We are well aware that volume at STP (0oC and 101.3kPa) of any gas is represented by 22.4L / mol. We will utilise this in our conversations. This is not new, but don’t worry, another concept was introduced today. It was a hot topic known as enthalpy.
From previous science studies, we know that reactions that release energy and exothermic. Ones that absorb energy are endothermic.
Let's try a few examples that involve volume and stoichiometry. In the decomposition of 5.0 g of NaCl, what volume of chlorine gas is produced?
Start with the balanced equation: 2NaCl --> 2Na + Cl2
Now we simply use conversion factors.
5.0g NaCl (1 mol NaCl / 58.5 g) (1 mol Cl2 / 2 mol NaCl) (22.4L / 1 mol Cl2)
= 0.96L Cl2
Let's try another one. If 3.0L of Br2 is consumed in the reaction 2Li + Br2 -->2LiBr, how much Li was used?
Simply, start with 3.0L and use appropriate conversion factors.
3.0L Br2 ( 1 mol Br2 / 22.4 L Br2) (2 mol Li / 1 mol Br2) ( 6.9 g Li / 1 mol Li)
= 1.8 g Li
If you require further practice, please refer to our previous examples and watch the video below.
With that now aside, we can have some fun. Enthalpy is described as the heat content of a system under a given pressure. It can also be defined as the energy stored in chemical bonds. There is also change in enthalpy, which is the amount of energy or heat absorbed in a reaction. Heat, on the other hand, is defined as the thermal energy in transit.
In class, we were given the following equation:
2C8H18 + 25O2 à 16CO2 + 18H2O + 5076kJ
Enthalpy has units of joules. Since the enthalpy is on the right side of the equation, it means that this process is exothermic. Had it been on the left, it would mean that the reaction required heat and would therefore be endothermic. An exothermic reaction has a negative change in enthalpy. An endothermic reaction has a positive change in enthalpy. The pictures below will demonstrate why this is so.
Example
Determine the amount of heat released when 1 mol of octane reacts.
Begin by making a balanced equation:
2C8H18 + 25O2 à 16CO2 + 18H2O + 5076kJ
1 mol C8H18 (5076 kJ / 2 mol C8H18) = 2538 kJ
Example
Using 2NH3 + 46.2kJ à N2 + 3H2, determine the maximum number of N2 moles produced in 712kJ of energy are used in the reaction.
712kJ (1 mol N2 / 46.2kJ) =15.4 mol N2
More on Enthalpy
The production of methane is shown in the equation below:
C(s) + 2H2(g) --> CH4 + 74kJ
This equation tells is that when we produce the one mole of CH4, we get a change of 74kJ.
The change in enthalpy of a system during a process that occurs under constant pressure, represented as /\H is equal to the head given off or absorbed by the system during that process. When talking about enthalpy, it is important to specify what state we are talking about. For example, the enthalpy of liquid water at 25C is different from gaseous water at 25C.
By convention the enthalpy change for a reaction /\Hrxn is taken to be the enthalpy of the system in going from reactants to products.
Hreaction = Sum (Hproducts) – Sum (Hreactants)
If the enthalpy of the products is less than that of the reactants, /\H will be negative in sign. If the products have a larger enthalpy than the reactants, /\ is positive and this is an endothermic process.
Example
Given the equation CH4 (g) + 2O2(g) à CO2(g) + 2H2O(g) + 802kJ, determine the change in enthalpy.
Since this is an exothermic reaction, we know that the change in enthalpy will be negative. Therefore, /\H = -802kJ per mole of CH4 reacted.As a consequence of the law of conservation of energy, the amount of heat associated with a reaction is directly proportional to the amount of substances involved. Thus, combustion of 2 mol of CH4 will produces 1604 kJ.
Example
How much heat is produced when 4.50g of methane gas are burned in a constant-pressure system?
To solve this, we will need to use stoichiometry.
802 kJ are produced when a mole of CH4 is burned. A mole of CH4 has a mass of 16.0g.
Heat produced = (4.50 g CH4)(1 mol CH4 / 16.0 CH4)(802 kJ/mol CH4)
=226 kJ
It is important to specify the states of the reaction. For example, if the product of methane were liquid water instead of gaseous, /\H would be -890kJ instead of -802kJ.
Hess’ Law of Constant Head Summation
According the this law, if a reaction can be carried out in a series of steps, /\H for the reaction will be equal to the sum of the enthalpy changes for each step (the enthalpy changes are additive).
· For example, the enthalpy change for the combustion of methane to form carbon dioxide and liquid water can be calculated from /\H for the combustion of methane giving gaseous water:
o CH4 (g) + 2O2(g) à CO2 (g) + 2H2O (g) /\H = -802 kJ
o (Add) 2H2O (g) à 2H2O (l) /\H = -88 kJ
o CH4 (g) + 2O2(g)+ 2H2Oà CO2 (g) + 2H2O (l) + 2H2O (g) /\H = -890 kJ
o Net equation:
o CH4 (g) + 2O2(g) à CO2 (g) + 2H2O (l) /\H = -890 kJ
To obtain the net equation, the sum of the reactants of the two equations is placed on one side of the arrow, and the sum of the products on the other.
* Since H2O (g) occurs on both sides of the arrow, it can be cancelled like an algebraic quantity that is on both sides of an equal sign.
The first law of thermodynamics, in the form of Hess’ law, teaches us that we can never expect to obtain more or less energy from a chemical reaction by changing the method of carrying out the reaction.
Heat of Formation
The heat of formation of a compound /\Hf is the enthalpy change involved in forming a mole of that compound from its constituent elements.
A standard enthalpy change /\Hfo is one that takes place with all reactants and products in their standard states.
By convention, the standard heat of formation of the stable form of any element is zero. (This will be shown in the following example.)
Example:
Find the heat of this reaction:
4NH3 (g) + 5O2 (g) à 4NO (g) + 6H2O (l)
1) Find heat of formation of the reactants.
Gaseous ammonia has a heat of formation of -45.90 kJ / mol. (Look on Wikipedia, etc)
Since oxygen is in the form it always takes, its heat of formation is zero.
Gaseous NO is +90.29 kJ / mol.
Lastly, the heat of formation of water is -285.83 kJ / mol.
2) Since heat of formation deals with “per mole” We need to multiply each heat of formation by the number of moles.
NH3 = -45.90 x 4 NO = 90.29 x 4 H2O = -285.83 x 6
3) Solve for /\Hf.
/\Hf = 90.29 x 4 + (-285.83 x 6) – (-45.90 x 4) + 0
= 361.16 – 1714.98 + 183.6
= -1170 kJ for this reaction
This is possible because of Hess’ Law, which tells us that no matter which path we take to find the heat of formation, it will be the same.
Fun eh? Let’s do one more.
C3H8 (l) + 5O2 (g) à 3CO2 (g) + 4H2O (l)
1) Find the heat of formation of the reactants.
2) Since heat of formation deals with “per mole” We need to multiply each heat of formation by the number of moles.
3) Solve for /\Hf.
Propane = - 104.7 kJ / mol
O2 = 0 kJ / mol
Carbon dioxide = -393.509 kJ / mol
Water = -285.82 kJ / mol
Propane = - 104.7 kJ / mol x1
O2 = 0 kJ / mol
Carbon dioxide = -393.509 kJ / mol x3
Water = -285.83 kJ / mol x4
/\Hf = /\products - /\reactants
= (-1180 + -1143) – (-104.7)
= -2218 kJ
Therefore, for every mole of propane that is combusted, 2218kJ will be produced.
C3H8 (l) + 5O2 (g) à 3CO2 (g) + 4H2O (l) + 2218kJ
Posted by Andrew.
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