Nitrogen in air reacts at high temperatures to form NO₂ according to the following reaction: N₂ + 2 O₂ → 2 NO₂
b. Estimate ∆H for this reaction (in kcal and kJ) using the bond energies from Table 7.1.

Question

Nitrogen in air reacts at high temperatures to form NO₂ according to the following reaction: N₂ + 2 O₂ → 2 NO₂

b. Estimate ∆H for this reaction (in kcal and kJ) using the bond energies from Table 7.1.

Accepted Answer

Welcome back everyone. Phosphorus trichloride is formed when di phosphorus reacts with chlorine gas. The reaction equation is diph phosphorus reacting with three moles of chlorine to form two moles of phosphorus trichloride. Calculate the entropy change delta H for the reaction in both kills. And in Klis using the bond association energies in the table below where we're given for each type of bond making up our reaction consisting of a phosphorus phosphorus phosphorus triple bond, a chlorine chlorine single bond and a phosphorus chlorine single bond the bond entropy in both units of kics per mole and kilojoules per mole. Let's begin by drawing out the structures making up our reaction. So beginning with our first reactant being di phosphorus, we need to calculate its total valence electrons. So we're calling that on our product table. Phosphorus is located down our group five. A recall that this corresponds to its number of valence electrons being five valence electrons given by the subscript of two, we would multiply by our two atoms of phosphorus which will yield a contribution of 10 valence electrons for di phosphorus. So in this case, we have our two phosphorus atoms and because phosphorus is located in the same group as nitrogen. We want to recall the bonding preference of nitrogen, which is to be bonded to or which is to have three bonds and a lone pair on itself. So, in this case, we'll test that bonding preference out on phosphorus and place a triple bond between our phosphorus atoms, which would use up a total of six of our 10 valence electrons leaving us with a balance of four in which we can place a lone pair on each of our phosphorus atoms, which would use up our last four V electrons leaving us with a balance of zero. And because we see that we have three directly attached val electrons in a covalent bond to the other phosphorus plus the two lone pair electrons on each phosphorus, we would have a total of five directly attached valence electrons on each phosphorus giving a net formal charge of zero for our structure. So we have a stable structure and we can move on to our next reactant which is our chlorine gas. So calculating total V electrons recall that chlorine on our product table is located in group seven a corresponding to seven valence electrons. We want to indicated by the subscript of two multiplied by our two atoms of chlorine, which will yield a contribution of 14 valence electrons total. So connecting our two chlorine atoms first with a single bond, we're then going to have a total of 12 v electrons remaining and sorry, that should be 14 minus two. Since we just have a single bond that we've drawn in leaving us with 12 ounce electrons in which in order to satisfy our formal charge of zero for stable chlorine atoms in the structure, we want a total of seven directly attached balance electrons. So we'll place a total of three lon pairs on each of our chlorine atoms, which would use up our balance of 12 ounce electrons leaving us with a balance of zero and a net formal charge of zero for the structure, meaning we have the correct blueish structure. So moving on to our product phosphorus trichloride, calculating total valence electrons beginning with phosphorus which as we stated before is located in group five a on the periodic table corresponding to five valence electrons. And then moving on to chlorine where we find as we stated above in group 7, 8 on the periodic table corresponding to seven valence electrons. And given by the subscript of three, we would multiply by three atoms in order to yield a contribution of 21 valence electrons by chlorine. Now taking the sum between our phosphorus and chlorine, we would have a total of 26 valence electrons total. Now because we only have one phosphorus that's going to be our central atom surrounded by three chlorine atoms in which we would make the base connections by making single bonds between our central atom phosphorus to our three chlorine, which would use up a total of six of our 26 films electrons leaving us with a balance of 20. So now we want to recall that each of our chlorine atoms currently only have one directly attached bound electron showed in a covalent bond to phosphorus. So we're going to need to place a total of three lone pairs around each chlorine atom in our structure, which would use up a balance of 18 of our 20 bounce electrons leaving us with two. And next, we would notice that our phosphorus currently only has three directly attached vals electrons shared in covalent bonds to chlorine. So to get its five vals electrons, we would use our last two electrons leaving us with a bounce of zero to place as a lone pair on phosphorus for our complete structure with a formal charge of zero. So now with each of our structures correctly drawn out for our reaction, we're going to write out our reaction in the form of structures where we have one mole of di phosphorus. Again connected by a triple bond reacting with three moles of chlorine gas connected by a single bond with again, a set of three lon pairs on each chlorine, which produces two moles of our phosphorus trichloride product where we have a lone pair on our phosphorus and three lone pairs on our chlorine atoms as we drew above. So with our structures drawn out, we're next going to recall that to calculate our entropy change of our reaction, we would take the sum of the bond entropy of our reactants subtracted from the sum of the bond entropy of our products. So plugging in based on our values above and based on our written reaction in terms of the structures, we would interpret that for our sum of bond entropy of our reactants, we have brackets where we have one mole of our phosphorus phosphorus bond, which is multiplied by our bond entropy for a phosphorus phosphorus triple bond. This is then added to our three moles of our chlorine chlorine single bond. So we would multiply by the bond entropy of a chlorine chlorine single bond. This completes the sum of bond entropy of our reactants. And now we're going to subtract from the bond entropy of our products in which we have a total of two moles of our phosphorus trichloride, where in this case, we have a total of three of our phosphorus chlorine single bonds making up our product. So because we have three moles of this phosphorus chlorine single bond, we're going to multiply by a coefficient of two, which would actually give us six moles multiplied by the bond entropy of our phosphorus chlorine single bond. And this completes the sum of bond entropy of our products. So in our next line plugging in our given values from our chart in the prompt, we have beginning our brackets for the bond entropy of our reactants, one mole multiplied by the bond entropy. For a phosphorus phosphorus triple bond being 117 kg cos per mole. Then added to this, we have our three moles multiplied by the bond entropy of a chlorine chlorine single bond given to us as 58 kg per mole. 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 the bon entropy of a phosphorus chlorine single bond given to us as 69 kill cows per mole. And this completes the sum of bond entropy of our product. So we would close off our brackets in our next line. We're going to plug in the result of our sum of bond entropy of our reactants, which in brackets is going to be 291 kg cows. After we cancel out our units of moles, then this is subtracted from the result of our sum of bon entropy of our products, which in brackets yields the result of 414 kg after we cancel out our unit of moles. So taking the difference between our two brackets, we would find that the entropy of our reaction in kills is negative 123 kg. And this would be our first answer for this example. And now we're going to need to calculate our entropy of our reaction in terms of kilojoules per mole. So just as before, we would take the sum of bond energies of our reactants subtracted from the sum of bond energies of our products. But this time in units of kilojoules per mole. So in this case, we would begin our sum of bond energies of our reactants with our bracket opening where we have one mole multiplied by the bond energy of a phosphorus phosphorus triple bond in kilojoules per mole, which is now 492 kilojoules per mole given from the prompt. This is then added to our three moles of our chlorine chlorine single bond multiplied by the bond energy of a chlorine chlorine single bond given in our prompt as 242 kilojoules per mole. And this will close off our brackets and complete the sum of bond energies of our reactants. Now subtracting from our sum of bond energies of our products, we would have minus and opening up our bracket six moles multiplied by the bond energy of a phosphorus chlorine single bond, which is given to us as 289 kilojoules per mole, which will complete the sum of bond energies of our products and we can close off our brackets. So in our next line, this will simplify so that we have the result of the sum of bond energies of our reactants equal to the value 1218 kilojoules per mole, which is then added to the result of the sum of bond energies of our products, where in the result of our bracket, we have 1,734 kilojoules. And just to actually be more accurate, we're going to need to have units of kilojoules so that we can cancel our units of moles for the entire calculation leaving us with kilojoules. And so in our next line from the difference here, we would find our entropy change of our reaction and kill jewels given as negative kilojoules, which will be our second final answer to complete this example. So our two final answers being our entropy change equal to negative 123 kg cos for the reaction and our entropy change equal to negative kilojoules for this reaction are our two final answers to complete this example. I hope this made sense and let us know if you have any questions.

Nitrogen in air reacts at high temperatures to form NO₂ according to the following reaction: N₂ + 2 O₂ → 2 NO₂
b. Estimate ∆H for this reaction (in kcal and kJ) using the bond energies from Table 7.1.

Verified video answer for a similar problem:

Key Concepts

Bond Energies

Enthalpy Change (∆H)

Stoichiometry of the Reaction

Nitrogen in air reacts at high temperatures to form NO2 according to the following reaction: N2 + 2 O2 → 2 NO2 b. Estimate ∆H for this reaction (in kcal and kJ) using the bond energies from Table 7.1.

Verified video answer for a similar problem:

Key Concepts

Bond Energies

Enthalpy Change (∆H)

Stoichiometry of the Reaction

Nitrogen in air reacts at high temperatures to form NO₂ according to the following reaction: N₂ + 2 O₂ → 2 NO₂
b. Estimate ∆H for this reaction (in kcal and kJ) using the bond energies from Table 7.1.