|Thermodynamics : Enthalpy|
Enthalpy (H) - The sum of the internal energy of the system plus the product of the pressure of the gas in the system and its volume:
After a series of rearrangements, and if pressure if kept constant, we can arrive at the following equation:
where H is the Hfinal minus Hinitial and q is heat
Enthalpy of Reaction (H)
- The difference between the sum of the enthalpies of the products and the sum of the enthalpies of the reactants:
In the above reaction, n and m are the coefficients of the products and the reactants in the balanced equation.
|Exothermic - Reaction in which a
heat to its surroundings.
H is negative (H < 0)
Ea is the activation energy which is discussed in
|Endothermic - Reaction in which a
heat from its surroundings.
H is positive (H > 0)
Let's distinguish various phase changes of water as either endothermic or exothermic.
1) The above reaction is EXOTHERMIC because heat is released when liquid water freezes to form ice.
2) The above reaction is ENDOTHERMIC because there must be an input of energy in order for water molecules in the liquid phase to have enough energy to escape into the gas phase.
3) The above reaction is ENDOTHERMIC because there must be an input of energy to break the bonds holding water molecules together as ice.
Standard-State Enthalpy of Reaction (H)
Three factors can affect the enthalpy of reaction:
- The concentrations of the reactants and the products
- The temperature of the system
- The partial pressures of the gases involved (if any)
The effects of changes in these factors can be shown relative to the standard-state enthalpy of reaction (H) which is the change in the enthalpy during a chemical reaction that begins and ends under standard-state conditions.
- The partial pressures of any gases involved in the reaction is 0.1 MPa.
- The concentrations of all aqueous solutions are 1 M.
Measurements are also generally taken at a temperature of 25C (298 K)
1940 - Germain Henri Hess
1) Nitrogen and oxygen gas combine to form nitrogen dioxide according to the following reaction:
Calculate the change in enthalpy for the above overall reaction, given:
This problem is very simple. If we simply add up the two reactions keeping all the reactants on the left and all the products on the right, we end up with the overall equation that we are given. Since we didn't make any changes to the individual reactions, we don't make any changes toH. If we add upH as well, we find the change in enthalpy:
Let's try one that is a bit more complicated.
2) From the following enthalpy changes:
calculate the value ofH for the reaction:
If we look at the final reaction, we see that we need 2 S atoms on the reactants side. The only reaction with S atoms is the third reaction, and in order to get 2 S atoms, we need to multiply the whole reaction by a factor of 2. The next reactant in the final reaction is 2 OF molecules. The only reaction with an OF molecule is the first reaction, and in order to get 2 OF molecules, we need to multiply the whole reaction by a factor of 2. On the products side of the final reaction, there is 1 SF4 molecule, and the only possible source of the SF4 molecule is the second reaction. However, the SF4 molecule is on the reactants side, which is not the side we need it on. So we'll have to FLIP the second reaction to get the SF4 molecule where we need it.
Now if we total up the reactions, we should end up with the given overall reaction:
Remember that everything we did to each reaction, we have to do to each respectiveH. So we have to multiply the first and thirdH values by a factor of 2. We also have to reverse the sign of the secondH. When we add these up we get:
Enthalpy of formation (Hf)
- The enthalpy associated with the reaction that forms a compound from its elements in their most thermodynamically stable states. These are measured on a relative scale where zero is the enthalpy of formation of the elements in their most thermodynamically stable states.
The standard-state enthalpy of reaction is equal to the sum of the enthalpies of formation of the products minus the sum of the enthalpies of formation of the reactants:
Sample enthalpy of formation calculation
Calculate the heat given off when one mole of B5H9 reacts with excess oxygen according to the following reaction:
In the reaction above 2 moles of B5H9 react with 12 moles of O2 to yield five moles of B2O3 and 9 moles of H2O. We find theHf by subtracting the sum of the enthalpies of the reactant from the sum of the enthalpies of the products:
- NOTE: The heat of formation of O2 is zero because this is the form of the oxygen in its most thermodynamically stable state.
- The energy required to break a bond. Bond energy is always a positive number because the breaking of a bond requires an input of energy (endothermic). When a bond is formed, the amount of energy equal to the bond energy is released.
The bonds broken are the reactant bonds. The bonds formed are the product bonds.
FindH for the following reaction given the following bond energies:
We have to figure out which bonds are broken and which bonds are formed.
2 H-H bonds are broken.
1 O=O bond is broken
2 O-H bonds are formed per water molecule, and there are 2 water molecules formed, therefore 4 O-H bonds are formed
Now we can substitute the values given into the equation:
Bond-dissociation enthalpy - The energy needed to break an X-Y bond to give X and Y atoms in the gas phase, such as in the following reaction: