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Problem 86

(a) The nitrogen atoms in an $\mathrm{N}_{2}$ molecule are held together by a triple bond; use enthalpies of formation in Appendix $\mathrm{C}$ to estimate the enthalpy of this bond, $D(\mathrm{N}=\mathrm{N}) .$ (b) Consider the reaction between hydrazine and hydrogen to produce ammonia, $\mathrm{N}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3 (g) .$ Use enthalpies of formation and bond enthalpies to estimate the enthalpy of the nitrogen- nitrogen bond in $\mathrm{N}_{2} \mathrm{H}_{4}$ . (c) Based on your answers to parts (a) and (b), would you predict that the nitrogen-nitrogen bond in hydrazine is weaker than, similar to, or stronger than the bond in $\mathrm{N}_{2}$ ?

Answer

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## Discussion

## Video Transcript

All right, guys. Regen problem number eighty six of chapter five in chemistry of Central Science. So the nation Adams and and two are held together by trip on use mantelpiece information appendix seeds as make the entropy of this pond. So we're going to do that. So we so remember the Delta Age information fella man. Terrence state like and to or h dude, that's going to be zero. So So dealt it. So remember, has law. Delta attraction is equal downstage of products myself. Day two reactions. The Delta H is going to be equal to zero minus two times for seventy two point seven. Still, Delta H is equal to mine minus nine. Forty five point four. Kill it, Jules. For me now for part B, they said concerned the reaction between hydrant hydrant produce ammonia and to a tour plus h to yield to N h. Story. He's m puppies, information and bond until peace to estimate the entropy of nature nation bond in tend to each board. All right, so for consistency sake, we're gonna flip this equation Asian so that we could show the information of hydrazine with an end with an N N bond. So first we're gonna find the Delta H information. I remember Delta H of age to that zero because elements Sanders State, they're gonna play within toward Calculator one eighty seven point seven eight. Killer Jules for a moment now. Now we're going to estimate the strength of that and and bun. So we're going to set the Delta H. We have here as thie as as Sandy equal to to have his law instead of using dealt information, we're going to use depth Delta Age of Age. Rigoni's the bond and will be because the the difference in the bonds and Toby from the process and reactive that's that's roughly equivalent to the Delta Agent formation Asian. For the rib action, that means going to set one eighty seven point seven eight. Kate. What are products? So we have for any three three we have. Let's just put it as X for R n an interaction. And then we have four and aged interactions two, three, ninety one, and then we're gonna put it for aged, too. So for thirty six, that's the age age bond and then minus for our product's going to be, too. The Times three and H Bonds three ninety one to them. We're gonna add it all together. Four times three ninety one was scored thirty six minus two times. Three times three. Ninety one? Yes. And if we put that into our calculators, we are going to get That's it. We're going to get one. Eighty seven point seven A is equal to thanks minus three. Forty six. Now we're gonna add that together. So X is gonna equal. Look, it's gonna equal as five. Thirty three point seven eight. So now, as a kill Jules Permal? Yes. Now we can. Now we can approximate it like this. This so we're gonna have this kind of reaction. Two. And guess feels and and And gas. So we're assuming it's an end bond. So we have have our bond and Toby five. Thirty three point seven, eh? Minus two times for seventy two point seven. So that's going to equal for eleven point six to kill a jewels from all? Yeah, as so we can see that this delta H is is more positive than this Delta H value. So we so based on that, we can see that as we can see that this is a form a trip on is and more spontaneous reaction and more favor reaction than forming just your double bun. So that so that means that our double So the answer party we see if we look att, the answers from A and B, we can see that the formation of a triple bond is more stable, the information of ADA, but because it has a a much lo, a much more negative and of change and entropy, which is a more favourable reaction.

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