# Homework Help: NaOH(s) -> NaOH(g)

1. Dec 23, 2008

### ghostanime2001

NaOH(s) --> NaOH(g)

$$\Delta H^{\circ}_{diss}$$ (O-O) = 251 kJ
$$\Delta H^{\circ}_{diss}$$ (O-H) = 465 kJ
$$\Delta H^{\circ}_{diss}$$ (H-H) = 435 kJ
$$\Delta H^{\circ}_{diss}$$ (Na-O) = 255 kJ

$$\Delta H_{sol}$$ = -46 KJ for $$NaOH_{(s)}$$

$$NaOH_{(s)} \rightarrow Na^{+}_{(aq)} + OH^{-}_{(aq)} \cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot$$$$\Delta H$$ = -46 KJ

$$\Delta H^{\circ}_{f}$$ = -427 KJ for $$NaOH_{(s)}$$

$$Na_{(s)} + \frac{1}{2}O_{2(g)} + \frac{1}{2}H_{2(g)} \rightarrow NaOH_{(s)} \cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot$$$$\Delta H$$ = -427 KJ

I forgot one part:
$$Na_{(s)} \rightarrow Na_{(g)} \cdot\cdot\cdot\cdot\cdot \Delta H$$ = 109 kJ

Predict $$\Delta H$$ for $$NaOH_{(s)} \rightarrow NaOH_{(g)}$$

Because im trying to find the enthalpy of one reaction, and i have other 'known' enthalpy reactions does this problem automatically turn into a Hess' Law type problem ? or is this still a bond energy/enthalpy reaction or just bond energy type problem ? Also i do not have a $$NaOH_{(g)}$$ in any of my reaction so how can i go about computing the targeted enthalpy ?

Last edited: Dec 24, 2008
2. Dec 24, 2008

### ghostanime2001

Re: NaOH(s)-->NaOH(g)

So does anyone know how to tackle this problem ?

3. Dec 31, 2008

### ghostanime2001

Re: NaOH(s)-->NaOH(g)

Okay here is what i tried doing, I know there might be something wrong and such but please please please i beg you indicate my mistake this is my last question regarding Reaction heats. Next up Equilibrium lol .... anyways.. *ahem*....

$$Na_{(s)} \rightarrow Na_{(g)} \cdot\cdot\cdot\cdot\cdot \Delta H$$=109 KJ
This means:

$$Na_{(g)} + \frac{1}{2}O_{2(g)} + \frac{1}{2}H_{2(g)} \rightarrow NaOH_{(s)} \cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot\cdot \Delta H^{\circ}_{f}$$ = -268 KJ

My question regarding the above question is when i convert Na(s)-->Na(g) does my product NaOH(s) become ---> NaOH(g) so that my target equation ends up NaOH(g) as a product ? or does it still remain a solid and my calculation or method is wrong and should be discarded because a better method is out there....... so please help. I did another part it seems to make sense to change NaOH(s) ---> NaOH(g) because im using the heat of dissociation and do those values mean to dissociate a given compound to its gaseous atoms or what ??. I know that those values are the same for any compound so does that also mean its the same values for the same compound but in a different state like with NaOH(s) and NaOH(g) ?????

I converted NaOH(s) into NaOH(g) to show my method but still i need a better explanation so i can understand and not just from theorizing "making things happen" which i dont like to do :( I also know this is sublimation lol thats my last point. I got -268 KJ for the second equation because im using bond energies or is this still not the correct method ? please help me.

$$NaOH_{(s)} \rightarrow Na_{(s)} + \frac{1}{2}O_{2(g)} + \frac{1}{2}H_{2(g)} \Delta H = 427 KJ$$
$$Na_{(g)} + \frac{1}{2}O_{2(g)} + \frac{1}{2}H_{2(g)} \rightarrow NaOH_{(g)} \Delta H = -268 KJ$$
==========================================
$$NaOH_{(s)} \rightarrow NaOH_{(g)} \Delta H = 159 KJ$$

You can do the cancellation yourself: Na, O2, and H2 get cancelled. This answer is correct on the back of my paper but can anyone suggest a better method ? of if this is somewhat the correct method then can anyone please clean up my explanation to make things clear ? Thanks a bunch. Now i go sleep ..... ZZZzz

4. Jan 1, 2009

### ghostanime2001

Re: NaOH(s)-->NaOH(g)

CAN somebody help me with this ?

5. Jan 4, 2009

### ghostanime2001

Re: NaOH(s)-->NaOH(g)

Helooooooooooooooooooooooo ?????????????? Is there anyone in this forum ??????????