Three hydrocarbons that contain four carbons are listed...

Three hydrocarbons that contain four carbons are listed here, along with their standard enthalpies of formation: $$ \begin{array}{llc} \hline \text { Hydrocarbon } & \text { Formula } & \Delta H_{f}^{0}(\mathrm{~kJ} / \mathrm{mol}) \\ \hline \text { Butane } & \mathrm{C}_{4} \mathrm{H}_{10}(g) & -125 \\ \text { 1-Butene } & \mathrm{C}_{4} \mathrm{H}_{8}(g) & -1 \\ \text { 1-Butyne } & \mathrm{C}_{4} \mathrm{H}_{6}(g) & 165 \\ \hline \end{array} $$ (a) For each of these substances, calculate the molar enthalpy of combustion to $\mathrm{CO}_{2}(g)$ and $\mathrm{H}_{2} \mathrm{O}(l)$ (b) Calculate the fuel value, in $\mathrm{kJ} / \mathrm{g}$, for each of these compounds. (c) For each hydrocarbon, determine the percentage of hydrogen by mass. (d) By comparing your answers for parts (b) and (c), propose a relationship between hydrogen content and fuel value in hydrocarbons.

Three hydrocarbons that contain four carbons are listed here, along with their standard enthalpies of formation:
$$
\begin{array}{llc}
\hline \text { Hydrocarbon } & \text { Formula } & \Delta H_{f}^{0}(\mathrm{~kJ} / \mathrm{mol}) \\
\hline \text { Butane } & \mathrm{C}_{4} \mathrm{H}_{10}(g) & -125 \\
\text { 1-Butene } & \mathrm{C}_{4} \mathrm{H}_{8}(g) & -1 \\
\text { 1-Butyne } & \mathrm{C}_{4} \mathrm{H}_{6}(g) & 165 \\
\hline
\end{array}
$$
(a) For each of these substances, calculate the molar enthalpy of combustion to $\mathrm{CO}_{2}(g)$ and $\mathrm{H}_{2} \mathrm{O}(l)$
(b) Calculate the fuel value, in $\mathrm{kJ} / \mathrm{g}$, for each of these compounds.
(c) For each hydrocarbon, determine the percentage of hydrogen by mass.
(d) By comparing your answers for parts (b) and (c), propose a relationship between hydrogen content and fuel value in hydrocarbons.

Key Concepts

Standard Enthalpy of Formation

This is the enthalpy change when one mole of a compound is formed from its elements in their standard states. It provides a baseline for calculating reaction enthalpies using Hess's Law, as it relates the energy content of substances to their formation from elemental forms.

Combustion Reactions

These are exothermic chemical reactions in which a hydrocarbon reacts with oxygen to produce carbon dioxide and water. Setting up a balanced combustion reaction is essential to determine reaction enthalpies, as it involves the stoichiometry of reactants and products.

Molar Enthalpy of Combustion

This is the energy released per mole of a substance when it is completely burned in oxygen. It is calculated using the standard enthalpies of formation of the reactants and products, providing a measure of the fuel’s energy output on a molar basis.

Fuel Value (Energy Density)

Fuel value refers to the amount of energy produced per unit mass of fuel, often expressed in kilojoules per gram. This concept is critical for comparing different fuels in terms of efficiency and energy content relative to their mass.

Percent Composition

This is the ratio of the mass of an element (such as hydrogen) to the total mass of the compound, expressed as a percentage. It is used to determine the contribution of each element to the overall formula, which in turn influences the energy released upon combustion.

Relationship between Hydrogen Content and Fuel Value

This concept examines the correlation between the percentage of hydrogen in a hydrocarbon and its fuel value. Since hydrogen typically contributes significantly to the energy released during combustion, a higher hydrogen content often correlates with a higher fuel value, thereby impacting the efficiency of the fuel.

Transcript

00:01 This is sort of a giant problem, so let's get ready.

00:05 We're given three hydrocarbons.

00:06 The hydrocarbons are butane, one butene, and one butane.

00:10 And i'm going to put the information that we're given into a table.

00:29 We are given the formula, and we are given the, you know that's formula, but i'll put that up there anyway.

00:49 And we're given the heat of formation in kilojoules per mole.

00:55 We are given 125 kilojoules per mole, negative 1, kilogels per more, and i've got the correct signs there.

01:07 And this, positive 165.

01:10 This is the information we're given.

01:14 Now for each one of these, we are asked to do the following.

01:19 We are asked to calculate the molar enthalpy of combustion, and that'll be in kilojoules per mole.

01:43 For part b, we are asked to convert that.

01:46 We're asked to convert kilojoules per mole to kilojoules per gram.

01:54 And for part c, we are asked to determine the percent hydrogen in each compound.

02:07 Once we've done that for all three, then we are asked to compare our answers for b and c and propose a relationship between hydrogen content and the fuel value.

02:47 Okay.

02:48 So what i'm going to do is i'm going to do a, b, and c for a, b, and c for each one separately.

02:55 I'm just going to do each one.

02:57 And then we'll compare them.

03:02 So that's what we're going to do.

03:03 Let me start with butane.

03:10 So for part a, what first is going to do first is write our balanced chemical equation.

03:34 And we're given that our products are co2 as a gas and h2o as a liquid.

03:41 This were given.

03:43 Now here we have a balanced chemical equation.

03:46 And if i was, or excuse me, an unbalanced chemical equation.

03:49 And if i was going to balance this, i would go ahead and put a 2, a 13, an 8, and a 10.

04:08 This is how i would balance this.

04:10 Now i'm going to divide each of these by two, so i'm going to get rid of these, and i'm going to have a one.

04:18 I'm going to put my one in, just because i can.

04:20 A 13 over two, a four, and a five.

04:29 These are going to be my values here.

04:32 So that's my first part of part a.

04:34 And then the second thing i'm going to do is i'm going to get my delta h by taking the summation of products, my delta h's for my products, minus the sum of the delta h's for the reactants.

04:53 Okay.

04:55 So my products are my co2, which be four times negative 3, 93 .5.

05:07 That's kilojoules per mole, but i'm going to leave it off so i can fit this all in one line.

05:11 Plus my coefficient of five times water, which is negative 285 .83.

05:26 And then we're going to subtract from that one times the value we're given for the butane was negative 125.

05:38 And for this one and this one only, i'm going to put my 13 .2 times zero, because that my oxygen will always be zero.

05:50 Finishing my work for this one, i got a negative 2 ,078, negative 2 ,878 and i'm just going to go, i guess i'll go 0 .15 because i think i use that in a subsequent calculation.

06:12 Killedjoules per mole.

06:15 That's too many sig -figs for a final answer, but that's not my final answer.

06:19 Then for b, we start with our negative 2878 .8 .15 kilojoules per mole.

06:36 And before i continue that, i'm going to go to c, which is our percent hydrogen.

06:47 I've got c4, i've got h10.

06:51 It's going to take four times 12 .01, it equals 48 .04.

07:01 And for hydrogen, i'm going to take 1 .008 equals 10 .08.

07:10 If i add these together i get 58 .12.

07:15 That's the number that goes right there, 58 .12 grams per mole.

07:25 And here i got negative 49 .52.

07:33 And that will be in kilojoules per gram.

07:37 Last but not least, i'm going to take this calculation and to figure that finished this up, i'm going to take 10 .08 divided by 58 .12 times 1 .12.

07:50 100 and i got 17 .34%.

07:57 Let you write this where i've got a little bit more room.

08:20 That's my percent of age.

08:21 So that's my answer for part c.

08:24 There's my answer for part b.

08:27 And i can't remember what i was doing for a.

08:31 Nope, i lied.

08:33 This is a this is b.

08:47 Nope, i was right.

08:49 I am losing my mind here.

08:56 So here's my answer for part a.

08:58 There's my answer for part a.