Bond Enthalpies Question

UPDATE: NVM I figured it out.

How can I calculate the enthalpy change of formation of CH3Br (g)?

C(s) → C(g) ∆H = 715 kJ mol-1
Br2(l) → Br2(g) ∆H = 15 kJ mol-1

Bond enthalpies:
H-H : 436
Br-Br: 193
C-H: 412
C-Br: 276

(edited 5 years ago)

Reply 1

Kian Stevens

16

The enthalpy change of formation requires one mole of product to be formed from its constituent elements, which are all in their standard states.
In this reaction, this can be shown using the balanced equation: C_(s) +

\frac{3}{2}

H₂_(g) +

\frac{1}{2}

Br₂_(l)

\rightarrow

CH₃Br_(g)

You want to start off by drawing a Hess cycle. Here's my extremely poorly drawn one, done using Paint...:

'H1', 'H2' and 'H3' correspond to

\Delta H_{1}

,

\Delta H_{2}

and

\Delta H_{3}

respectively.
The important thing to note is that you'll be starting from the bottom rectangle, as you're going from a bunch of elements in their standard states to form the starting reactants, and these reactants then form bonds which results in the formation of bromomethane.
This process' enthalpy change is

\Delta H_{3}

, and is calculated using:

\Delta H_{3} = \Delta H_{1} + \Delta H_{2}

.

You're given some bond enthalpies, so you can work out some values for

\Delta H_{1}

and

\Delta H_{2}

:

•

$\Delta H_{1}$ is the sum of each average bond enthalpy which is present. However, with the way average bond enthalpies are determined, the result will be negative. There are three C-H bonds and one C-Br bond, hence this is equal to: (412 x 3) + 276 = -1512 kJ mol^-1.

•

$\Delta H_{2}$ is the sum of each enthalpy change from elements to starting materials. The trick here is that hydrogen is also being converted into something too, and this will have its own enthalpy change which isn't given, unlike the ones for carbon and bromine. This enthalpy change is just a sum of each elementary enthalpy change, based on the stoichiometry in the balanced equation right at the start of this reply. Hence, this will be equal to: 715 + ( $\frac{3}{2}$ 436) + ( $\frac{1}{2}$ 15) = 1376.5 kJ mol^-1.

Now we have everything to work out the enthalpy change of formation for bromomethane.
As said before, this is calculated using:

\Delta H_{3} = \Delta H_{1} + \Delta H_{2}

. So, it's just a case of plugging some numbers in.
Hence,

\Delta H_{3}

= -1512 + 1376.5 = -135.5 kJ mol^-1.

(edited 5 years ago)

Reply 2

Pixelsgamer

1

Original post by Kian Stevens

The enthalpy change of formation requires one mole of product to be formed from its constituent elements, which are all in their standard states.
In this reaction, this can be shown using the balanced equation: C_(s) +

\frac{3}{2}

H₂_(g) +

\frac{1}{2}

Br₂_(l)

\rightarrow

CH₃Br_(g)

You want to start off by drawing a Hess cycle. Here's my extremely poorly drawn one, done using Paint...:

'H1', 'H2' and 'H3' correspond to

\Delta H_{1}

,

\Delta H_{2}

and

\Delta H_{3}

respectively.
The important thing to note is that you'll be starting from the bottom rectangle, as you're going from a bunch of elements in their standard states to form the starting reactants, and these reactants then form bonds which results in the formation of bromomethane.
This process' enthalpy change is

\Delta H_{3}

, and is calculated using:

\Delta H_{3} = \Delta H_{1} + \Delta H_{2}

.

You're given some bond enthalpies, so you can work out some values for

\Delta H_{1}

and

\Delta H_{2}

:

•

$\Delta H_{1}$ is the sum of each average bond enthalpy which is present. However, with the way average bond enthalpies are determined, the result will be negative. There are three C-H bonds and one C-Br bond, hence this is equal to: (412 x 3) + 276 = -1512 kJ mol^-1.

•

$\Delta H_{2}$ is the sum of each enthalpy change from elements to starting materials. The trick here is that hydrogen is also being converted into something too, and this will have its own enthalpy change which isn't given, unlike the ones for carbon and bromine. This enthalpy change is just a sum of each elementary enthalpy change, based on the stoichiometry in the balanced equation right at the start of this reply. Hence, this will be equal to: 715 + ( $\frac{3}{2}$ 436) + ( $\frac{1}{2}$ 15) = 1376.5 kJ mol^-1.

Now we have everything to work out the enthalpy change of formation for bromomethane.
As said before, this is calculated using:

\Delta H_{3} = \Delta H_{1} + \Delta H_{2}

. So, it's just a case of plugging some numbers in.
Hence,

\Delta H_{3}

= -1512 + 1376.5 = -135.5 kJ mol^-1.

I belive your answer is incomplete. You have not taken the energy needed to break the Br-Br bond once it is in gaseous form. The answer should, therefore, be (1/2 193) - 135.5 = -39 kj mol-1

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