Hess's+law

Hess’s law states that the enthalpy change on converting a given set of reactants to a set of particular products is always constant, irrespective of the reaction pathway taken. For example a reaction that converts A to B. The reaction can be done directly involving a direct conversion of A to B or it can be done indirectly by converting A to C and then to B. Irrespective of what path you decide to take, the enthalpy change for the production of B from A is always the same.



In other words ΔHx = ΔHy + ΔHz Therefore the enthalpy change for the formation of a product from a given reactant is only dependent on the difference in enthalpies of the products and the reactants.

Example:

Calculate the enthalpy change (ΔH f ) for this reaction:

2C(s) + 2H 2 (g) ---> C 2 H 4 (g)

Given the following possible reactions:

C(s) + O 2 (g) ---> CO 2 (g) ΔH 1 = -395 KJ

H 2 (g) + 1/2 O 2 (g) ---> H 2 O (l) ΔH 2 = -287 KJ

C 2 H 4 (g) + 3O 2 (g) ---> 2CO 2 (g) + 2H 2 O (l) ΔH 3 = - 1416 KJ

We first have to reverse the third reaction so that C 2 H 4 is in the products side( this is because in the original equation C 2 H 4 is on the products side).

Reversing the equation reverses the sign of the of ΔH 3 (the enthalpy of the reaction)

Thus we have:

2CO 2 (g) + 2H 2 O (l) ---> C 2 H 4 (g) + 3O 2 (g) -ΔH 3 = +1416 KJ

Since the third equation involves two moles of CO 2 and H 2 O reacting we double equations 1 and 2 as well so that we have two moles of CO 2 and H 2 O in these reactions as well.

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">However keep in mind that doubling the equation also doubles ΔH.

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">2C(s) + 2O 2 (g) ---> 2CO 2 (g) 2ΔH 1 = 2x(-395) KJ

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">2H 2 (g) + O 2 (g) ---> 2H 2 O (l) 2ΔH 2 = 2x(-287) KJ

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">All of these equations can be represented on a diagram.

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">And according to Hess’s law:


 * <span style="color: red; font-family: Tahoma,sans-serif; font-size: 14pt;">ΔH f = 2ΔH 1 + 2ΔH 2 – ΔH 3 **


 * <span style="color: red; font-family: Tahoma,sans-serif; font-size: 14pt;"> = 2(-395) + 2(-287) + 1416 **


 * <span style="color: red; font-family: Tahoma,sans-serif; font-size: 14pt;"> = +52 KJ mol -1 **

<span style="font-family: Tahoma,sans-serif; font-size: 14pt;">ΔH is positive which shows that the reaction is endothermic.


 * By the end of this lesson you should be able to: **
 * ** Find the enthalpy change of a reaction which is the sum of two or three other reactions with known enthalpies **


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