consider-the-combustion-of-ethane-mathrmc_mathrmh-mathrmh_6g-a-write-a-balanced-equation-for-the-rea

Question

Consider the combustion of ethane, $\mathrm{C}_{\mathrm{H}} \mathrm{H}_{6}(g)$ 
a. Write a balanced equation for the reaction. 
b. Draw an energy diagram for the forward reaction. Label the reactants and the products. Use the table of heats of combustion from Lesson $103 :$ Fuelish Choices to label the heat of reaction.

David Collins · Accepted Answer

this question is a conceptual question that reviews a little bit of balancing chemical reactions. The first thing that it asks you to dio is right. The balanced chemical reaction for ethane. It&#x27;s a combustion reaction, so it&#x27;s ethane, combining with oxygen to produce carbon dioxide and water. The first thing we&#x27;ll do is we&#x27;ll balance the carbons. We have two carbons, so we needed to hear. Then we&#x27;ll look at the hydrogen. We have six hydrogen, so we need a three here. Then we recognize that we have two times two is four plus three that seven Oxygen&#x27;s to get seven Oxygen&#x27;s from 02 We need to multiply it by seven halves. But that won&#x27;t work because we want integers so we will multiply everything by two in order to get our balanced chemical reaction. We also produce some heat. So potential energy diagram is going to look like this where the reactant or higher in potential energy than the products and the amount of energy that is released will be the difference between where we started with the reactant and where we end up with the products and going back to lesson one or three. We can look up this value. It&#x27;s 15 1560 killer jewels per mole, but it&#x27;s negative because this is the amount of heat that is released.

Kevin Zaborsky · Answer

in part a of this problem. We want to determine the change in entropy of the overall combustion reaction of ethane and in part B, we want to find the same or benzene. We need to determine the Delta h of reaction using two different methods. The first way is to write out the thermal chemical equations corresponding to the mental p of formation reactions for the appropriate chemical species in the overall combustion reaction from their elements. We then need to combine these thermo chemical equations using Hess&#x27;s law to calculate the overall change in entropy of the entire reaction. The second way we&#x27;re going to determine Delta h of reaction is by using the equation that incorporates the n tal piece of formation. So starting with part A, we know that we&#x27;re combusting ethane, which means we&#x27;re reacting it with oxygen to produce urban dioxide and water. So the first thing that we need to do is balance this combustion reaction. If we start with one mole of ethane, we see that we need two moles of carbon dioxide to balance out the carbons. However, you 21 oxygen atom being present and water that would lead to in odd number of Oxygen&#x27;s on the right side, and we would have to use fractions in order to balance that equation. So instead we put a to here so that we have four carbons total on the left side, and so we would need a four here have four CEO twos, and now we see that we have a total of 12. Hydrogen is on the reactant side, and so we wouldn&#x27;t need a six year balance out the hydrogen. Now we can see on the product side that we have a total of eight oxygen atoms from CO two plus six oxygen atoms from H 20 And so that&#x27;s a total of 14 owes on the right side, which needs to be balanced with 14 owes on the left side. So that is the balanced combustion reaction for the combustion of everything. The first way that we need to find the Delta h of reaction, its buy rating at the thermal chemical equations corresponding to the the formation reactions of each of the chemical species. Starting with ethane, we see that it forms from the elements of carbon and hydrogen and so solid carbon and die atomic hydrogen exists naturally, and we can react those to form one mole of C two h two. We go through the other species, but we knew that 02 is exists naturally and therefore has a an entropy of formacion value of zero. So we ignore that when rating at the thermo chemical equations and move onto the next one, CO two involves carbon and oxygen. So we react solid Kurban in di atomic oxygen to form one mole of CO two and finally for water. Hydrogen and oxygen are going to react, and they both exist naturally, is die atomic gases in for all the Until p information thermo chemical reactions. We are producing one mole of the desired compound, and so we have to balance out the left side of each reaction appropriately, with proper sticky metric coefficients to produce one mole of each product in each case. Now we can look up the entropy of formation values for each one of these thermo chemical equations in the appendix. So now that we have everything that we need to use Hess&#x27;s law to rearrange the thermal chemical equations, you resembled the desired combustion reaction. We see that we need to moles of C two h six as a reactant. And here we have one mole of C two h six as a product. So to make it a reactant, we have to flip it or multiply about a negative value. It needs a coefficient of two. So we want to play reaction A Are you negative too? Now we see that we need four moles of CO two in the products here week we form one mole of co two is a product, so we need to multiply the second thermo chemical equation by four. Finally, we need six miles of water. As a product, we conform one mullah of water as a product. But we need to multiply that by six to get the appropriate striking metric coefficient. So now we we multiply those three chemical reactions by those appropriate coefficients. And when we do that, we also have to multiply the delta, each values by this by the same amount. So if we multiply a negative 83.8, where the formation of ethane, you have to multiply that by negative too to get 167.6 and or the 2nd 1 we multiply negative 393.5 by four, the last only multiplying negative, 285.8 by six. And that&#x27;s how we get these values. And now we can see that using Hess&#x27;s law, we can cancel out like terms to produce the desired combustion reaction. We have four moles of carbon on each side which cancel out and six miles of water or ah, hydrogen gas on each side. And also cancel out. We have four plus three moles of 02 both on the reactant side, which combined to produce a total of seven moles of oxygen for the combustion reaction. And now we see that this matches the overall combustion reaction that we balanced to begin. And so now we just add all of the adjusted entropy values, which are given in units of killing Joules per mole were producing one mole in these formation reactions to a corresponds to kill a JAL&#x27;s total energy. And when we do all that math, we get a Delta H value of negative 3000 121.2 killer jewels. When we had those three adjusted dental piece together. Now the second way to determine the delta h of reaction is using this equation where we add up all of the entropy of formation values multiplied by the number of moles in the combustion reaction for each species in the products and subtracted the same. From then, we subtract. The B reactant is from the value of the products so that the total entropy change of the combustion reaction equals a total entropy change of the products minus the total entropy change of the reactors. Looking at our equation, we start with the product side with So we have four moles of co two, six moles of water. And we used those entropy of formacion values for co two and water that we looked up for the first part of this problem. We know that oxygen forms naturally said as an entropy information value of zero. So for the reactant portion of the equation, we only have to worry about C two h two. We see that we have two moles of that we want to play by its entropy information value and now we subtract the little entropy change of the products minus the total entropy change of the reactant and we see that we get a delta H of reaction of negative 3121 point to kill it, Jules, which is which matches what we got before solving it, using the manipulation of the thermo chemical equations. And now we solve part. Be using the same algorithm. We place it to here, see that we need a coefficient of 12. Balance out 12 curb ins on each side. This is for the combustion of benzene. And now we have 12 hydrogen atoms, which we need. We need six miles of water to balance those out. And finally we have 24 oxygen&#x27;s from CO two plus six oxygen&#x27;s from water for a total of 30 oxygen&#x27;s, which needs to be balanced out with 15 moles of 02 And now we write out the thermal chemical equations for the information of these chemical species, starting with C six h six. We need carbon and hydrogen to form one mole of C six h six and we balance. We balance it appropriately so that we produce one mole. We go through with the other species and we also list out there thermo chemical equations as we did in part A, and we can look up their entropy information values, as we did in part a. Now we can rearrange them to resemble our desired combustion reaction. We see that we want to moles of benzene as a reactant, and we conform one mole of benzene as a product, so we need to flip it to make it a reactant. Multiply it by two to match the story key metric coefficient. We don&#x27;t worry about oxygen because it forms naturally, and we see we want 12 moles of CO two. As a product, we conform one mole of CO two&#x27;s of products, so we need to we need him out to play reaction, be by 12 and we want six moles of water. As a product we conform. One mole of water is a product, so we need to multiply the third reaction by six. And now, when we do that and combine the equations via Hess&#x27;s law, we cancel out 12 urban&#x27;s on each side of the equation, and six hydrogen is on each side of the equation, and we add the 12 and three oxygen&#x27;s to get 15 oxygen&#x27;s and the combustion of benzene, which matches our target, our target reaction. And now, after we multiply the first still to H value by negative to the 2nd 1 by 12 the 3rd 1 by six. These are the values that we end up with and when we add them together we get a delta h of reaction for the combustion of benzene to be negative 6535 killer jewels. And now we verify that solution using the entropy of reaction equation involving NTP&#x27;s information starting with the products you see that we have 12 months 12 moles of co two in six moles of water. In the products we multiply each of those by their respective entropy information values to get the total entropy change of the products and then for the reactant Swe have two moles of C two h six and we don&#x27;t worry about Oh too. Do you want to play those two moles? And the reactant is by its value for gentle P information to get the total entropy change of the reactant and then we subtract the entropy change of the reactions from the entropy change of the products you get a total delta h of reaction of negative 6535 killer JAL&#x27;s, which matches what we calculated using the manipulation of thermo chemical equations.

David Collins · Answer

Let&#x27;s start again by writing the chemical reaction. It&#x27;s going to be ethane. See for each six, burning and oxygen so reacting with oxygen to produce water and carbon dioxide than to balance it. We recognize that we have two carbons here, so we&#x27;re gonna need two carbons here, so we&#x27;ll put our two in front of the carbon dioxide and we have six hydrogen. So we need six hydrogen. So will put a three there in front of the hydrogen, and then we need to balance the oxygen&#x27;s. We have seven. Oxygen&#x27;s over here, so seven have would be what we were crying or in front of the oxygen. But we don&#x27;t like to have a balanced chemical reaction with fractions, so we&#x27;ll multiply everything through by two, and this will then be are balanced chemical reaction, the products and of course, our water and carbon dioxide to calculate the mass of oxygen will start with the massive ethane that reacted, covered the massive ethane into molds of ethane, using the molar mass of ethane, then use historic geometry to go from moles of ethane, two moles of oxygen, knowing that it&#x27;s 2 to 7 when we have moles of oxygen We convert to grams of oxygen by multiplying by the molar mass of oxygen which is 32 we get 50.7 grams of oxygen. The total mass of products equals the total massive reactant. So this is the law of conservation of mass. So if we know the total mass of the, um oxygen that reacted in the total massive everything that that reacted that will give us the total mass of reactant switch will also be equal to the total massive product.

Benjamin Angeles · Answer

everybody. So this question is asking for several things in A is asking for the balanced equation for the combustion of methane and in B the balanced equation for freezing water from liquid. And both of them are also asking for the anthem Pete diagrams. So it&#x27;s starting with a will start for the combustion of methane. So ethane has the chemical formula C two H six and the comm bun shin is its reaction with oxygen. So we have nothing gas. We&#x27;re gonna reacted with 02 g and we know the products of a combustion are carbon dioxide and water. So to balance this, I&#x27;m gonna first start off with trying to match up the carbons and the high trillions according to the thing. So I think has two carbons. So I&#x27;m gonna put a coefficient of to by the carbon dioxide. And Ethane also has six hydrants. So I&#x27;m gonna put a coefficient of three to get that six by the H 20 Now that I have those coefficients, I&#x27;m going to try to match up the oxygen&#x27;s. Since I have three plus four, oxygen&#x27;s on the right so that seven total oxygen&#x27;s I&#x27;m gonna have a coefficient of seven halves by this left oxygen. And when I multiplied this seven halfs through, I&#x27;ll get 702 on the left and then seven oxygen on the right. So since I don&#x27;t want to have any fractions in my final balanced equation to get rid of that seven halves, I&#x27;m gonna multiply this whole thing by two. And when I do that, my final balanced equation is too. C two h six gas plus seven go to gas and we&#x27;re left with for co two and six water molecules. So that is our balanced equation moving on to the entropy diagram. We know that this reaction is going to be exo thermic, So that means that the reactant since which all did note by this lying here are going to have a higher energy than the products which had to know by this line here, draw your connecting line for that. And then I&#x27;m going to label the parts of this diagram. So we have the reactant right here and the products right here. And we know that this change in energy from reactions to products is our negative. Don&#x27;t h which indicates that the reaction is exo thermic, moving on to the second part, Part B. We have not a reaction, but just the physical change of water from liquid to solid. So we have Htoo liquid and we&#x27;re going to convert that toe ice. We&#x27;re gonna say that&#x27;s H 20 solid. And when we freeze things, we know that that&#x27;s going to release heat. So we&#x27;ll put heat in the products here. We&#x27;re going to do the same thing Since he is in our product side, we know that this process is exo thermic So our reactant are going to have a higher energy that our products have are indicated lines here and then our connective line here. And then I&#x27;ll label the reactant on the left and our products on the right. And we know from the last example that this right here is our negative don&#x27;t age indicating that this process is also exit thermic

Charotte Mcadam · Answer

way. Just have violence Equations for the combustion off East pain. First example as e saying, as you can see it, c two h six and then our second example is ethanol with C two h 50 H. We have alcohol functional group present and so moving on we&#x27;re looking at the end will be of combustion of each of our staff ting material So the entropy often question off. E thing is negative is negative 47.51 killer jewels background Next Moving onto ethanol or NCP off combustion is negative 26.82 Killer jewels program So ethanol is partially oxidized by And so we&#x27;re looking at how this effects are entropy off combustion. So the entropy change for the combustion of ethanol is less negative than the Bali for anything. So particularly our oxidizing e thing to form ethanol does decreased A quantity of energy program available from the outline combustion reaction

Jesse Leeder · Answer

we have to write the reaction for the combustion of benzene. So when we combust a hydrocarbon, we will react it with 02 in the gas fees and the products of this reaction for part a will be H 20 in the liquid phase and CO two in the guest thieves. Now we have to balance this reaction and start with our carbons. Um, left. We have six carbons on the right. We have one carbon completely six year to bounce a on the left. We have six hydrogen ins on the right. We have to hide regions so you can put a three there receptive balance our oxidants that we can add up All the oxygen&#x27;s on the right. We have three times one from the water, plus six times to from the carbon dioxide gives us 15. Since we already have a subscript of to for the oxygen guests, the coefficient will be 15 hands. And here&#x27;s our balanced reaction. Remember, do you find the Delta G free overall reaction? You take the sum of all of the Delta Jeez of the products times their story key metric coefficients minus these some of the delta. Jeez, of the reactions times their story. Key metric coefficients for our products. We had a six times negative. 394 0.359 It&#x27;s our Delta G for gas. Carbon dioxide three times our delta G for water in liquid phase negative. 237.129 on the React Inside we have, 02 in the gas phase. It&#x27;s Delta G will be zero because this is its standard state in our delta. G for Benzie will be 124.5. Put this into a calculator, Get the Delta G to be negative. 3000 202 killer jewels Effort Part B. We could just repeat the same calculation because it&#x27;s the same balanced equation. But instead of liquid water, we&#x27;re now using gashes water. So we have six times negative. 394.359 plus three times our new Delta G for gas water. It&#x27;s negative. 228.572 minus our benzene still 124.5. Well, give us approximately negative 3176 killer jewels

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Consider the energy diagrams below for the combus…

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