(a) The nitrogen atoms in an N2 molecule are held together by a triple bond; use enthalpies of formation in Appendix C to estimate the enthalpy of this bond, D(N‚N). (b) Consider the reaction between hydrazine and hydrogen to produce ammonia, N2H41g2 + H21g2¡2 NH31g2. Use enthalpies of formation and bond enthalpies to estimate the enthalpy of the nitrogen– nitrogen bond in N2H4. (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 N2 ?

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welcome back everyone in this example we have a diatonic bromine gas molecule where the bromine atoms are bonded by double bond and to estimate the entropy of this bond using entropy of formation. So we should first recall that entropy is represented by the term H. Where we would have delta H. As our change in entropy. Whereas entropy of formation is going to be describing describes our reaction entropy for the formation of our compound of di atomic bombing from its elements in their most standard states. So we want to write out the reaction to show the formation of bromine gas. Di atomic roman gas which will break down into two moles of our mon atomic roaming gas. And so at this point we want to refer to our textbooks or online to get our values for the entropy of formation Of our mon atomic bombing gas and the entropy of formation of our di atomic bombing gas. So we would see that for our di atomic bombing gas we have an entropy of formation value of 30.9 with units of Kayla jewels Parimal. Whereas for our mon atomic burning gas we have an entropy of formation Of 111.9 kg jewels per mole. And so to calculate our change in entropy to form the bond between our bromine atoms, we would say that we take the some of the entropy of formation of our products subtracted from the sum of the entropy of formation of our reactant And so plugging in what we know from our reaction above. We would say that we have two moles of our bromine atom multiplied by its entropy of formation being 111.9 kg joules per mole. And so I'll say two br gas and then this is then subtracted from our Ethiopia formation of our diatonic burning gas which we stated above. We just have one mole of where we have a entropy of formation of 30.9 kg joules per mole. So just to be extra clear, let's expand this and say we have one mole of br two multiplied by its entropy of formation. And here we would say two moles of romaine mon atomic gas multiplied by its entropy of formation, meaning that when we cancel or actually we don't have to cancel out units here because entropy should be in units of joules per mole. So we would just type in our difference between our products and our calculators. And this is going to yield a change in entropy of 192.9 kg jewels per mole. And we want to round this to our smallest number of sig figs from our entropy of formation of di atomic bombing which is three sig figs. So 23 sig figs we would say this is about 193 kg joules per mole. As our change in entropy And this would be our final answer as the entropy change of the bond between our atoms of bromine for our di atomic bromine molecule using entropy of formation, I hope everything I review was clear. If you have any questions, leave them down below and I'll see everyone in the next practice video.

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Key Concepts

Enthalpy of Formation

Bond Enthalpy

Triple Bond Characteristics