Acetylene (H–C≡C–H) is the fuel used in welding torches.
b. Estimate ∆H for this reaction (in kJ/mol) using the bond energies listed in Table 7.1.

Question

Acetylene (H–C≡C–H) is the fuel used in welding torches.

b. Estimate ∆H for this reaction (in kJ/mol) using the bond energies listed in Table 7.1.

Accepted Answer

Welcome back, everyone. Propine is being explored as a rocket fuel for spacecrafts determine the entropy change for the combustion of propine in kilos per mole. Using the bond association energies in a table below, we're given our various bonds that make up the reaction and their associated average bond entropy and units of kilojoules per mole. We're going to need to recall that for the combustion of propine, we would take our propine represented by the chemical formula C three H four, which reacts with oxygen and recall that our products for combustion are always going to be carbon dioxide and water. So this is our combustion reaction and we need to make sure that this is balanced. Notice that we have a total of three moles of carbon on the reactant side and just one mole of carbon on the product side. So we're going to place a coefficient of three in front of our product carbon dioxide, which will now change our number of oxygens on the product side to three times our subscript of two and CO2, which would give us six and then we would have seven from the oxygen and water. So in order to balance this out, we're going to make a total of eight oxygens on the product side by placing a coefficient of four on the product side in front of oxygen, which will give us eight oxygens on the reactant side. And this will then be balanced by a coefficient of two in front of our product water, which will give us eight oxygens on both sides of our reaction. And now we have a balanced combustion reaction. So now we're going to need to write out the structures of each of our reactants and products. Beginning with our reactants, we have C3H4 for propine in which we want to recall that our structure for propine consists of a total of four carbon hydrogen single bonds, which we can clearly see in our given formula of or sorry bond line structure of propine where we have C H three and then C H giving us four total C H single bonds. We also can observe that we have one carbon carbon triple bond in our bond line structure as well as one carbon carbon single bond in our bond line structure of propine. Next, for our reactant oxygen, we would recall that we have two oxygen atoms which are singly bonded to one another or rather bonded with a double bond or a pi bond between them. Now, from our balanced equation, we have a coefficient of four in front of oxygen. So we're going to multiply by four, which will yield a contribution of four moles. Whereas based on our coefficient of one in front of our propine, we have just one mole of our carbon carbon triple bond and one mole of our carbon carbon single bond. But we have four moles of our carbon hydrogen single bond which make up our structure in propine. Now, moving forward to our products beginning with carbon dioxide, recalled that our structure of carbon dioxide appears as follows where we have oxygen doubly bonded to our central carbon, which is also doubly bonded to our second oxygen in the structure. On the other side. And within the structure, we can observe that we have two moles of our carbon oxygen double bond, which based on our coefficient and our balanced equation in front of CO2, we would multiply by three to give us a contribution of six moles. Next, our 2nd product is water. So recall that in our structure of water, we have each hydrogen sin bonded to a central oxygen atom. So we have a total of two moles of our oxygen hydrogen single bond, which based on our coefficient from our bounced equation should be multiplied by a coefficient of two which would yield a contribution of four moles. Now, let's recall that to calculate entropy change for a reaction represented by delta H, we would take the sum of the bond entropy of our reactants subtracted from the sum of the bond entropy of our products. So beginning with the sum of bond entropy of our reactants, we would open up brackets. And our first reactant being propine, we would state that we have four moles multiplied by the bond energy of a carbon hydrogen single bond, which is then added to one mole multiplied by the bond entropy of a carbon carbon triple bond. Then added to this, we have one mole multiplied by the bond entropy of a carbon carbon single bond. Then for our second reactant, we're going to add four moles multiplied by the bond entropy of an oxygen, oxygen double bond. This will complete the sum of bond entropy of our reactants. And now we're going to subtract from the sum of bond entropy of our products. We're opening up our brackets. We would have two moles multiplied by or sorry, we stated three moles. So three moles. No, sorry. That's a mistake again. So looking at our CO2, which is our first product, we had two moles of our carbon oxygen double bond, but we had to multiply by our coefficient of three, which gave us six moles multiplied by our bond entropy of an oxygen carbon double bond, which is then added to the contribution of four moles multiplied by our bond entropy of an oxygen hydrogen single bond. And this will close off our brackets and complete the sum of bond entropy of our products. Now plugging in our values from our chart above beginning with the sum of bond entropy of our reactants. We open up our brackets and we have four moles multiplied by the bond entropy of a carbon hydrogen single bond. Given us 413 kilojoules per mole, which is then added to one mole multiplied by the bond entropy of a carbon carbon triple bond. Given in our chart as 839 kilojoules per mole, which is then added to one mole multiplied by the bond entropy of a carbon carbon single bond. Given us 347 kilojoules per mole. Then added to this, we have four moles multiplied by the bond entropy of an oxygen, oxygen double bond given us 498 kilojoules per mole and closing off our brackets. This completes the sum of bond entropy of our reactants. Now subtracting from the sum of bond entropy of our products. We have six moles multiplied by our bond entropy of an oxygen carbon double bond. Given in our chart as 799 kilojoules per mole. Then added to this, we have our four moles multiplied by the bond entropy of a oxygen hydrogen single bond given in our chart as 467 kilojoules per mole. So this completes our brackets for the sum of bond entropy of our products. And now in our next line, we would simplify so that the result of the bon entropy of our reactants is equal to the value 4830 kilojoules because we will be able to cancel our unit of moles. And then for the result of our sum of bond entropy of our products, we would have the result being kill the jewels after we cancel our unit of mose. So taking the difference between these two terms, we'd find our entropy change of our reaction equal to negative 1832 kilojoules. Now this will be per mole of our reaction. So although we canceled our units of moles, we will include it as kilojoules per mole of our reaction. And because this calculation included numerous steps of multiplication and addition, as well as subtraction, we're just going to assume that based on each of our entropy values or bond energy values, sorry being about three significant figures that our final answer should be rounded to about three significant figures. And so to write this by three significant figures, we can use scientific notation so that we move our decimal from the end three places which would give us 1.8, 3 times 10 to the negative or sorry to the positive. Since we moved to the left third power kill a jewels per mole of reaction. And this would be our final answer to complete this example as our entropy change of our reaction for the combustion of propine using the bond association energies given. So our final answer is going to be again, negative 1.83 times 10th of the third power Kala jewels per mole of reaction. I hope this made sense and let us know if you have any questions.

Acetylene (H–C≡C–H) is the fuel used in welding torches.
b. Estimate ∆H for this reaction (in kJ/mol) using the bond energies listed in Table 7.1.

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Key Concepts

Bond Energies

Enthalpy Change (∆H)

Welding and Acetylene