How do I find delta S of 2 Na(s) + 2H2O(l) -> 2NaOH(aq) +H2(g)?

Question

I tried to find it using this....
^Srxn= 2S(NaOH) - S(H2) -2S(Na) - 2S(H2O)
= 2(48.1)-130.7 - 2(51.21) - 2(69.95)
= 96.2 - 130.7 - 102.42 - 139.9
= -279.82
Where did I go wrong??

Answer 1

My tables aren't quite the same as your (but I have an old old text). But
dSrxn = (n*dSproducts) - (n*dSreactants)
dSrxn = (2*NaOH + H2) - (2*Na + 2H2O)
What stands out is the -S(H2) you have. Shouldn't that be a +. If this doesn't fix the problem please repost this (at this place is more convenient) and copy the numbers you have in your table. You're going about this the right way. It's just a matter of getting the right numbers and right signs in the right place and punching in the right numbers on the calculator.