation of liquid water, will then subtract off the delta h of formation of n -a -o -h and hcl, and this gives us a standard delta h for the reaction of negative 177 .8 kilojoules.

02:55 We'll then use this equation up here to solve for delta e...

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