Write an equation for the formation of each compound from
its el...

Question

Write an equation for the formation of each compound from 
its elements in their standard states, and find ΔH °rxn for each in 
Appendix IIB. 
d. CH3OH(l )

Accepted Answer

welcome back everyone in this example. We need to provide the formation equation of the nitrogen tetroxide gas from its elements in their standard state. We need to determine what the entropy of the reaction is for this example. So we have two things to solve for. This is going to be our first part of our question and this is going to be part two of our question. So for part one we want to recall what a formation equation is and this is going to be for the formation of one mole of our compound. So in this case this is our nitrogen tetroxide some it's elements that make the molecular formula up in their standard state. And so because we see that we have the elements nitrogen in our molecular formula and oxygen, these are going to be the elements that we want to write out in their standard state. So for the standard states beginning with nitrogen, we want to recall that nitrogen exists as a diatonic molecule on its own. And so what that means is we should recall the bonding preference of nitrogen which is to have three bonds and one lone pair. So we would have our two nitrogen atoms because we have a di atomic molecule which will appear as a triple bond between one another and sorry about that. So we have a triple bond between these two nitrogen atoms and they each just have one lone pair moving on to our oxygen atom. We have the standard state of oxygen which we should recall also will exist as a diatonic molecule on its own meaning we would form 02 and recalling the bonding preference of oxygen. We would recall that oxygen prefers to have two bonds and two lone pairs. So we would have two oxygen atoms bonded to one another with a double bond and they each have two lone pairs. So now that we understand how these atoms or elements exist in their standard states as di atomic molecules, we can incorporate this to write out our formation equation. And so for part one we can say that our formation equation is going to be the combination of our di atomic nitrogen gas reacting with oxygen gas which is going to form our product. Given in a prompt di nitrogen tetroxide gas. Our next step is to make sure that this is balanced. And so we would want to see that our nitrogen atoms, we have two of them on the reacting side and two on the product side. So nitrogen is balanced. But looking at oxygen we have two atoms on the reactive side and four on the product side. In order to balance out oxygen, we're going to place a coefficient of two so that we have it multiplied by the subscript of two. To give us four atoms of our oxygen gas which will be balanced with the four atoms on the right hand side. And so what this means is that with that subscript our equation is now balanced. So this would be our first final answer for part one, which is to list our formation equation for di nitrogen tetroxide. So moving on to part two where we have to calculate the entropy of our reaction, we should recognize that because our di nitrogen tetroxide is formed from our individual atoms which are nitrogen and oxygen. We can recall that because that is true. Our entropy of our reaction is going to be equal to the entropy of formation of our di nitrogen tetroxide. And so we would say that when we look up in our textbooks the entropy of formation of our product die nitrogen tetroxide. We would see that in our textbooks or online we have a value of 9.16 kg joules per mole. And because we understand that by nitrogen tetroxide is formed from Our individual atoms nitrogen and oxygen which exists as di atomic molecules. We would say that therefore our entropy of our reaction is equal to the entropy of formation. So also 9.16 kg joules per mole. And so our second and final answer is going to be this value here for the entropy change of our reaction. So what's highlighted in yellow are our final answers? I hope that everything I explained was clear. But if you have any questions leave them down below and I will see everyone in the next practice video

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