Enthelpy of NaOH(s) -> NaOH(g)

[ghostanime2001]

Enthalpy of NaOH(s) --> NaOH(g)

Homework Statement

Given:
1. [itex]Na_{(s)} \rightarrow Na_{(g)}[/itex] [itex]\Delta H = 109 kJ[/itex]

2. [itex]\Delta H^{\circ}_{diss} = 251\:kJ\:for\:O_{2}[/itex]

3. [itex]\Delta H^{\circ}_{diss} = 435\:kJ\:for\_{2}[/itex]

4. [itex]\Delta H^{\circ}_{diss} = 465\:kJ\:for\:O-H[/itex]

5. [itex]\Delta H^{\circ}_{diss} = 255\:kJ\:for\:Na-O[/itex]

6. [itex]\Delta H_{soln} = -46\:kJ\:for\:NaOH_{(s)}[/itex]

7. [itex]\Delta H^{\circ}_{f} = -427\:kJ\:for\:NaOH_{(s)}[/itex]

Predict [itex]\Delta H[/itex] for [itex]NaOH_{(s)} \rightarrow NaOH_{(g)}[/itex] ?

Homework Equations

see above

The Attempt at a Solution

I understand the standard states of all substances for #1-3 but I am unsure of #4,5,7. I don't understand what the standard states are for #4,5 (both reactants and products). I don't understand what the physical state of sodium is for #7 (is it solid or gas ?). Also how will the aqueous ions produced from dissociation of sodium hydroxide in #6 cancel out ? and how will covalent bonds of [itex]Na-O[/itex] or [itex]O-H[/itex] as reactants cancel out ?

Thanks! :)

 

[Borek]

You can assume bond dissociation enthalpy doesn't depend on the state. In #7 you start with elements in the standard state and end with NaOH in the standard state.

 

[ghostanime2001]

How will I cancel aqueous hydroxide ion or sodium cation ? So the standard state of elemental sodium in #7 is solid ? Also, after O--H bond or Na--O bond dissociates do the Na or O or H assume a gaseous state ?

 

[ghostanime2001]

Any new thoughts from my previous reply ?

 

[DeeNos]

I would like to clarify a few things before providing a response:

1. The standard state for a substance is the most stable form of that substance at 1 atmosphere pressure and a specified temperature, usually 25°C. For example, the standard state for oxygen is O2 gas, for hydrogen is H2 gas, and for sodium is solid Na.

2. The enthalpy of formation, denoted by ΔHf°, is the enthalpy change when one mole of a substance is formed from its elements in their standard states. For example, the ΔHf° for NaOH(s) is -427 kJ/mol, which means that 427 kJ of energy is released when one mole of solid NaOH is formed from solid sodium and gaseous oxygen and hydrogen.

3. The enthalpy of dissolution, denoted by ΔHdiss°, is the enthalpy change when one mole of a substance dissolves in water. For example, the ΔHdiss° for NaOH(s) is -46 kJ/mol, which means that 46 kJ of energy is released when one mole of solid NaOH dissolves in water.

Now, to answer the question, we need to consider the following reactions:

1. Na(s) + 1/2O2(g) + 1/2H2(g) -> NaOH(s) ΔHf° = -427 kJ/mol
2. NaOH(s) -> Na+(aq) + OH-(aq) ΔHdiss° = -46 kJ/mol
3. Na+(aq) + OH-(aq) -> NaOH(aq) ΔHsoln = -46 kJ/mol

To calculate the enthalpy of the reaction NaOH(s) -> NaOH(g), we need to use Hess's Law, which states that the enthalpy change of a reaction is the sum of the enthalpy changes of the individual steps that make up the reaction.

Step 1: Na(s) + 1/2O2(g) + 1/2H2(g) -> NaOH(g)
This step requires breaking the covalent bonds in O2 and H2 and forming ionic bonds in NaOH. The ΔH for this step can be calculated as follows:

ΔH1 = 109 kJ/mol (from given information) + 1/2(251 k