Enthalpy?

[zebra]

Standard enthalpy of formation:

 

Species Enthalpy (kJ/Mol)

 

Br(g) 111.9

Br2 (l) 0

Br2 (g) 30.9

 

What is the standard enthalpy of vaporization of bromine?

What is the energy neede for the reaction Br2 (g) ---> 2Br (g), i.e. the Br -Br bond energy?

[jake.com]

Standard enthalpy of formation:

 

Species Enthalpy (kJ/Mol)

 

Br(g) 111.9

Br2 (l) 0

Br2 (g) 30.9

 

What is the standard enthalpy of vaporization of bromine?

What is the energy neede for the reaction Br2 (g) ---> 2Br (g), i.e. the Br -Br bond energy?

 

29.96 kJ mol-1 of Br2

Br-Br 228

[hermanntrude]

29.96 kJ mol-1 of Br2

Br-Br 228

 

Jake, please read our policy on homework questions and in particular note that we DON'T give out the actual answer but help our members to learn it for themselves.

 

I always tell my students they are welcome to copy each other, but that they should copy the method not the answer. that way they only have to copy once.

[jake.com]

Jake, please read our policy on homework questions and in particular note that we DON'T give out the actual answer but help our members to learn it for themselves.

 

I always tell my students they are welcome to copy each other, but that they should copy the method not the answer. that way they only have to copy once.

 

sorry, my bad.

[hermanntrude]

you didn't even tell the original poster HOW you did it.

 

zebra:

 

This question is based on Hess's law:

 

[math]\Delta[/math]H° = [math]\sum[/math]([math][math]\Delta[/math]H[math]^{o}_{f}[/math] [math]_{products}[/math]) - \Delta[/math]H[math]^{o}_{f}[/math] [math]_{reactants}[/math])

 

What this means is that you must take the heat of formation (for each product

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Merged post follows:

Consecutive posts merged

you didn't even tell the original poster HOW you did it.

 

zebra:

 

This question is based on Hess's law:

 

[math]\Delta[/math]H° = [math]\sum[/math]([math]\Delta[/math]H[math]^{o}_{f}[/math] [math]_{products}[/math]) - ([math]\Delta[/math]H[math]^{o}_{f}[/math] [math]_{reactants}[/math])

 

What this means is that you must take the heat of formation ([math]\Delta[/math]H[math]^{o}_{f}[/math] for each product, multiply each one by the stoichiometric coefficient for the product in the equation (the stoichiometric coefficient is the big number before any compound in an equation), add them all up and subtract the same thing for the products.

 

In your first question, for the vaporization of bromine, the reaction is as follows:

 

[ce]Br2(l) -> Br2(g)[/ce]

 

now you can see that the enthalpy change for this reaction is simply the heat of formation for gaseous bromine subtract the heat of formation for liquid bromine.

 

for the second question, you need to take the heat of formation for 2 moles of bromine atoms (2 x 111.9) and subtract the heat of formation for one mole of gaseous Br2

 

Since you've been given the answers for this question, I must reccomend that you try this out on some more complicated questions. Your textbook will have a few like this, with answers as well so you can check. But just reading this post will NOT help you learn this technique.