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Enthalpy Coursework Definition

Unit 2: Week 7: Session Objectives: 1 Define and briefly explain the role of entropy, and enthalpy in biochemical reactions Enthalpy (∆H)  is the change in chemical bond energy in a reaction (bond energy of products minus the bond energy of reactants). Also called HEAT  CONTENT.  Entropy (∆S)  is a measure of disorder. It is often negligible in reactions  (ATP hydrolysis) in which the number of reactants and products are equal (and  no gas is formed). 2 Give examples of biochemical, mechanical and transport work. Mechanical work ->  the high energy phosphate bond of ATP is  converted into movement by changing the conformation of a protein (ex. myosin  head group) Transport work (active transport) ->  ATP is used to transport a  compound against its [ ] gradient. (EXAMPLE: ATPASES) Biochemical Work->  formation of chemical bonds. would be unfavorable  without using ATP and the removal of products (intermediates that continue  down the pathway) and accumulation of more substrates (EXAMPLE ENGERY  OF REACTION) 3 Calculate the overall delta G of a series of reactions if given the delta G for each individual reaction Just add all of them together, summative 4 Describe the role of ATP as an energy carrier, specifically the role of the high energy phosphate bonds (Figure 19.2)-B onds between the phosphate groups are called Phosphoanhydride  bonds.-ATP unstable because the closely grouped negative phosphate groups  (repel and strain)-When those bonds are broken (Hydrolysis), energy is released  (products->ADP and phosphate are more stable)-Release energy as HEAT. BUT in the cell it is NOT hydrolyzed directly  (heat is inefficient for transfer to energy requiring processes. INSTEAD  enzymes transfer the phosphate group directly to metabolic intermediates or proteins (phosphoric transfer reaction) to drive their reactions forward. 5 Compare the redox coenzymes, FAD, NAD and NADPH as the cell's electron carriers (Figure 19.9, 19.10)-NAD ⁺  accepts two e  as a HYDRIDE ion (H¹ ) to form NADH, and a  ⁻ ⁻ proton  is released into the medium. Usually involved in oxidation of  ALCOHOLS and ALDEHYDES.-FAD  accepts two e  as hydrogen atoms (w/1 e  each), which are  ⁻ ⁻

Experiment to Compare the Enthalpy Changes of Combustion of Different Alcohols

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Experiment to Compare the Enthalpy Changes of Combustion of Different Alcohols


Introduction: This plan will try to outline how the experiment of
comparing changes of combustion of different alcohols will be
conducted and what results are expected.

Background

When chemical reactions take place they are often accompanied by
energy changes.

Chemical reactions most frequently occur in open vessels. That is,
they take place at constant pressure. Enthalpy refers to energy at
constant pressure (volume may vary).

Enthalpy:

An example is best to illustrate to show enthalpy works. Methane -
how much energy does its molecules contain? The first thing needed is
the amount of methane present = 1 mole (16 g). What ever its value,
the total amount of energy in a given amount of a substance (sometimes
called the Heat energy content) is known as the enthalpy, denoted H.

Methane is a fuel to get energy from it, react it with oxygen.

CH4(g) + 2O2(g) CO2(g) + 2H2O(l)

The above chemical equation shows that 2 moles (64 g) of oxygen
molecules are required to burn 1 mole of methane. Again, it is
impossible to know the total enthalpy (heat energy content) of the
oxygen. Likewise, we can't know the total heat energy content of 1
mole of CO2 and 2 moles of H2O (the products).

Enthalpy Change

H = (HCO2 + 2HH2O) - (HCH4 + 2HO2)

In general,

H = Hproducts - Hreactants

But remember, this is theoretical; it is not possible to determine the
absolute value of the enthalpy of a chemical element or compound.
However, H values for chemical reactions can be obtained. They can be
measured experimentally, or calculated using Hess's Law (see later),
or worked out in other ways.

Exothermic and Endothermic Reactions

When chemical reactions take place they are often accompanied by heat
changes. The system (the reactants which form products) may give out
heat to the surroundings, causing them to warm up. In this case the
reactants have more stored energy (greater total enthalpy) than the
products. Such chemical reactions are said to be exothermic. The
system may take heat from the surroundings, causing them to cool down.
In this case the reactants have less stored energy (less total
enthalpy) than the products. Such chemical reactions are said to be
endothermic.

Exothermic reactions give out energy to the surroundings.
Endothermic reactions take energy from the surroundings.

Most reactions take place at constant pressure...

It is possible to measure changes in heat energy that accompany
chemical reactions. Most reactions take place in vessels that are open
to the atmosphere, that is, they take place at constant pressure
(volume may vary). The special name given to a change in heat energy
content measured at constant pressure is enthalpy change.

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This change
in enthalpy is denoted by H. The value of H (often expressed in kJ,
or kJ mol-1 when appropriate) is given a negative sign for exothermic
reactions and a positive sign for endothermic reactions, indicating
whether the system loses or gains energy as a result of the reaction.

The value of H is given a negative sign for an exothermic reaction.
The value of H is given a positive sign for an endothermic reaction.

Precise thermochemical measurements are made in a closed vessel of
fixed volume, such as a calorimeter. For a reaction involving a change
in volume of gases there is a small but real difference in the
measured heat change. You can read more about this here.

Enthalpy Level Diagrams...

Enthalpy level diagrams can be used to illustrate overall exothermic
and endothermic changes.

For an exothermic reaction the total enthalpy of the products is less
than that of reactants. For an endothermic reaction the total enthalpy
of the reactants is less than that of the products. For each, the
difference in these total enthalpies is equal to the overall enthalpy
of the reaction, H.

Temperature and pressure matter...

As well as the amounts of substances reacting (molar amounts are
taken), the precise value of H depends on both the temperature and
pressure at which it is measured. For this reason H values are
expressed at standard conditions (normally 298 K and 1 atm.). A
standard enthalpy change is written as H°298. You can read more about
temperature and pressure here.

Thermochemical data for chemical reactions can be found in chemical
data books.

CH4(g) + 2O2(g) CO2(g) + 2 H2O(l) H°298 = - 890.3 kJ mol-1
½H2(g) + ½I2(s) HI(g) H°298 = + 26.5 kJ mol-1

Note the following changes:

CO2(g) + 2 H2O(l) CH4(g) + 2O2(g) H°298 = + 890.3 kJ
H2(g) + I2(s) 2HI(g) H°298 = + 53.0 kJ

Activation Enthalpy

Methane and oxygen do not react spontaneously when mixed. An input of
energy, such as a flame, is required to get the reaction started,
after which its exothermic nature will sustain it. With regard to the
collision theory of reaction rates, molecules react only if in a
collision they possess between them energy equal to or greater than a
certain critical value. This is called the activation enthalpy, Ea. An
enthalpy profile diagram illustrates this:

Definitions:

Standard Molar Enthalpy of Combustion, H°c,298. The heat evolved when
1 mole of an element or compound completely burns in oxygen, measured
at 298 K and 1 atm. pressure.

Standard Molar Enthalpy of Formation, H°f,298. The heat change when 1
mole of a compound forms from its elements in their standard states,
measured at 298 K and 1 atm. pressure. From the definition, the
enthalpy of formation of an element at 298 K and 1 atm. is zero.

Here is a very useful equation that follows from the definition for
enthalpy of formation. It can be used to calculate overall enthalpy
changes when the enthalpies of formation of the reactants and products
are given. It is

H = Hf (products) - Hf (reactants)

A calorimeter is used to measure heat changes directly (with
adjustments made for standard conditions at constant pressure).
However, this is not possible for some reactions. Their enthalpy
changes can be calculated with the application of Hess's Law.

Hess's Law says that the overall enthalpy change accompanying a
chemical reaction is independent of the route taken in going from
reactants to products (provided that in each case the same initial and
final states of temperature and pressure apply to the reactants and
products).

Energy can neither be created nor destroyed.

Specific heat capacity is used in some thermochemical calculations. It
is the amount of energy required to raise the mass of 1 gram of a
substance by 1 °C.

q = specific heat capacity x mass x T.

The specific heat capacity of water is 4.184 J g-1 K-1 . This means
that 4.184 J of energy are required to raise each gram of water by
each °C.

Heat capacity can also be used in calculations. It is the amount of
energy required to raise the whole mass of a body by 1 °C.

q = heat capacity x T.

Why do energy changes accompany chemical changes?

The answer involves the chemical bonds that hold atoms together in
molecules. In chemical reactions the rearrangement of atoms involves
breaking chemical bonds in reactant molecules and forming new bonds in
product molecules. The atoms are themselves not created or destroyed,
but are simply rearranged. Breaking chemical bonds requires energy;
bond breaking is an endothermic process. Conversely, when chemical
bonds form, energy is given out; bond formation is an exothermic
process.

The energy needed to break a bond and the energy given out when a bond
forms are definite and characteristic for each bond. The energy of the
product molecules may therefore be greater or smaller than the energy
of the reactant molecules. Thus a chemical reaction is accompanied by
an overall enthalpy change. This is the enthalpy of reaction, ï„H

Sources:

· Salters Horners Chemical Ideas

· http://www.innovescent.com/gallery/design/illustrations.html

Apparatus needed for experiment:

* Spirit burner
* Alcohols e.g. methanol Hexanol
* Thermometers
* 100cm3 of coldwater.
* Digital scales
* Ruler
* Copper calorimeter able to hold 100cm3 of liquid.
* Splint
* Metal stand + clamp
* Safety goggles
* Tin foil
* Measuring cylinder
* Bunsen burner.

Aim: To determine the different enthalpy changes of combustion of
different alcohols and to explain the trends and patterns.

Prediction:

As the amount of carbon atoms in the alcohol increases, the higher the
enthalpy of combustion will be. I have made this prediction, using the
values for the enthalpy change of combustion for each alcohol,
calculated using bond enthalpies and Hess' law.

Methanol's molecular formula is CH3OH. This is the basic structure for
all the alcohols, then to make the larger ones an extra carbon is
added to the existing carbon each time and the oxygen-hydrogen
molecule gets added to the atoms added to the new carbon atom

When methanol combusts in air, it reacts with oxygen molecules to from
water and carbon dioxide. The balanced equation fort this is:

CH3OH (l) + 1.5O2 (g) CO2 (g) + 2H2O (l)

This means that the bonds broken are; 3 carbon- hydrogen, 1
carbon-oxygen, 1 oxygen-hydrogen and 1.5 oxygen- oxygen (double bond)
and the bonds broken are; 2 carbon- oxygen (double bond) and 4
oxygen-hydrogen.

Constructing a Hess' law cycle will show how these are linked
together:

(l) + 1.5O2 (g) CO2 (g) + 2H2O (l)

C (g) + 4H (g) + 4O (g)

If a calculation for the amount of energy needed to break the bonds is
made and then the amount of energy given out from bond formation, the
resultant energy difference (negative because the reaction is
exothermic) is the enthalpy change of combustion.

Average bond enthalpies for elements in their gaseous states
(kJmol-1):

Carbon - Carbon (C-C) = +347
Carbon - Hydrogen (C-H) = +413
Oxygen - Hydrogen (O-H) = +464
Carbon - Oxygen (C-O) = +358
Carbon - Oxygen double bond (C=O) = +805
Oxygen - Oxygen double bond (O=O) = +498

Energy absorbed when bonds are broken (positive): (E=Energy)

= E 3(C-H) + E (C-O) + E (O-H) + E 1.5(O=O)
= 3(413) + 358 + 464 + 1.5(498)
= 2808kJmol-1

Energy given out when bonds are made (negative): (E=Energy)

= E 2(C=O) + E 4(O-H)
= 2(805) + 4(464)
= -3466kJmol-1

Enthalpy change of combustion = Energy absorbed + energy given out

= 2808 + -3466
= -658kJmol-1

The combustion of methanol gives out -658kJ of energy for every mole
of fuel burnt. This is only an approximate result and the actual value
in my experiment may be different because, firstly the bond enthalpies
are an average and they may vary in different molecules and secondly
because the value are worked out assuming that the reactants and
products are in a gaseous state, when in practise the water and
alcohol are liquids. This means that my values for enthalpy change of
combustion are likely to be different, but as long as the same method
is used for each alcohol the pattern can be seen. I will now use the
same bond enthalpy values to work out the estimation of enthalpy
change of combustion for each alcohol.

Ethanol:

C2H5OH (l) + 3O2 (g) 2CO2 (g) + 3H2O (l)

Energy absorbed when bonds are broken (positive): (E=Energy)

= E 5(C-H) + E (C-C) + E (C-O) + E (O-H) + E 3(O=O)
= 5(413) + 347 + 358 + 464 + 3(498)
= 4728kJmol-1

Energy given out when bonds are made (negative): (E=Energy)

= E 4(C=O) + E 6(O-H)
= 4(805) + 6(464)
= -6004kJmol-1

Enthalpy change of combustion = Energy absorbed + energy given out

= 4728 + -6004
= -1276kJmol-1

Propan1-ol:

C3H7OH (l) + 4.5O2 (g) 3CO2 (g) + 4H2O (l)

Energy absorbed when bonds are broken (positive): (E=Energy)

= E 7(C-H) + E 2(C-C) + E (C-O) + E (O-H) + E 4.5(O=O)
= 7(413) + 2(347) + 358 + 464 + 4.5(498)
= 6648kJmol-1

Energy given out when bonds are made (negative): (E=Energy)

= E 6(C=O) + E 8(O-H)
= 6(805) + 8(464)
= -8542kJmol-1

Enthalpy change of combustion = Energy absorbed + energy given out

= 6648 + -8542
= -1894kJmol-1

Butan1-ol:

C4H9OH (l) + 6O2 (g) 4CO2 (g) + 5H2O (l)

Energy absorbed when bonds are broken (positive): (E=Energy)

= E 9(C-H) + E 3(C-C) + E (C-O) + E (O-H) + E 6(O=O)
= 9(413) + 3(347) + 358 + 464 + 6(498)
= 8568kJmol-1

Energy given out when bonds are made (negative): (E=Energy)

= E 8(C=O) + E 10(O-H)
= 8(805) + 10(464)
= -11080kJmol-1

Enthalpy change of combustion = Energy absorbed + energy given out

= 8568 + -11080
= -2512kJmol-1

Energy given out when bonds are made (negative): (E=Energy)

= E 6(C=O) + E 8(O-H)
= 6(805) + 8(464)
= -8542kJmol-1

Enthalpy change of combustion = Energy absorbed + energy given out

= 6648 + -8542
= -1894kJmol-1

Alcohol (carbon atoms) Enthalpy change of
Combustion (kJmol-1)

Methanol (1) -658
Ethanol (2) -1276
Propan1-ol (3) -1894
Butan1-ol (4) -2512

As you can see, as the number of carbon atoms (the larger the
molecule) in the molecule increase, the enthalpy change of combustion
(energy given out) increases. This is because there are more higher
value bond enthalpies made, giving out more energy. The increase is a
constant 618kJmol-1 per carbon atom or for every 2 hydrogen atoms, so
the structure of the alcohol proportionally affects the enthalpy
change of combustion. So the larger alcohols will give out more heat
energy per mole and the temperature of the water will rise more while
less of the fuel is burnt leading to a higher enthalpy change of
combustion.

Safety-Hazard assessment:

These are the safety considerations I will take into account when
doing my experiments-

? The alcohols are flammable, so they need to be handled carefully and
never exposed to a naked flame

? The experiment produces heat, so I need to take care when handling
hot equipment i.e. wait until they've cooled down

? Alcohol vapours can catch fire at very low temperatures, so whenever
possible keep the lid on the spirit burners and re-fill them in a fume
cupboard

? The fumes have a toxic effect if inhaled-make sure the room is well
ventilated. Seek medical attention if they are inhaled

? They are irritating to skin so if split on skin or eyes, rinse with
water and seek medical attention

? If they are any spillages, clear them up.

? Wear safety goggles at all times to protect my eyes and will cover
my skin where possible

? Take care with other people's experiments as well, trying not to
expose their alcohols to a naked flame

Method:

Check that all apparatus needed is available.

1. 100cm3 of water will be measured using a measuring cylinder and
then poured into a clean copper calorimeter and then its
temperature will be recorded using a thermometer.

2. A spirit burner filled with the first alcohol to be used
methanol will be weighed, with its cap on to stop vapour
escaping from the burner, using a set of digital scales. A clamp
stand and clamp will be set up above the spirit burner.

3. To stop draughts affecting the alcohols flame when burning and
therefore producing anomalous results, a foil wind block will be
set up using tin foil wrapped into a cylinder and placed around
the spirit burner.

4. The spirit burner will set up under the clamp stand and clamp,
the copper calorimeter will be clamped 2cm above the tip of the
wick of the burner, measured using a ruler.

5. The tin foil wind break will be placed around the spirit burner
and the clamp stand placed over the burner again. A flame will be
taken with a splint from a Bunsen burner, the cap from the spirit
burner will be removed and the wick lit. A stop watch will be
started and the temperature of the water observed using a
thermometer in the water.

6. When the water has risen in temperature by 50oC, the flame will
be extinguished by blowing it out.

7. Remove foil and weigh spirit burner. Record how much the spirit
burner has decreased in mass using a table of results.

8. Repeat experiment using another spirit burner with the next
alcohol to be tested, otherwise repeating all relevant steps.
Record all results in a table.



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