- Created by: LouiseG
- Created on: 01-06-16 17:25
Enthalpy and Enthalpy Change
Enthalpy is a measure of how much energy there is in a system; whether this is kinetic, potential, heat.... This cannot be measured directly. Instead, enthalpy change is measured -
"Enthalpy change is the heat energy transferred between a system and its surroundings at constant pressure during a reaction"
Note constant pressure - (A common multiple choice question) this is to ensure we are always comparing like with like.
The change can be either exothermic or endothermic -
EXOTHERMIC - Releases heat to the surroundings, so the surroundings warm up: but this leaves the system itself with less energy. So the enthalpy change is negative. This kind of reaction is observed when there is more bond making (which releases heat) than breaking.
ENDOTHERMIC - Takes in heat from the surroundings (and so cools them down) in order to provide energy for breaking bonds. By taking in energy, the products are left with more energy than the reagents - enthalpy change is positive.
Remember BENDY MEXICANS (Breaking = Endothermic Making=Exothermic (bonds))
To compare every reaction fairly, we look at enthalpy changes under standard conditions - the conditions that exist at everyday temperature and pressure, in the lab. There are:
100kPa of pressure (Note: old spec used 101kPa, (1 atmosphere) but IUPAC (and more importantly for you, Edexcel) use and now only accept 100k Pa)
298oK temperature (this is 25oC)
To indicate an enthalpy change happened under these standard conditions, the symbol (that appears like a circle crossed through) is used in notation:
Notation for Enthalpy
Enthalpy change is written as a number, either positive or negative, with the units kJmol-1 (energy per mole of the reaction as stated)
The enthalpy change referred to is denoted in this form: "delta H" meaning "change in enthalpy" and the small, subscript letter(s) referring to the type of reaction:
"f" - formation
"c" - combustion
"neut" - neutralisation
"r" - reaction
Standard Enthalpy of Combustion
"The enthalpy change measured at 100kPa and 298K, when one mole of a substance is completely burnt in oxygen"
Note that the first part of this definition (red) is applied to all "standard enthalpy of..." definitions, because these are the standard conditions. it is the last part in blue which is specific to combustion. Note that "one mole" is always somewhere in there! (Here, it refers to the fuel).
E.g.: CH4 (g) + 2O2 (g) --> CO2 (g) + 2H2O (l)
Remember, as it's under standard conditions, state symbols are important: they should be what the substance is at r.t.p. - e.g., water is a liquid, not a gas. This is one reason why standard enthalpies are rarely reproduced and recorded in the lab, because sometimes the products aren't in these states - water often boils off into steam in combustion.
Also, remember as it's complete combustion you can't form any CO, HCs or C(s). And, when balancing the equation, you can't add any coefficients to the fuel (here CH4) because 1 MOLE must be burnt!
This is the method that enthalpy of combustion can be found for in the lab.
- A spirit burner is weighed.
- It is used to heat a known value of water in a copper can, and the temperature change recorded using a thermometer, which continually stirs the water.
- The spirit burner is reweighed to see how much fuel has been burnt.
- The energy transfer is found from Q = mct , where Q=energy (J) m=Mass of water (where 1ml = 1g) c = specific heat capacity of water, 4.18 and t = change in temperature
- The enthalpy of combustion is found from -Q/n; where Q is your last value, in kJ (so divide by 1000); and n is the number of moles of fuel burnt (so it's energy per mole, or kJmol-1). Note that the value will always be negative for combustion as they're always exothermic.
- (To calculate moles: Find the mass of fuel burnt (from the difference in the masses of the burner) and divide by the molar mass (Mr )of the fuel).
Sources of error
There's always a question about why your lab value differs from the data book value! Here are some experimental sources of error:
- Heat energy is lost in heating the air and the can, not the water. Solution: insulate/use a lid and a draught excluder
- Some of the fuel doesn't burn completely and forms soot/CO instead (No solution, except ensure there's plenty of oxygen available)
- The conditions are not standard. Water vapour is often formed, not H2O(l)
- As the experiment takes a long time, the water may cool down which isn't taken into account. Solution: limit the length of the experiment/ensure insulation(A "bomb" calorimeter is the most accurate and finds a close to databook value)
Standard Enthalpy of Neutralisation
"The enthalpy change measured at 100kPa and 298K when one mole of water is produced from a reaction between an acid and an alkali"
Again, it is only the latter part that is really specific to this definition. Notice this time that the "one mole" part refers to the water formed, not the acid or alkali or salt. For example:
1/2H2SO4(aq) + NaOH(aq) --> 1/2Na2SO4(aq) + H2O(l)
Notice you can't multiply this through by two because then you'd have two, not one mole of water.
For all strong acid-base reactions, the standard enthalpy is very similar. Why? Because the same ionic reaction essentially happens, if we take out the spectator ions and look at the ionic equation:
OH-(aq) + H+(aq) ---> H2O (l)
Therefore the same numbers and types of bonds are being broken and made, so they all have a standard enthalpy value around -57.8 kJmol-1 . If you're ever doing calculations, check your answer comes out somewhere around here! (But it may often be less exothermic, (see sources of error page); and weak acids will not behave quite like this - see A2 topic 12).
Enthalpy of Neutralisation experiment
This is slightly different from combustion, because you have two starting products:
- Measure out the acid and alkali into two separate polystyrene cups, with one in slight excess so that the reaction will be complete (e.g., an alkali of slightly higher concentration so all the acid reacts).
- Measure both their starting temperatures (find a mean if they're different).
- Add them together (carefully!) and record the temperature change.
- Calculate Q in the same way as before. Remember now the mass being heated is both the acid and the alkali. Assume their specific heat capacity is the same as water's.
- For the moles part (-Q/n), remember it's one mole of water (whereas for combustion it was the fuel you were concerned with). So calculate the moles of water - which could be the same as the acid or alkali, but it depends on the reaction!
- Use moles = concentration x volume for the acid/alkali (whichever WASN'T in excess; and the moles ratio to find the moles of water).
Standard Enthalpy of Formation
"The enthalpy change measured at 298K and 100kPa when one mole of a substance is made from its elements in their standard states."
2C(s,graphite) + 3H2(g) + 1/2 O2(g) ---> C2H5OH (l)
Notice that carbon is said to be graphite in its standard state, as this is its most common elementary form. Once again, when balancing don't change anything on the right-hand side: only 1 mole of the product can form.
- Note: If another product is formed alongside, it's NOT a formation reaction. ONLY one product, made from its elements in standard states, can be made if it is said to be standard enthalpy of formation.
These reactions are difficult or impossible to measure enthalpy directly because they don't usually happen - ethanol is not made by reacting lumps of carbon with hydrogen and oxygen gas at room temperature! So, Hess' Law can be used instead....
Hess' Law and Cycles
Hess' Law basically follows the principles of conservation of energy. It states that:
"The enthalpy change of a reaction is independent of the path taken to get there, provided the conditions remain constant"
This means that, even if there is an intermediary stage before the reactants become products, the final delta H will still be the same.
This is why, in drawing a catalysed reaction profile diagram, the position (on the energy axis) of the products and reactants don't change. Even if a different reaction path is provided by the catalyst, the overall energy drop or rise between the two cannot be altered.
This idea can be used when calculating H for "impossible" reactions.
Constructing the Hess' Law Cycle
The "Cycle" is a common way to find the enthalpy of a reaction, say the formation of CO from C and O, when other data is known - like the combustion data. The data you are given, and the type of reaction it represents, is just as important as the one you are looking for as it dictates how you form your cycle...
- 1. Write your reaction you're looking for in a line.
- 2. Check what data you are given - look at the subscript! If it's COMBUSTION, write the combustion products of everything in your reaction below the line. If it's FORMATION, write out the elements involved below the line. If it's NEUTRALISATION, write out the salts and water for all the products below the line. Balance this.
- 3.Draw arrows between your reaction and the "intermediary" stage, in the direction the reaction would happen - for example, in combustion, your reactants will combust to become CO2, H2O etc. so the arrows should go from the reaction to the intermediary.
- 4.In contrast, for formation; the substances are made from their elements, so the arrows point upwards towards each molecule or compound. (NOTE: ELEMENTS have no formation data as they are already elements, so ignore them for formation).
- 5.Write on the arrows your enthalpy data, making sure to balance (e.g. if you have 2CO2 in your reaction, multiply the enthalpy of formation data by two). CONT>>>>
6. Go along your arrows, from reagents to products. If you have to move backwards down an arrow, REVERSE the enthalpy sign: so if it's -ve, add it on as a positive number, etc.
Here's all that information in a picture:
- The arrows point UP as the data is from FORMATION
- You would go AGAINST the left-hand arrow, and up the right-hand arrow to get the enthalpy change, i.e 483.6 - 393.5 = 90.1 - the 483.6 has become positive as you go "the wrong way".
- In this image, the individual data have been combined into just two arrows but remember if you have separate data that you'd need 2(enthalpy formation of H2O) and 2(Enthalpy formation of H2) as there are two moles of these.
Some more Hess' Law!
Some things to remember about Hess' Cycles:
- You're not ALWAYS given formation and combustion data. Sometimes, enthalpy of reaction is used - which can be any reaction! It is done in the same way, with whatever the products of these reactions are placed below and the arrows placed in the correct direction.
- Remember to balance before you start adding!
- Remember to change the signs if you go against an arrow. This means the direction of the arrow really is the most important thing - the intermediaries themselves are not actually that important/used!
- When adding up the enthalpies, you're going from reagents -> products so always move left to right across your page!
Above are links to some short clips that may help to explain this better step-by-step. (Copy and paste into a new page, or follow link in the comments )
"The enthalpy change when one mole of a bond in the gaseous state is broken".
Bond enthalpy is another way to calculate enthalpy change, where you look at the individual enthalpies of the bonds themselves as they're made and broken.
No more "standard conditions"! These are measured always as a gas (like ionisation energy).
Therefore when you calculate enthalpies with bond data, the values can vary from experimental or Hess' Law data - as the states are all different.You're always given MEAN data because each bond in a different environment is slightly different - the C-H bond in methane, for example, may not have the same enthalpy value as a C-H in ethanoic acid. So an average is taken out across ALL the C-H bonds.
Bond Enthalpy again
Bond enthalpy values are always POSITIVE - that means, it's the energy put IN to BREAK the bonds. However, the energy RELEASED in MAKING the bonds is the same, but negative.
So, to calculate enthalpy change, do:
Sum of bonds broken - Sum of bonds made
Therefore, you'll always be doing the sum of the reagents' bonds - the sum of the products' bonds, because it's the reagents that you're breaking up into the products, which make new bonds.
- 1. Draw a displayed formula for the reaction, so you can see all the bonds
- Calculate the total enthalpy on both sides of the reaction. So, if you have methane and oxygen as reagents, you'll want at least 4(C-H) and however much (O=O) to react with it.
- Take away the right-hand side products from the left-hand side reagents. In an EXOTHERMIC reaction, this will leave a negative number because more energy is released in making the new bonds than had to be put in.
- Do a common sense check, if possible: if your reaction is combustion, the answer should ALWAYS be negative, because they're always exothermic reactions.
Enthalpy Level diagrams
In an exothermic reaction, the products have less energy than the reactants, and vice versa for endothermic. This is shown on an energy level diagram, which tells us how the total enthalpy changes:
Notice you don't need to show the activation energy, or label the x-axis. However, for an enthalpy PROFILE diagram, this is essential. Tips:
- Make sure your arrow always points towards the products
- An enthalpy sign should be included next to the arrow, as either -ve or +ve (the enthalpy change of the reaction)
- If you do label the x-axis, label it "time" or "progress of reaction"