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.) $$ \text { (a) } 2 \mathrm{C}_{2} \mathrm{H}_{2}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g) $$ (b) $\mathrm{C}_{2} \mathrm{H}_{2}(g)+5 \mathrm{~N}_{2} \mathrm{O}(g) \longrightarrow$ $$ \begin{aligned} & 2 \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)+5 \mathrm{~N}_{2}(g) \\ \text { (c) } \mathrm{NH}_{4} \mathrm{Cl}(s) \longrightarrow \mathrm{NH}_{3}(g)+\mathrm{HCl}(g) & \end{aligned} $$ (d) $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CO}(l)+4 \mathrm{O}_{2}(g) \longrightarrow$ $$ 3 \mathrm{CO}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(g) $$

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.)
$$
\text { (a) } 2 \mathrm{C}_{2} \mathrm{H}_{2}(g)+5 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{CO}_{2}(g)+2 \mathrm{H}_{2} \mathrm{O}(g)
$$
(b) $\mathrm{C}_{2} \mathrm{H}_{2}(g)+5 \mathrm{~N}_{2} \mathrm{O}(g) \longrightarrow$
$$
\begin{aligned}
& 2 \mathrm{CO}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g)+5 \mathrm{~N}_{2}(g) \\
\text { (c) } \mathrm{NH}_{4} \mathrm{Cl}(s) \longrightarrow \mathrm{NH}_{3}(g)+\mathrm{HCl}(g) &
\end{aligned}
$$
(d) $\left(\mathrm{CH}_{3}\right)_{2} \mathrm{CO}(l)+4 \mathrm{O}_{2}(g) \longrightarrow$
$$
3 \mathrm{CO}_{2}(g)+3 \mathrm{H}_{2} \mathrm{O}(g)
$$

Key Concepts

Gas-Phase Reaction Work Correction (PV Work)

For reactions involving gases, the difference between ?H° and ?E° arises from pressure–volume work done during the reaction. The correction accounts for the work associated with changes in the number of moles of gas, and is calculated using the relation ?E° = ?H° ? ?nRT, where ?n is the change in moles of gas.

Standard Enthalpy Change (?H°)

Standard enthalpy change is the heat absorbed or released during a reaction carried out under standard conditions (usually 1 atm pressure and 25°C). It is typically determined using tabulated standard heats of formation for the reactants and products, applying Hess’s law to combine formation reactions into the overall reaction.

Standard Internal Energy Change (?E°)

Standard internal energy change represents the change in the total energy of the system (excluding kinetic and potential energy contributions) for a reaction under standard conditions. It is related to the enthalpy change through the work done by or on the system, particularly the pressure–volume work in reactions involving gases.

Hess’s Law

Hess’s law states that the total enthalpy change of a reaction is the same regardless of the number of steps or the pathway taken, as long as the initial and final states are identical. This principle is used to calculate reaction enthalpies by summing standard heats of formation or combining known reaction enthalpies.

Standard Heats of Formation

Standard heats of formation are the enthalpy changes that occur when one mole of a compound is formed from its elements in their standard states. These values are essential for calculating the overall reaction enthalpy by subtracting the sum of the heats of formation of the reactants from that of the products.

Transcript

00:01 The first thing we need to do for each part of this problem is calculate the standard delta h for the reaction.

00:06 Then knowing the standard delta h for the reaction and the change in the number of moles of gas, we can use this equation here to solve for the standard change in internal energy, where delta ng is the change in moles of gas.

00:21 Here's our first chemical reaction, acetylene reacting with oxygen to make carbon dioxide and water.

00:28 So the standard delta h for this reaction will be four times, the standard delta h of formation of carbon dioxide, plus two times the standard delta h of formation of gaseous water, and then minus five times the standard delta h of formation of oxygen, which is zero, and still minus two times the standard delta h of formation of acetylene.

00:55 This gives us a standard delta h for this reaction of negative 2 ,511 .1 kilojoules.

01:04 Delta e standard will then be a standard, our delta h standard that we just calculated minus delta n delta n is going to be the moles of product of gas which is six minus the moles of gaseous reactant which is seven six minus seven is negative one we'll multiply that by our r value 8 .314 multiplied by the kelvin temperature and then from there we have joules so we need kilojoules so that we can add it to this kilojule value, we'll divide by 1 ,000, and we get negative 2 ,508 .6.

01:48 You'll notice in most cases the delta e and the delta h values are close.

01:54 So the next reaction, we have acetylene reacting with dinitrogen oxide to produce carbon dioxide, water, and nitrogen gas.

02:04 To calculate the delta h standard of this reaction, will take five times the delta h of formation of nitrogen, which is 0, multiplied by the delta h of formation of gaseous water, negative 241 .8, plus two times the delta h of formation of carbon dioxide.

02:25 We'll then subtract off the delta h of formation of acetyling, positive 226 .75, and then five times the delta h of formation of n2o.

02:39 This gives us a delta h standard for this reaction of negative 1 ,663 .4 kilojoules.

02:46 Now to get delta e standard, we'll take our delta h standard minus off delta n.

02:52 So we've got 5, 6, 7, 8, moles of gas...

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