# Homework Help: Calculating NaOH dissociation entalphy problem

1. Aug 28, 2014

### tester.dla.mmc

NaOH dissociation is exothermic so ΔH<0

But when I have a reaction
NaOH(s) => Na+(aq) + OH-(aq)

and count ΔrH using standard entalphy change of formation:
ΔfH (NaOH) = -425 kj/mol
ΔfH (OH-(aq)) = -230 kj/mol
ΔfH (Na+(aq)) = +242 kj/mol
Data from Atkins phisical chemistry basics

ΔrH= ΔfH (products) - ΔfH (substrates) = (-230 + 240,12) - (-425) = +435,12 kj/mol
From this calculation it seems that reaction is pretty endothermic and that's surely not true.

Can someone tell me where am I making mistake ?

2. Aug 29, 2014

### Staff: Mentor

And not -242 kJ/mol?

Generally speaking hydration of small ions is quite exothermic, so this positive value looks suspicious.

BTW: you count fingers, but you calculate amount. Number of fingers - liczba palców, amount of water - ilość wody, palce są przeliczalne, ilość jest nieprzeliczalna.

3. Aug 29, 2014

### tester.dla.mmc

Thanks for so fast response.

ΔfH (Na+(aq)) value in my book was given without sign (probably author just forgot about it) and I assumed that it must be positive since losing electron is usually positive (didn't take into consideration that is't (aq) ion) but now everything makes much more sense.

Thank you