Learn what catalysts are and how they affect the activation energy and reaction rate of a chemical reaction.

Catalysts and Catalysis

A catalyst is a chemical substance that affects the rate of a chemical reaction by altering the activation energy required for the reaction to proceed. This is called catalysis. A catalyst is not consumed by the reaction and it may participate in multiple reactions at a time. The only difference between a catalyzed reaction and an uncatalyzed reaction is that the activation energy is different. There is no effect on the energy of the reactants or the products. The ΔH for the reactions is the same.

Positive and Negative Catalysts

Usually when someone refers to a catalyst, they mean a positive catalyst, which is a catalyst which speeds up the rate of a chemical reaction by lowering its activation energy. There are also negative catalysts or inhibitors, which slow the rate of a chemical reaction or make it less likely to occur.

Promoters and Catalytic Poisons

A promoter is a substance that increases the activity of catalyst. A catalytic poison is a substance that inactivates a catalyst.

How Catalysts Work

Catalysts permit an alternate mechanism for the reactants to become products, with a lower activation energy and different transition state. A catalyst may allow a reaction to proceed at a lower temperature or increase the reaction rate or selectivity. Catalysts often react with reactants to form intermediates that eventually yield the same reaction products and regenerate the catalyst. Note that the catalyst may be consumed during one of the intermediate steps, but it will be created again before the reaction is completed.

This reaction is a summary of an ecologically friendly process for halogenating alkynes. The reaction works with both hydrobromic and hydrochloric acid, and produces water as its only waste product. It also gives a good yield of the halogenated product.
Instead of undergoing anti-Markovnakov addition of HBr, the alkynes are halogenated. This is due to the way the reactants are mixed. Mixing a hydrohalogenic acid with a solution of t-butylhydroperoxide (TBHP) and hydrogen peroxide will oxidize the halogens, causing them to become positively charged. The charged halogens will then attack the alkynes, and a halogenation reaction will occur.

Salicylic acid reacts better with acetic anhydride than acetic acid, so acetic acid will provide the acetyl group which will react with the alcoholic -OH group on the salicylic acid. (The reaction is on the top of the post.)

Chemicals needed for the reaction: Salicylic acid, Acetic anhydride, and Concentrated sulfuric acid.

Equipment: 250 mL Erlenmeyer flask, Hot water bath, Ice bath, Buchner funnel and filter paper, Glass stirring rod, and Electronic pan balance and weighing boat.

butane: C₄H₁₀ (CH₃CH₂CH₂CH₃)
butene: C₄H₈ (CH₂=CHCH₂CH₃)

If the double bond is not terminal (if it is on a carbon somewhere in the center of the chain) then the carbons should be numbered in such a way as to give the first of the two double-bonded carbons the lowest possible number, and that number should precede the “ene” suffix with a dash, as shown below.

correct: pent-2-ene (CH₃CH=CHCH₂CH₃)
incorrect: pent-3-ene (CH₃CH₂CH=CHCH₃)
The second one is incorrect because flipping the formula horizontally results in a lower number for the alkene.

If there is more than one double bond in an alkene, all of the bonds should be numbered in the name of the molecule – even terminal double bonds. The numbers should go from lowest to highest, and be separated from one another by a comma. The IUPAC numerical prefixes are used to indicate the number of double bonds.

octa-2,4-diene: CH₃CH=CHCH=CHCH₂CH₂CH₃
deca-1,5-diene: CH₂=CHCH₂CH₂CH=CHCH₂CH₂CH₂CH₃

Note that the numbering of “2-4″ above yields a molecule with two double bonds separated by just one single bond. Double bonds in such a condition are called “conjugated”, and they represent an enhanced stability of conformation, so they are energetically favored as reactants in many situations and combinations.

6.00kJ + H2O(s) –> H2O(l)
delta H = +6.00kJ

! In some textbooks delta H is written as a product or reactant !

The preceding is based upon the Law of Conservation of Energy (James Joule, 1818-1889, Joule also developed the First Law of Thermodynamics): energy is neither created nor destroyed in ordinary chemical or physical changes.
2. Quantitative delta H
delta H = qreaction mixture (at constant temperature only)

q = (m)( delta t)(Cp)

q = heat absorbed by the water in joules (J)
m = mass of substance
delta t = tfinal – tinitial
Cp = specific heat of water = 4.184 J/g oC

When using moles, molar heat capacity is used. The units are kJ/mol K

1 cal = 4.184 J

B) 2. Calorimetry

1. Coffee-cup calorimeter (only used for reactions in solution, must be at constant pressure)
qreaction=-qwater
2. Bomb calorimeter (reaction gases, and must have constant volume)
qreaction=-(qwater+qbomb)
qbomb=C delta t, where C is the calorimeter constant (Cv of bomb x mass of bomb, really same equation)
3. delta H vs. delta E for chemical reactions
delta H=qp since delta E=qp-P delta V
substituting gives delta H= delta E+P delta V
where P will usually be in atmospheric pressure, and delta V is volume change at that pressure.

C) 3. Laws of Thermochemistry
1. The magnitude of delta H is directly proportional to the amount of reactant or product.
-Thus delta H can be used as a conversion factor in a balanced equation to obtain amounts of reactant/product or delta H itself. (mole to mole ratio’s).
2. delta H for a reaction is equal in magnitude but opposite in sign to delta H for the reverse reaction.

Problems 1: delta H Calculation

1. When 1 mol of methane is burned at constant pressure, 890.3kJ of energy is released as heat. Calculate delta H for a process in which a 5.8 gram sample of methane is burned at constant pressure.
CH4(g) + 2O2(g) –> CO2(g) + 2H2O(l) + 890.3 kJ
= 320 kJ
delta H = -320 kJ

2. For the reaction of methane with oxygen given in the notes, calculate the delta H in kJ if 5.8 grams of oxygen are consumed in the process.
= 81 kJ
delta H= -81kJ

3. Ammonium nitrate, NH4NO3, is commonly used as an explosive. It decomposes by the following reaction:

NH4NO3 –> N2O(g) + 2 H2O(g) + 37.0kJ

delta H = -37.0 kJ

If 72.0 grams of H2O are formed from the reaction, how much heat was released?
= 73.9 kJ
3. Hess’ Law: The value of delta H for a reaction is the same whether it occurs directly or in a series of steps (state function).

delta Htotal = delta H1 + delta H2

often used to calculate delta H for one step, knowing delta H for all steps and for the overall reaction.
**All of the laws of thermochemistry follow from the fact that the enthalpy H of a substance is one of its properties.**

D) 4. Heats of Formation
Molar heat of formation ( delta Hf) is equal to the enthalpy change, delta H when one mole of the compound is formed from the elements in their stable forms at 25oC and 1 atm is delta Ho (pronounced ‘delta h naught’). delta Ho of a solution is of a 1M solution, at 1 atm and 25 oC.

Heats of formation are usually negative quantities.

delta H = sigma delta Hf products – sigma delta Hf reactants To apply the above relation, use the following rules:

1. The contribution for each compound is found by multiplying the heat of formation in kJ per mole by the number of moles of compound, given by its coefficient in the balanced equation.

Heats of formation can be found in appendix A-4

2. Any element in its stable form is omitted.
Can also apply to heats of formation to ions.

Arbitrarily assign H+ ion to be zero. delta Hf H+(aq) = 0

Having established the above, a scale can be established with Hydrogen ion as the base.
1. Calculate delta H0rxn for the following reaction.

2 C3H6(g) + 9 O2(g) –> 6 CO2(g) + 6 H2O(l)
*Appendix 4*
delta Hrxn = sigma delta Hf products – sigma delta Hf reactants
delta Hrxn = [ 6 H2O(l) + 6 CO2(g) ] – [ 2 C3H6(g) + 9 O2(g) ]
delta Hrxn = [ 6(-286 kJ) + 6 (-393.2 kJ) ] – [ 2(20.9 kJ) + 9(0)]
delta Hrxn = [ -1716 kJ + -2361 kJ ] – [41.8 kJ ]
delta Hrxn = -4118.8 kJ / 2 mol = 2059 kJ / mol

2.
1. Calculate delta H0 for 2Al(s) + Cr2O3(s) –> Al2O3(s) + 2Cr(s).
2. Compare this reaction to sample exercise 6.10, “thermite” reaction.

Which reaction yields more energy per gram of metal formed?

1. delta Hrxn = [ Al2O3(s) + 2 Cr(s) ] – [ 2 Al(s) + Cr2O3(s) ]
delta Hrxn = [ (-1676 kJ) + 2(0 kJ) ] – [ 2(0 kJ) + -1128 kJ ]
delta Hrxn = [ -1676 kJ ] – [ -1128 kJ ]
delta Hrxn = = -10.1 kJ / g Al
2. delta Hrxn = = -15.75 kJ / g Al
“Thermite” reaction releases more energy per gram of metal formed.