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The standard heat of formation of CaBr2 is -675 k…
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Eugene S.
University of Minnesota - Twin Cities

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Problem 130 Hard Difficulty

Use average bond energies together with the standard enthalpy of formation of C(g) (718.4 kJ>mol) to estimate the standard enthalpy of formation of gaseous benzene, C6H6(g). (Remember that average bond energies apply to the gas phase only.) Compare the value you obtain using average bond energies to the actual standard enthalpy of formation of gaseous benzene, 82.9 kJ>mol. What does the difference between these two values tell you about the stability of benzene?

Answer

260.4 $\mathrm kJ/mol$ very stable

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Video Transcript

So this question wants us to use average bond energies as well as thie heat of formation off gashes carbon in order to determine our estimate. Rather, the heat a formation off gashes, benzene and then compare that to the actual value. And, ah, make a statement about the stability of benzene. And so the first thing we do is write out the reaction. So we have six see gas plus three H to gas forms. See six h six gas, which is benzene. And so the first thing we want to do is find the overall Delta H, which is the sum of Bond's broken minus thie. Some of bonds formed, and that would be the H of Bond's broken and h of bonds formed. And so that's going to be. The bonds broken were breaking three h h bronze and then Bonds formed. We're going to break three C C double bonds and three cc single bonds as well as six ch bonds, and that's because benzene looks like this. We have six carbons in a ring that is only five there. We have six carbons in a ring, and each of them has a hydrogen. Ah, bonded to it. So we have one two, three C C double bonds, one two three cc single bonds and six CH single bonds. And so plugging in values. Here we have three times for thirty six minus. Hey, three times six eleven plus three times three, forty seven plus six times for fourteen. And these are all the average bond energies of these respective bonds. And so that gives us thirteen oh eight minus fifty three, fifty eight. And therefore we have adult h of forty fifty killer Jules Permal. And so now that we have that, we can determine the Delta H of formation off benzene. By doing the equation, Delta H equals the Delta H formation off benzene see six h six gas minus the delta H of formation of see gas six times that and then minus three times the delta H of formation of H two gas, which will go to zero. And so we have Ah, negative. Forty fifty. Killer Jules Permal equals Delta H of formation of benzene. What we're looking for minus six times what we're given, which is seven eighteen point four killa Jules per mall. Ah, we're given in the problem as the Delta H of formation of sea gas. And so that gives us an estimate of the Delta H of formation off benzene to be to sixty point for killer Jules Permal. And so, ah, the actual value given is around eighty to kill a Jules, and so the estimated value is much higher. And what that tells us is that benzene is very stable because it's predicted that the formation is much higher in energy than it actually is, meaning it takes much less energy than we predict. In order to form it on, that generally means that it is very stable.

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