Use average bond enthalpies from Table 5.4 to estimate Δ𝐻 for the following gas-phase reaction of ethylene, (C2H4), oxygen, and hydrogen to form ethylene glycol (C2H6O2), which is the principal component of automotive antifreeze:

Question

Accepted Answer

Welcome back everyone. If methanol was used as a fuel, it would burn in the following manner where two moles of methanol liquid would react with three moles of oxygen gas to form two moles of carbon dioxide gas and four moles of gaseous water or water vapor. What is the entropy of the reaction using average bond energies? And below, we're given a chart between bonds that make up our entire reaction and their corresponding bond energies and units of Kjos Permal. Now, in order to determine the types of bonds that make up our chemical reaction, we're going to need to draw the structure of each of our reagents. So let's begin with our first reactant, which is our C H 30 H or in other words, our compound Methanol, we need to begin by calculating total valence electrons in which because carbon is located in group four A that will correspond to four valence electrons, then we have hydrogen, which we should recall is in group one, a corresponding to one valance electron which we want to multiply by our three atoms of hydrogen given by the subscript of three in our formula for methanol, which would give a contribution of three valence electrons then moving on to oxygen, which we should recall is in group six, a corresponding to six valence electrons. And then we have again, 1/4 atom of hydrogen. So apologies for my mistake, but we would multiply our one valence electron per hydrogen times the four atoms of hydrogen that we have total in methanol, which would actually give a contribution of four valence electrons for hydrogen. And so now we want to just add up our total of electrons and this would give us a total of 14 valance electrons total for methanol. Next, we want to draw our lewi structure for methanol in which we would have our central atom as carbon. Because recall that although we want our least electron negative atom to typically be in the center. In this case, we want hydrogen to remain as terminal atoms. So we would surround this central carbon by three of our terminal hydrogens. And then we want to include to the right of carbon, our oxygen, which is an alcohol group. So the oxygen should be bonded to a hydrogen. Now, let's make our base connections by creating bonds between each of our atoms. We would have three bonds between our carbon and hydrogen atoms. And then we would have a single bond between carbon and oxygen and then a single bond between oxygen and our fourth hydrogen. Now using up these bonds, we would count a total of 2468, 10 of our 14 valence electrons total used leaving us with a bounce of four valence electrons. And as we stated, oxygen being in group six a to be to be neutral and have a formal charge of zero so that it's stable, it should have six directly attached valence electrons in any structure. So in this case, because this oxygen currently has two covalent bonds in our structure, one to carbon and one to our hydrogen making up our alcohol group, it currently only has two directly attached valence electrons. So we're going to need to add our last four electrons has two lone pairs on our oxygen, so that it now has a total of six directly attached valence electrons. And so this would complete our structure for methanol. So we'll separate the page or the screen so that we have our second reactant that we're considering which is oxygen gas o sub two in which to calculate total valence electrons, we would take oxygen being in group six A and corresponding to six vals electrons which we want to multiply by our two atoms of oxygen, which would yield a contribution of 12 valence electrons. And that would give us our 12 valence electrons total making up our molecule of oxygen. And in this case, we have just two atoms. So we would have our two oxygens next to one another and recall the bonding preference of oxygen, which is that oxygen should be bonded with two bonds and two lone pairs on itself. So we would do that for each of our oxygens place two single covalent bonds between them or rather a double bond between our oxygens. And then we would have two sets of lone pairs on each of our two oxygens. And counting our electrons used, we would count 2468, 10, 12 of our total valence electrons used, using up our balance and leaving us with a bounce of zero. Meaning we have completed our structure for oxygen. And next, we would move to our products. So we'll make another separation on our screen where for our first product, we have carbon dioxide. And beginning with carbon, as we stated, being in group four A that yields a contribution of four bounds electrons. And then oxygen being in group six A yields a contribution of six bounce electrons multiplied by our two atoms of oxygen giving a contribution of 12 vans electrons for oxygen, taking up our total number of vals electrons. We would have 12 plus four, which would give us a total of 16 vols electrons for a structure of carbon dioxide. And so writing out our structure, we have our least electron negative atom which would be carbon in the center surrounded by our two oxygen atoms. And as we stated before, oxygen has a bonding preference of two bonds and two lone pairs and carbon, we recall has a bonding preference of having four bonds. So between our first oxygen and our central carbon, we would place a double bond and then between our right oxygen and central carbon, we would place another double bond using up a balance of eight of our 16 v electrons total. And then we would use the last eight of our advanced electrons as two sets of lone pairs on each of our oxygen atoms. And this would complete our lower structure so that we have an overall formal charge of zero and a staple structure. Now, moving on to our last product in which we have our molecule of water, we're going to again begin by calculating total valence electrons. And so for our total v electrons of water, we begin with hydrogen, which as we stated, being in group one, a yields a contribution of one valance electron. However, we have two atoms of hydrogen given by the two, the subscript of two, which would yield a contribution of two valence electrons from hydrogen. And next, we have oxygen, which as we stated is in group six A on our periodic table corresponding to six val electrons. And that would give us six plus two, which would give us a total of eight foun electrons total for our structure of water. And in our structure of water, we would have hydrogen as our terminal atom. So then oxygen is our central atom, we would make our base connections of our hydrogens to oxygen. So that oxygen will have two bonds. And then we just would place two lone pairs on our oxygen to use up our balance of eight bounce electrons. And this would leave us with a formal charge of a net formal charge of zero forest structure as our stable complete Lewis structure. Now, combining all of our written structures together, we would have the following reaction where we have our two moles of methanol given by again, our structure with carbon in the center bonded to oxygen with two lone pairs, it bond to hydrogen to from the oxygen and then carbon bonded to our three terminal hydrogens. This is our methanol reactant reacting with three moles of oxygen gas, which we have the following structure. Then on our product side, we have our two moles of carbon dioxide gas from our structure we determined above. And then our second product is our form moles of water based on our structure that we've drawn above. And now we can clearly see the types of bonds making up all of our re agents in our chemical reaction. And we can move on to calculate our entropy change delta H of our reaction in which we should recall, this is calculated by taking the bond energy of our reactants subtracted from the bond entropy or energies of our products. So in this case, we want to begin with the bond energies of our reactants. Now, looking at our first reactant methanol, we have and we'll start with brackets, we have three carbon hydrogen bonds within methanol and multiplying by our coefficient of two, that would give us six moles of our carbon hydrogen single bonds making up methanol. And also within our structure of methanol, we would add our one mole of our carbon oxygen single bond in which multiplying by our coefficient of two would actually give us two moles of our carbon oxygen single bond. Then adding to this, we have just one of our oxygen hydrogen single bonds making up methanol, which multiplied by a coefficient of two would give us two moles of our oxygen, hydrogen single bond. And then adding to this, we have our second reactant in which we have our oxygen oxygen double bond. Where based on our coefficient, we would say we have plus three moles of again, our oxygen oxygen double bond. This would close off our brackets for the bond entropy of our reactants or bond energies of our reactants subtracted from in our next bracket, our bond energies of our products. So beginning off our bond energies of our products. Our first product we stated is carbon dioxide. And we can see that we have two of our oxygen carbon double bonds which multiplied by our coefficient of two. For carbon dioxide would give us four moles of our carbon oxygen double bonds, which is then added to our second product in which we have two of our oxygen, hydrogen single bonds. Which multiplied by a coefficient of four for water in our product side would give us a total of eight moles of our oxygen, hydrogen single bonds. And this would complete our sum of bond entropy or energies of our products. So in our next line, we're going to simplify by plugging in our values given from our chart above in which for our reactant methanol, we have, as we stated, six moles of our or multiplied by our bond entropy of our carbon hydrogen single bond, which from our chart is a value of kilojoules per mole. This is then added to our two moles of our carbon oxygen single bond in methanol multiplied by its bond energy Of 143 kilojoules per mole for a carbon oxygen single bond. This is then added to our two moles of our bond energy of an oxygen hydrogen single bond given in our prompt as 97 kilojoules per mole. And then added to this, we have our three moles multiplied by our bond energy of an oxygen, oxygen double bond in oxygen gas, which according to our tart is 121 kilojoules per mole. So closing off the brackets for our sum of bond energies of our reactants, we will then subtract from the sum of bond energies of our products. So beginning off the bracket, we have four moles of our bond energy of a carbon oxygen double bond in carbon dioxide given in our chart as 120 kilojoules per mole added to our eight moles multiplied by the bond energy of an oxygen, hydrogen single bond given in our chart as kilojoules per mole in our molecule of water. And this will complete the sum of bond energies of our products. And now simplifying in our next line, we will have the results of our sum of bond energies of our reactants equal to a value of 1,515. And notice that our units of moles will cancel out, leaving us with units of kilojoules. And this would be the closing off of our brackets for the sum of bond energies of our reactants. Now subtracting from the sum of bond energy of our products, we're going to begin our bracket and get a result of 1,256 kilojoules after we cancel out our units of moles. And from this difference between our sum of bond energies of our reactants minus our sum of bond energies of our products. In our next line, we would have our answer for the entropy change of our reaction equal to our value kilojoules. Now, based on our least precise measurement being our bond energy of our oxygen, hydrogen single bond with two sig figs as 97 kilojoules per mole. We want to round our final answer to a value of or to about two sigfig which would give us 2.6 times 10 to the positive second power kills. And this would be our final answer as the entropy of our reaction corresponding to choice D in the multiple choice as our final answer. So I hope that this made sense and let us know if you have any questions.

Use average bond enthalpies from Table 5.4 to estimate Δ𝐻 for the following gas-phase reaction of ethylene, (C2H4), oxygen, and hydrogen to form ethylene glycol (C2H6O2), which is the principal component of automotive antifreeze:

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Key Concepts

Bond Enthalpy

Enthalpy Change (ΔH)

Gas-Phase Reactions