La molécula de metano, CH4, tiene la geometría que se muestra en la figura 2.21. Imagine un proc

step 1 This is an interesting problem. We are asked to consider the methane molecule, which has the geo

La molécula de metano, CH4, tiene la geometría que se...

La molécula de metano, $\mathrm{CH}_4$, tiene la geometría que se muestra en la figura 2.21. Imagine un proceso hipotético en el cual la molécula de metano se "expande" extendiendo simultáneamente los cuatro enlaces $\mathrm{C}-\mathrm{H}$ hasta el infinito. Entonces, tenemos el proceso $$ \mathrm{CH}_4(\mathrm{~g}) \longrightarrow \mathrm{C}(\mathrm{g})+4 \mathrm{H}(\mathrm{g}) $$ (a) Compare este proceso con el inverso de la reacción que representa la entalpia estándar de formación. (b) Calcule el cambio de entalpía en cada caso. ¿Cuál proceso es más endotérmico? ¿Qué explica la diferencia en los valores de $\Delta H^{\circ}$ ? (c) Suponga que $3.45 \mathrm{~g} \mathrm{de}_4(\mathrm{~g})$ reaccionan con $1.22 \mathrm{~g} \mathrm{de}_2(\mathrm{~g})$ para formar $\mathrm{CF}_4(\mathrm{~g})$ y $\mathrm{HF}(\mathrm{g})$ como únicos productos. ¿Cuál es el reactivo limitante en esta reacción? Suponiendo que la reacción se efectúa a presión constante, ¿cuánto calor se desprende?

La molécula de metano, $\mathrm{CH}_4$, tiene la geometría que se muestra en la figura 2.21. Imagine un proceso hipotético en el cual la molécula de metano se "expande" extendiendo simultáneamente los cuatro enlaces $\mathrm{C}-\mathrm{H}$ hasta el infinito. Entonces, tenemos el proceso
$$
\mathrm{CH}_4(\mathrm{~g}) \longrightarrow \mathrm{C}(\mathrm{g})+4 \mathrm{H}(\mathrm{g})
$$
(a) Compare este proceso con el inverso de la reacción que representa la entalpia estándar de formación. (b) Calcule el cambio de entalpía en cada caso. ¿Cuál proceso es más endotérmico? ¿Qué explica la diferencia en los valores de $\Delta H^{\circ}$ ? (c) Suponga que $3.45 \mathrm{~g} \mathrm{de}_4(\mathrm{~g})$ reaccionan con $1.22 \mathrm{~g} \mathrm{de}_2(\mathrm{~g})$ para formar $\mathrm{CF}_4(\mathrm{~g})$ y $\mathrm{HF}(\mathrm{g})$ como únicos productos. ¿Cuál es el reactivo limitante en esta reacción? Suponiendo que la reacción se efectúa a presión constante, ¿cuánto calor se desprende?

Key Concepts

Thermochemical Processes at Constant Pressure

When reactions occur at constant pressure, the heat exchanged is equal to the change in enthalpy of the system. This concept is vital for interpreting experimental data and comparing the energy changes of different reaction pathways, as it directly links the measured heat flow to the chemical transformations taking place, including both endothermic and exothermic processes.

Hess’s Law

Hess’s Law states that the total enthalpy change of a reaction is the same, no matter how it is carried out, as long as the initial and final conditions are identical. This principle allows for the calculation of reaction enthalpy changes by combining known enthalpies of formation or bond energies from a series of steps, providing a powerful tool for evaluating both direct and reverse reaction pathways.

Reaction Stoichiometry and Limiting Reactants

Reaction stoichiometry involves the quantitative relationships between reactants and products in a chemical reaction, which is essential to determine how much product forms and which reactant will run out first. The concept of the limiting reactant is particularly important when dealing with reactions involving multiple reagents, as it limits the extent of the reaction and dictates the final energy change based on the amounts available.

Standard Enthalpy of Formation

This concept refers to the enthalpy change when one mole of a compound is formed from its elements in their standard states under standard conditions. It is crucial for comparing different reaction pathways, as reversing a formation reaction changes the sign of the enthalpy, allowing one to analyze how energy is absorbed or released when breaking down or forming chemical bonds.

Bond Dissociation Energy

Bond dissociation energy is the amount of energy required to break a specific chemical bond in a molecule, converting it completely into its separated atoms in the gas phase. In processes where bonds are broken, such as extending bonds to infinity, bond dissociation energy is a key factor in determining the total endothermicity of the reaction.

Transcript

00:01 This is an interesting problem.

00:03 We are asked to consider the methane molecule, which has the geometry shown, and here i'm going to try to draw it.

00:10 So methane, i'm going to make my carbon black, and then i'm going to go this way, this way, and this way, and try to put the vertices of a tetrahedron on here.

00:37 There.

00:44 Imagine a hypothetical process in which methane, which is a gas, is expanding by extending all of the four ch bonds.

00:59 So they're talking about is making these bonds all longer.

01:09 And that would be analogous to doing this.

01:25 Okay.

01:27 We're going to compare this process, compare this to the reverse of the reaction that represents the standard enthalpy of formation.

02:28 Okay.

02:33 So here is the standard, the enthalpy, standard, standard entropy of really, standard enthalpy of formation will be c, obviously, plus four, excuse me.

02:49 I'm going to go to h2, you'll see h4.

03:01 And the reverse reaction of this one, if i do the reverse of this, obviously, again.

03:12 And you'll notice that i've got a different state for my carbon, and i have diatomic hydrogen.

03:21 So the states are different, and hydrogen is listed as h2 or h, both gases.

03:48 Let me see if they want anything else in this problem.

03:52 I think that's good for that part.

03:54 Let's see what b is.

03:57 I've got to make my, give me a second here.

03:59 I've got to make my textbook a little bigger so i can see it better.

04:05 And then we're going to calculate the enthalpy change in each case.

04:23 Anthopy change for each.

04:31 And after we do that, we're going to say which is more endothermic.

04:44 And what accounts for the difference in the delta in the heats of formation? i should say heats of reaction.

05:11 Okay.

05:14 So let's begin our next part.

05:34 I feel like i've lost my internet connection here.

05:36 I can't see.

05:37 There we go.

05:38 Sorry about that.

05:44 So our first reaction was, let me do a different color for this one.

05:52 Our first reaction was, i'm going to go with brighter colors.

05:57 C .h4.

06:08 And remember to do our delta h, it's always, i'm only going to write this once for our reaction is equal to the sum of the products, delta h for the products, minus the sum of the sum of, the delta h for the reactants.

06:31 So we can look these up.

06:33 I'm going to make a little note.

06:34 I'll make this an orange in the tables in the back of your book.

06:46 And don't forget to multiply by coefficients.

06:50 So for this problem, our delta h, whoops, change back, for the reaction is going to equal carbon and for this first one i'll just put a one in.

07:06 We can look up the value of 718 .4 plus now i'm going to take my 4 times hydrogen which is 217 .94.

07:26 And for that i'm going to subtract one again because that's my coefficient for methane and negative 74 .8.

07:39 And i forgot to put my kilojoules here.

07:41 So i'll just add it down there.

07:44 When i do my math on this one, i get 1 ,665 kilojoules.

07:54 My second reaction is ch4.

08:04 Let's see this time it's a solid plus 2h2 which is a gas.

08:15 So my heat of reaction this time is going to equal 0 for my carbon plus zero for my oxygen minus my negative 74 .8 kilojoules.

08:36 So the this will be positive 74 .8 kilojoules.

08:46 Okay, we can see that our first reaction is more exothermic.

08:52 So i'm going to write purple is more exothermic, or excuse me, more endothermic.

09:07 And what accounts for the difference? the big difference here is because of our standard states for c, s, and h2g.

09:26 In the green one in this reaction.

09:32 Okay.

09:34 Let me see what my next question is, so i can write it down nicely.

09:38 C, suppose 3 .45 grams of methane gas reacts with 1 .22 grams of fluorine gas, forming carbon tetrafluoride gas.

10:33 Keep wanting to put a parenthesis there.

10:38 And hydrogen fluoride gas.

10:46 They want us to find the limiting reactant and they want us to find heat evolved.

11:03 So it's going to be a rather longish but not too bad.

11:07 And whenever i'm doing these things i always cheat and i just look up the molar masses so i don't have to sit and figure them out...

Please give Ace some feedback