If hydrogen were used as a fuel, it could be burned according to this reaction: H 2 (g) + 1/2 O 2 (g) → H 2 O(g) Which fuel yields more energy per gram?

welcome back everyone in this example, we're told that methane burns according to the given reaction. We need to calculate the entropy of the reaction for the combustion of methane and then for the combustion of ethane and then we need to identify which of these fuels produces more energy Permal upon combustion. So we have a three part question. Part one is finding our entropy of methane. Part two is finding the entropy of ethane and then part three is figuring out which of these two fuels produces more energy. So beginning with part one, we want to recall how to find our entropy of our reaction. And we can recall that. We would calculate this by taking the sum of our entropy of our bonds that are broken on our reactant side, which is then subtracted from the sum of the entropy of our bonds that are formed on the product side. So to make this simple, we'll draw out the structural formulas of our Compounds making up our reaction in the combustion of methane. And beginning with methane we have Ch four. So we have a carbon that is bonded to four hydrogen atoms in a single bond. And according to our reaction, we just have one mole of this type of or of our methane molecule. And so counting the types of bonds in this molecule, we see, we have four of these carbon hydrogen single bonds, One here, one here, another here and another here. So we'll write that down as a note four of our carbon hydrogen single bonds then added on to this. We have our second reactant where we have two moles of our oxygen di atomic molecule in gaseous form. And so we would recall that oxygen has a bonding preference of having two bonds. So we have two oxygen atoms double bonded to one another. And even though we have just one type of this oxygen double bonded to oxygen bond here, because we have two moles, we have two times one. So we would say two moles comes one of our oxygen oxygen double bond, and that's going to give us two of our oxygen oxygen double bond. So then on our product side where our bonds are formed, we have first our carbon dioxide where we should recall that carbon in this molecule is going to be double bonded to two oxygen atoms. And so we would be able to count two of these carbon oxygen double bonds. So making note of that below, we have two of our carbon oxygen double bonds. And then for our last bonds formed from our gaseous water molecule, we have our two moles of this molecule formed where we have oxygen single bonded to two hydrogen atoms. And because we count two of these oxygen hydrogen single bonds making up our molecule, we would say we have two oxygen hydrogen single bonds, times two moles. And that is going to give us two of our oxygen or sorry for of our oxygen hydrogen single bonds. So calculating our entropy change for the combustion of methane, we would say. And we'll actually use the color purple to keep things consistent, we would say that we have four multiplied by the entropy of our carbon hydrogen single bonds. Then added onto this, we have two multiplied by the entropy of our oxygen oxygen double bond. This completes our some of our bonds formed. And then on our product side we would subtract so that we have two multiplied by the entropy of the carbon oxygen double bond, which is then added to four times the entropy and will make more room here. So four times the entropy of our oxygen hydrogen single bonds. And so what we need to do is go to our textbooks or online to get these values of the entropy of these types of bonds that we have listed out here. And so we would say that our entropy of our reaction for the combustion of methane is equal to for multiplied by in our textbook the entropy of a carbon hydrogen single bonds, which is a value of 413 Kayla Jules Permal. This is then added to two, multiplied by the entropy of an oxygen oxygen double bond, which has a value in our textbooks of 495 kg joules per mole. Then we would be subtracting from this, our entropy some of our entropy of our bonds formed where we have two multiplied by the entropy of our carbon oxygen double bonds, which have a value of 799 kg joules per mole in our textbooks which is then added to four, multiplied by our entropy of our oxygen hydrogen single bonds which have a value in our textbooks of 467 kg joules per mole. And so in our calculators we're going to get a result for entropy of a reaction for the combustion of methane equal to a value of negative kg joules per mole. Now, because we have a negative value for entropy here, we should recall that our combustion of methane is going to be there four eggs a thermic. So for our first answer, we have found our combustion or our entropy for our combustion of methane. So this is for methane here and now we want to move on to part two of our prompt which is finding the combustion of methane. So for part Two we'll do that below here. So, beginning by writing out our reaction, we have ethane C two H six which reacts with oxygen gas to form in a combustion carbon dioxide and water in both gaseous form. And we want to recall that we need to bounce this equation. So we're going to place a coefficient of 7/2 in front of our oxygen A two in front of our carbon dioxide and a coefficient of three in front of our gaseous water. Now, our next step is to draw out our structural formulas so that we can see the type of bonds that may pick up our equation. So we have the thing where we have two carbon atoms single bonded and sorry about that. So these two carbon atoms are single bonded to one another and are also bonded to a total of six hydrogen. So we have three hydrogen is here, and then three hydrogen is bonded to our second carbon in methane. And then for our second reactant, we have again our oxygen double bonded to an oxygen making up our di atomic oxygen molecule. We have a coefficient of 7/2. So we would say we have 7/2 moles multiplied by our one oxygen oxygen double bond. And that's going to give us 7/2 of our oxygen oxygen double bond. And going back to the Thane, we would be able to see that we have just one mole of our Ethan, but we have a total of six of our carbon hydrogen single bonds making up methane. Moving on to our product side, we can see that we have carbon dioxide and water where we have carbon double bonded to two oxygen atoms. We have two moles of this formed and so we would say we have two moles multiplied by our two carbon oxygen double bonds. And that's going to give us a total of four of our carbon oxygen double bonds. And then our second product is going to be our three moles of our water gaseous molecules. So we have two oxygen hydrogen single bonds where we would say we have three moles multiplied by r two oxygen hydrogen single bonds. And that's going to give us a total of six of our oxygen hydrogen single bonds. And so going into our calculation for the change in entropy of our reaction for the combustion of methane, we would say that we have and correction. We also need to write down the one carbon carbon single bond making up our ethane molecule in the center here. So going back to our entropy of our reaction calculation and we'll put an answer in here, we have the sum of our bonds that are broken on our react inside. So we have the entropy of our carbon carbon single bond which is added to six times the entropy of our carbon hydrogen single bond in Ethan. And then added to this, we have 7/2 of our oxygen or 7/ multiplied by our entropy of our oxygen double bond, oxygen. Then subtracted from this on our product side, we have four multiplied by the entropy of our carbon double bonded to oxygen bond, which is then added to six times our And we'll scoot this over. So six times the entropy of our oxygen hydrogen single bonds that are broken or formed on the product side. So plugging in our values from our textbooks, we would say that the entropy change of our reaction for the combustion of methane is going to be first. Our entropy of our carbon carbon single bond which in our textbooks has a value of 347 kg joules per mole, which is then added to six, multiplied by the entropy of our carbon hydrogen single bonds in our textbooks, which has a value of 413 kayla jules Permal. And then added on to this we have 7/2, multiplied by the entropy of our oxygen oxygen double bond, which have a value in our textbooks of 495 kg joules per mole. Subtracted from this, we have four multiplied by the entropy of our carbon oxygen double bond in our textbooks, which is a value of 799 kg joules per mole. And then added to this, we have six multiplied by the entropy of our oxygen hydrogen single bonds in our textbooks, as the value 467 kg joules per mole. So what we're going to get is our entropy change of our reaction for the combustion of methane equal to a value of negative 1440 kayla jewels per mole. And because we recall that when we have a negative entropy value that this therefore means we have an extra thermic reaction, which we recall means energy is released. We would say that thus Ethan Which is our C2H6, produces more energy Permal. And so for our last two answers, we have our entropy for the combustion of methane highlighted in yellow here and then we have the fact that the Thane will be the one that produces energy Permal based on its entropy of its combustion, which is a more negative value negative 1440 kg joules per mole, signaling that we release that amount of energy in that reaction. And so what's highlighted in yellow represents all of our three final answers to complete this example. So I hope that everything that I reviewed was clear. If you have any questions, please leave them down below and I will see everyone in the next practice video.

Ch.10 - Chemical Bonding I: The Lewis Model

All textbooks

Tro 6th Edition

Ch.10 - Chemical Bonding I: The Lewis Model

Problem 111d

Previous problem

Next problem

Chapter 10, Problem 111d

If hydrogen were used as a fuel, it could be burned according to this reaction: H₂(g) + 1/2 O₂(g) → H₂O(g) Which fuel yields more energy per gram?

Verified step by step guidance

Identify the enthalpy change (\( \Delta H \)) for the combustion of hydrogen gas (\( \text{H}_2 \)) using standard enthalpy of formation values.

Calculate the molar mass of hydrogen gas (\( \text{H}_2 \)) to convert the energy per mole to energy per gram.

Compare the energy yield per gram of hydrogen with other common fuels, such as gasoline, by calculating their energy yield per gram using their respective combustion reactions and enthalpy changes.

Consider the energy density (energy per unit mass) of hydrogen and compare it with other fuels to determine which has a higher energy yield per gram.

Discuss the implications of using hydrogen as a fuel in terms of energy efficiency and environmental impact.

Verified Solution

Video duration:

12m

This video solution was recommended by our tutors as helpful for the problem above.

Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Combustion Reaction

A combustion reaction is a chemical process in which a substance reacts rapidly with oxygen, releasing energy in the form of heat and light. In the case of hydrogen, it reacts with oxygen to form water, producing a significant amount of energy. Understanding the stoichiometry of this reaction is essential for calculating energy yields.

Energy Density

Energy density refers to the amount of energy stored in a given system or region of space per unit volume or mass. For fuels, it is typically expressed in joules per gram or kilojoules per liter. Comparing the energy density of hydrogen with other fuels helps determine which fuel is more efficient for energy production.

Enthalpy of Combustion

The enthalpy of combustion is the heat released when one mole of a substance is completely burned in oxygen. It is a crucial concept for evaluating the energy output of different fuels. For hydrogen, the enthalpy of combustion is high, indicating that it releases a large amount of energy per gram when combusted, making it a potent fuel source.