Question

Calculate Delta n (change in moles of gaseous substances) for each of the following balanced chemical equations...

Answer 1

\[2C _{6}H _{6}O(l) + 17O _{2} --> 12 CO _{2}(g) + 12 H _{2}O(l)\]
\[2SO _{2}(g) + O _{2}(g) --> 2SO _{3}(g)\]
\[N _{2}(g) + O _{2}(g) --> 2NO(g) \]
\[2Na(s) +Br _{2}(l) --> 2NaBr(s) \]

On the 1st 3 I have
12 -17 = -5
2 - 3 = -1
2 - 2 = 0
but on the last one there is not gaseous substance?? .. So now I'm confused.

Answer 2

Delta n=0 then

Answer 3

Ok. That's what I was thinking, but it seemed too easy!! Lol... So then if delta n = 0 it is neither endo or exothermic right?

Answer 4

delta n is nothing to do with exo or endo

Answer 5

dont confuse with deltan n and delta H

Answer 6

I know that .. the next question asked for if it's endo or exothermic .. & my TA told me that if the gaseous substance on the product side is less than the reactant side then it is exothermic and vice versa...............

Answer 7

I know what delta H is .. but this is how my TA told me to answer the endo/exothermic ?'s? ...Which I also thought was strange.

Answer 8

me also strange to wt ur madam told
it may be true but not universal

Answer 9

Well, he gives 100% credit as long as we try. LOL so I guess it doesn't matter., but I like to understand it

Answer 10

humnn me dont much of it LOL
ask ur TA only

Answer 11

He's pretty unhelpful!! But maybe my actual teacher can explain

Answer 12

Thanks anyways

Answer 13

ur welcome

Answer 14

cntrywtr, go look up the relationship between Delta H and n. Its got other terms.

Answer 15

I already know the difference I don't need to look it up. Thanks.............