Physics EK1001 #374

[rjns11]

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The heat of formation of water vapor is:

Answer according to the back of book: negative, but less negative than the heat of formation of liquid water.

Obviously, I don't correctly understand heat of formation. I thought the vaporization of water was endothermic, giving it a positive delta(H). Does the heat of formation require H2(g) and O2(g) to dissociate, and then H2O(g) to form?

Thanks for any help you can provide.

 

[Phantastic]

The vaporization of water: H2O(l) --> H2O(g) is endothermic, because we need to give the molecules energy to be liberated into the gas.

deltaH(f) = H(products) - H(reactants)

2H2(g) + O2(g) --> 2H2O(g) is the formation reaction of the gas, I believe. All reactions require some energy (activation) to get going, but this one will produce energy in the form of heat (deltaH < 0) which came from the energy differential of the products and reactants.

I'm not sure exactly why the energy differential is larger for the formation of the liquid water.

 

[rjns11]

The vaporization of water: H2O(l) --> H2O(g) is endothermic, because we need to give the molecules energy to be liberated into the gas.

deltaH(f) = H(products) - H(reactants)

2H2(g) + O2(g) --> 2H2O(g) is the formation reaction of the gas, I believe. All reactions require some energy (activation) to get going, but this one will produce energy in the form of heat (deltaH < 0) which came from the energy differential of the products and reactants.

I'm not sure exactly why the energy differential is larger for the formation of the liquid water.

Thank you.

If you want to show the heat of formation of ice, would it look like this:

2H2(g) + O2(g) ---> 2H2O(g) [ (-)delta(H) ]
2H2O(g) ---> 2H2O(l) ---> 2H2O(s) [ (+)delta(H) ]

Overall, the heat of formation of ice is endothermic with a positive delta(H).

Is that all correct?

Thanks again.

 

[Phantastic]

The heat of formation of ice would be ice forming from solid hydrogen, and solid oxygen, since the heat of formation is the product in its standard state and the reactants in their standard states--not from the heat of formation of water vapor, and then condensation followed by fusion.

Condensation and fusion (gas-->liquid-->solid) should produce energy, as our products are more stable than our reactants (lower energy). That means negative deltaH when going from gas-->solid.

 

[rjns11]

Let me revise my last post. I think the heat of formation of water is less negative than water vapor because H2(g) and O2(g) would initially form water, and then you have to apply heat to vaporize that water.

2H2(g) + O2(g) ---> 2H2O(l) + heat [ (-)delta(H) ]
2H20(l) + heat ---> 2H2O (g) [ (+)delta(H) ]
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overall reaction = (-)delta(H), but less negative than H2O(l) formation


2H2(g) + O2(g) ---> 2H2O(l) + heat [ (-)delta(H) ]
2H2O(l) ---> 2H2O(s) + heat [ (-)delta(H) ]
------------------------------------------
overall reaction = (-)delta(H) which is less than that of water or water vapor's.

Again, please tell me if I'm correct and I'll move on!

 

[rjns11]

The heat of formation of ice would be ice forming from solid hydrogen, and solid oxygen, since the heat of formation is the product in its standard state and the reactants in their standard states--not from the heat of formation of water vapor, and then condensation followed by fusion.

Condensation and fusion (gas-->liquid-->solid) should produce energy, as our products are more stable than our reactants (lower energy). That means negative deltaH when going from gas-->solid.

I didn't see this before posting my last response. Aren't hydrogen and oxygen both gases in their standard states?

 

[Phantastic]

Ah, yes. That extra energy could be from going to gas-->liquid. That would explain the more negative delta H for the liquid.