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Enthalpy Changes - Bond Enthalpies

Hey Guys,

I need some clarification of a few questions, I don't know where to go with it :frown:

Q)
Calculate the

\Delta H_f

for bromomethane,

CH_3Br

, given the following data:

C_(s)-----> C_(g)

\Delta H = 715 Kj/mol

Br_2(l)----> Br_{2(g)}

\Delta H = 15 kj/mol

Br-Br = 193
H-H = 436
C-H = 412
C-Br = 276

A)
The equation for formation of bromomethane will look like this:

2C_(g)

3H_{2(g)}

Br_{2(g)}

----->

2CH_3Br

The bond enthalpy for Hydrogen will be (436 x 3)= 1308 (it is given)
C will be (2 x 715) = 1430 seeing as we are converting 2 moles of carbon into gas from solid.
Now for

Br_2

I am having trouble, it is in liquid so we need it in gas, and that will need +15 and in the table, it tells us that the Br-Br bond enthalpy is 193.
So which one is it that I should use? Do I add both of them together?
If so, that will give me 208 kj as the bond enthalpy..

After I know that, I can work out

\Delta H_f

by doing
(total bond enthalpy of reactants) - (total bond enthalpy of product)

I just need someone to clear my doubt about the

Br_2

, what value do I use for it?

Also................(another doubt)
2) Does

\Delta H_c

O_2

= 0

3)
Use the following data to calculate the average C-S bond energy on

CS_2(l)

S(s)-------->S(g)

\Delta H = 223

C(s)-------->C(g)

\Delta H = 715

\Delta H_f

CS_2(l)

= 88

CS_2(l)

---------->

Unparseable latex formula:

CS_2_(g)

= 27

I am totally lost on this one...could someone please explain this step by step :s-smilie:

Reply 1

Zygroth

I'm not really understanding Q1 :s-smilie:

Surely you break 1.5 H₂, 0.5 Br₂ and make 3x C-H bonds along with a C-Br bond? I'd ignore the gas thing, seems irrelevant to me.

Similar problem with Q3; the enthalpy of formation of CS₂ is 88 which is the formation of 2 C-S bonds with no bonds breaking so...

Reply 2

raceforthefishman

for the first question I'm fairly sure you add the values together for your bond enthalpies and your turning into gas values *theres a name for that im sure....*

Reply 3

Wenzel

1) Like Zygroth said, I'm pretty sure you ignore the whole solid/gas/liquid thing, can't see how it's relevant. :s-smilie:

To make CH₃Br you're breaking 1.5x H-H bonds and 0.5 Br-Br bonds and forming 3x C-H and 1x C-Br, so:

Energy of bonds broken - Energy of bonds formed =
(1.5*436+0.5*193)-(412*3+276) = -761.5kJ/mol

2) Yes

3) Pretty sure it's just 88/2 = 44.