# The Naked Scientists Forum

#### Garabato

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« on: 27/06/2007 23:28:59 »
If Im not wrong, the reaction between H2 and I2 to form HI is exothermic and reversible.

Wich reaction have the greater activation energy, foward or reverse?

Wich reaction will be increased by more raising the temperature?

Thanks

PS: Im new here, do you guys only answer theoretical questions? Or can I ask for engineering problems solving as well? (Like book´s hard excersices and that kind off stuffs).

#### Cut Chemist

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« Reply #1 on: 29/06/2007 06:05:36 »
As you said the reaction is reversible

Forward:
H2 + I2 --> 2HI

or

Reverse:
2HI  -->  H2 + I2

Understanding activation energy is easier if you look at a reaction coordinate diagram.
Here's one shown below.

The forward reaction is exothermic.  It gives off heat, therefore it must go from high energy to low energy.  This corresponds to going from reactants to products (left to right) in the reaction coordinate diagram shown above.  In the forward reaction the activation energy is labeled Ea

In reverse you use the same diagram, but this time you'll be going from products to reactants (right to left.)  Since the forward reaction is exothermic, the reverse reaction must be endothermic.  It is, because going from products to reactants means going from low energy to high energy.  The activation energy for the reverse reation would be equal to (Ea + ΔGo).  So the activation energy for the reverse reaction is much higher.

Raising the temperature = raising the kinetic energy
The effect of temperature on the rate of a reaction is determined by the arrhenius equation.

http://en.wikipedia.org/wiki/Arrhenius_equation

Since the forward reaction is exothermic and has a lower activation energy its rate will be less effected by raising the temperature. Once the reaction is already going its rate will be fast and difficult to stop, and raising the temperature won't make it go much faster.

Since the reverse reaction is endothermic, the rate of the reverse reaction at low temperatures will be very slow, because it needs heat.  Increasing the temperature will have a significant effect on its rate.
« Last Edit: 29/06/2007 06:13:25 by Cut Chemist »

#### Garabato

• Jr. Member
• Posts: 17
« Reply #2 on: 01/07/2007 07:51:10 »
Thank you very much, but this leads me to another question.

In an equilibrium reaction (a reversible one), is there always going to be an endothermic and an exothermic reaction?...why?...and is there any way to predict (without direct experimenting) wich one is the exothermic one?

In case that there is no way to tell without experimenting...what kind of experiment can be done in order to find out wich is wich ?

Thank you.

#### Cut Chemist

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« Reply #3 on: 02/07/2007 05:52:21 »
If a reaction is reversible, then yes one direction will be endothermic and the other direction will be exothermic.  It has to do with the total energy of the products and reactants.  The energy of the products will never be exactly the same as the energy of the reactants, so there will always be either a net gain or a net loss of energy.

This requires some understanding of thermodynamics.

To determine which direction is exothermic and which is endothermic, you need to calculate ΔH for the reaction, or ΔHorxn

if ΔHorxn is positive then the reation is endothermic
if ΔHorxn is negative then the reaction is exothermic

To calculate ΔHorxn all you have to do is look up the heat of formation (ΔHof) for each product and each reactant.  These values are posted in the CRC for almost every compound that exists.  Here's a link to that book online...

http://www.hbcpnetbase.com/

Once you have those values

1.  Multiply the ΔHof of each product by its stoicheometric coefficient
3.  Multiply the ΔHof of each reactant by its stoicheometric coefficient
5.  Subtract the reactants from the products

Here is the full equation as you would see in a text book...

ΔHorxn = Σ n ΔHof[products] - Σ n ΔHof[reactants]
« Last Edit: 02/07/2007 05:54:19 by Cut Chemist »

#### Bored chemist

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« Reply #4 on: 02/07/2007 20:52:16 »
It's also possible to calculate the energy change from the equilibrium constant.

#### Cut Chemist

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« Reply #5 on: 03/07/2007 05:21:17 »
If you have the equillibrium constant, then you can calculate the change in Gibbs free energy (ΔG) not change in enthaply (ΔH.)

You would then need the change in entropy and the temperature to calculate ΔH

-The Gibbs energy determines if the reaction is spontaneous or not
-The Enthalpy determines if the reation is endothermic or exothermic

#### lightarrow

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« Reply #6 on: 03/07/2007 23:01:33 »
If Im not wrong, the reaction between H2 and I2 to form HI is exothermic and reversible.

From these data, the reaction should be endothermic:
http://webbook.nist.gov/cgi/cbook.cgi?Formula=HI&NoIon=on&Units=SI&cTG=on

ΔH°f(HI) = +26.4KJ/mol

So, for what Cut Chemist has already written, the direct reaction should have the greater activation energy.
« Last Edit: 03/07/2007 23:09:20 by lightarrow »

#### Cut Chemist

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« Reply #7 on: 04/07/2007 18:55:44 »
It depends...

I2(s), ΔHof = 0 kJ
I2(g), ΔHof = (+)62.44 kJ

H2(g), ΔHof = 0 kJ
HI2(g), ΔHof = (+)26.48 kJ

(These values came from an Atkins Physical Chemistry Text)

H2(g)  +  I2(s)  -->  2HI(g)

ΔHorxn = 2(26.48 kJ) = (+)52.96 kJ

H2(g)  +  I2(g)  -->  2HI(g)

ΔHorxn = 52.96 kJ - 62.44 kJ = (-)9.48 kJ

I think this reaction is normally carried out with both species as gases.  This is from wikipedia...

Quote
Additionally HI can be prepared by simply combining H2 and I2. This method is usually employed to generate high purity samples.

H2 + I2 → 2 HI
For many years, this reaction was considered to involve a simple bimolecular reaction between molecules of H2 and I2. However, when a mixture of the gases is irradiated with the wavelength of light equal to the dissociation energy of I2, about 578 nm, the rate increases significantly. This supports a mechanism whereby I2 first dissociates into 2 iodine atoms, which each attach themselves to a side of an H2 molecule and break the H -- H bond:[3]

H2 + I2 + 578 nm radiation → H2 + 2 I → I - - - H - - - H - - - I → 2 HI
« Last Edit: 04/07/2007 19:04:48 by Cut Chemist »

#### Bored chemist

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« Reply #8 on: 04/07/2007 20:03:24 »
"If you have the equillibrium constant, then you can calculate the change in Gibbs free energy (ΔG) not change in enthaply (ΔH.)

You would then need the change in entropy and the temperature to calculate ΔH

-The Gibbs energy determines if the reaction is spontaneous or not
-The Enthalpy determines if the reation is endothermic or exothermic   "

Oops, sorry, I forgot, you need the equilibrium constant at 2 temperatures to get both G and H changes. Of course, photochemistry gives rise to a different set of kinetics too.

#### lightarrow

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« Reply #9 on: 04/07/2007 20:53:30 »
It depends...

I2(s), ΔHof = 0 kJ
I2(g), ΔHof = (+)62.44 kJ

Right.

#### The Naked Scientists Forum

« Reply #9 on: 04/07/2007 20:53:30 »