Chemical reactivity - level 2 chem PDF

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Chemical reactivity -

If the forward reaction dominate, then we say that the equilibrium lies to the right and there would be a higher concentration of products than reactants in the mixture. If the reverse reaction dominates, then we say the equilibrium lies to the left and there would be a higher concentration of reactants than products.

Equilibrium constant:

aA + bB ⇌cC +dD

Lower case letters = moles Kc =

[ C]ᶜ [ D ] ᵈ [ A] ᵅ [ B ] ᵇ

upper case letters = compound

P/R

Solids and liquids are not included in the equilibrium constant equation (can’t be made more concentrated or diluted) If Kc is greater than 1, the forward reaction is favoured as we have more products present when compared with the reactants. Large/small

lots of products formed and fewer reactants

If Kc is equal to 1, the forward and reverse reaction are approximately the same so there are equal quantities of products and reactants. P=R

neither reaction formed

If Kc is less than 1, the reverse reaction is favoured as we have no more reactants than products. Small/large

lots of reactants and fewer products formed

Kc only gives information on which reaction is favoured and by how much. It provides no info on the rate of the reaction (how fast it proceeds). Le Chatelier’s principle If a closed system is opened and a change is made to the system in equilibrium, the system in equilibrium will act in such a way as to undo the change. Three ways that can change an equilibrium system: -

Concentration Temperature Pressure

Adding catalyst will effect both the forward and reverse reactions so will not change the equilibrium system Concentration: If you increase the concentration of a reaction, the system will minimise the change by promoting the reaction that uses up the extra reactant. The forward reaction will be favoured. If you increase the concentration of a product, the reverse reaction will be favoured to remove the extra products.

If you take away some of the products, the system will favour the forward reaction so as to produce more product, to replace the amount that you took away.

−¿ ( aq ) ¿ +¿ ( aq ) +Cl ¿ NaCl ( s ) ⇌ Na -

If we increase Na⁺ The opposite is decreased Na⁺ Backwards reaction Decrease in Cl⁻ Increase in NaCl

Pressure: Increasing the pressure means that you have decrease the amount of space available - it has become more crowded in order to decrease the amount of crowding the system needs to provide fewer gas molecules. Since gas molecules have the most kinetic energy, they take up the most space. The reaction that produces the least number of moles of gas molecules will be favoured.

2 SO 2 ( g ) +O 2 ⇌ 2 SO 3 (g) 3 moles

2 moles

Increase pressure: -

Decrease pressure Moving it to least number of moles Forward reaction More products formed Less reactants

Decrease pressure: -

Increase pressure Moving it to most number of moles Backward reaction Less products More reactants

Decreasing the pressure will favour the reaction that produces the most gas molecules – in order to get back to the same amount of crowding that the system had before the change E.G

PCl3 ( g )+ Cl2 (g) ⇌ PCl5 ( g)

Decrease pressure: try to increase pressure, moving to more number of moles, backward reaction, less PCl₅, more PCl₃ and Cl₂. Temperature When you increase the temperature of the system, you are giving it extra heat energy. To minimize the change, that the energy needs to be used up. An endothermic reaction will use up the extra heat energy so will be favoured.

If you decrease the temperature, the exothermic reaction will be favoured, as it will produce extra heat to replace the hate that was removed. Decrease temperature: -

Increase temperature Heating up Exothermic Forward reaction Heat is exiting (+) Bonds forming

Increase temperature: -

Decrease temperature Cooling down Endothermic Backwards reaction Heat is entering (-) Bonds breaking

Temperature will alter the Kc equilibrium constant. If Kc increases, there will be more products i.e. the forward reaction is favoured. If the Kc decreases there will be more reactants i.e. the reverse reaction is favoured. Example: Sulfur dioxide reacting with oxygen. 2SO2(g) + O2(g) ⇌ 2SO3(g) ∆H is -ve ∆H is -ve for the forward reaction. There are 3 mole of gas molecules on the left. There are 2 mole of gas molecules on the right. The equilibrium expression is: Kc = [SO3]2 [SO 2]2[O2] Temperature A decrease in temperature causes the reaction to speed up in the exothermic direction. This releases heat to oppose the temperature decrease. An increase in temperature causes the reaction to speed up in the endothermic direction. This absorbs the increased heat supplied to the system. In order to promote the formation of SO3 we must carry out the reaction at low temperatures, as the forward reaction is exothermic. The low temperature allows the heat energy from the system to be absorbed by the surroundings. Doing the reaction at high temperatures would favour the reverse process. Pressure Low pressure promotes the reaction that has the greater number of gaseous particles. High pressure promotes the reaction that has the lower number of gaseous particles. In order to promote the formation of SO3 we must carry out the reaction at high pressure. This simply reduces the amount of space available in the reaction vessel and encourages the reaction in a direction which leads to the lesser number of gas particles. Low pressures would favour the reverse.

Change in Condition

Direction of Shift in Equilibrium Position

Temperature Increase Decrease

Endothermic Direction is Favoured Exothermic Direction is Favoured

Pressure Increase Decrease

Towards the side with the least number of moles of gas Towards the side with the greater number of moles of gas

Reactant Concentration Increase Decrease

Forward Reaction (to the right) Backward Reaction (to the left)

Product Concentration Increase Decrease

Backwards Reaction Forwards Reaction

Catalyst added

No change in equilibrium position or in Kc, but the equilibrium is reached more quickly

Acids and bases Acid: Acid donates proton to bases. An acid is a substance that releases H⁺ ions in solution. Hydrogen atom consists of one proton and one electron so the H⁺ ion consists a naked proton - hence H⁺ ions are known as protons. When a proton is in water it is always associated with a water molecule and is known as a hydronium ion (H₃O⁺). Solutions containing H₃O⁺ are acidic solutions. If we think of a general acid (HA) then it will react with water as follows:

−¿ ¿ HA+ H 2 O ⇌ H 3 O + A

A = can be any substance with a proton on it

In this reaction the acid has donated a hydrogen ion to the water – it has donated a proton to water. All acids are defined as proton donors. Bases: Bases accept protons from acid (bases dissociate to give OH- ions). A base is a substance that accepts H⁺ ions/protons. If we think of a general base (A-), it will react with water as follows:

−¿ ¿ −¿+H 2 O⇌ HA+O H ¿ A We say that A- has accepted a proton, so it is acting as a base. The H₂O molecule is acting as an acid – it has donated a proton, leaving behind the OH-. All bases can be defined as proton acceptors. Amphiprotic: some of the substances can act as both proton acceptors or donors e.g. water. Strong acids and bases

A strong acid is one that dissociates completely in water – it is a strong proton donor and donates all of its protons to the water molecules.

−¿ ¿ +¿+Cl e.g. ¿ HCL+H 2 O → H 3 O the HCl completely breaks up so that all that is left in solution are the hydronium ions and the chloride ions. NOTE: we do not use the equilibrium arrow. Examples of strong acids: hydrochloric acid (HCl), nitric acid (HNO₃) and sulfuric acid (H₂SO₄). A strong base is one that ionizes in water and contains hydroxide ions. It is also a strong proton acceptor.

−¿ ¿ +¿+OH e.g. ¿ NaOH ⟶ Na Examples of strong bases: sodium hydroxide (NaOH) and potassium hydroxide (KOH), and any hydroxide. Weak acids and bases A weak acid does not dissociate significantly in water, so very few protons are released

+¿ ¿ −¿+ H3 O e.g. ¿ CH 3 COOH + H 2 O ⇌C H 3 CHOO The ethanoic acid only partially breaks up and equilibrium is established. Examples of weak acids: ethanoic acid (CH₃CHOO) and ammonium ion (NH₄⁺). A weak base is one that does not dissociate significantly in water. Examples of weak bases: ethanoate ion (CH₃COO⁻) and ammonia (NH₃)

pH and reactivity: a solution of a strong acid compared to that of a weak acid of the same concentration: the strong acid will have a lower pH (is more acidic) as a greater concentration of H₃O⁺ ions will be released when compared to a weak acid. The strong acid solution will react with metals and carbonates faster than a weak acid as the strong acid contains more ions, so there is a greater chance of collision. Conductivity For a solution to conduct electricity, it must have charged particles present – in solution, there will be ions. The greater the concentration of ions present, the greater its conductivity. Strong acids which ionise completely, will be better conductors then weak acids (of the same concentration) which only partially dissociate. Concentrated and dilute acids and bases A concentrated acid will contain more H₃O⁺ ions than the same volume of a dilute acid. Remember that the container of dilute acid contains a lot more water than that of the concentrated acid. A concentrated base will contain more OH- ions than the same volume of a dilute base – again, because there is more water in the dilute version. As a guide, a solution of acid or base with a concentration > 6mol/L⁻¹ is described as concentrated. A solution with a concentration < 6mol/L⁻¹ is described as dilute – regardless of whether it is a strong or weak acid or base: A concentrated solution will be: -

Better conductor of electricity because there are more ions present React faster with metals and other reactants because there are more collisions.

−¿ ¿ + ¿+ A ¿ HA + H 2 O ⇌ H 3 O

e.g.

the pH scale pH gives a measure of hydronium ion concentration. It is defined as the negative log of the hydronium ion concentration in a solution. Calculating pH:

H3 [ ¿O] pH =−log¿ Calculating conc.: shift log – [pH]

pOH= -log [OH-] pH + pOH = 14

Rates of reaction:  

Reaction rate can vary from a fraction of a second to hundreds of years. Reaction rate is measured by the change in concentration of a reactant or product per unit of time, e.g. second, minute, hour, day... Above Absolute Zero (-273°C), all particles move. Absolute Zero, particles have no energy so don’t move and no reactions occur.

 

In solids, particles are fixed in place so any movement is restricted to vibration only. In liquids, particles stay in contact with other particles but now have enough energy to slide past each other as well as vibrate. In gases, the particles have enough energy to break free from other particles and move with speed. Liquid and gas particles move about constantly, resulting in collisions.

 

How a reaction occurs:    

Not all collisions result in reactions. Molecules react only if they collide at the correct angle and with enough force to break any existing bonds. Increasing the frequency of collisions will increase the rate of the reaction. Molecules must have a minimum amount of energy to start a reaction. This is known as the Activation Energy (Ea).

There are four ways we can increase the number of successful collisions and hence the rate of a reaction: 1. Increasing the concentration of reactants. Increasing the number of particles, increases the likelihood of a collision. If the reactants are gases, the relative concentration can be increased by increasing the pressure. Greater chance it will be more successful  increase rate

2.

Increasing the temperature. Particle speed increases with temperature, they have more kinetic energy, resulting in more collisions, AND more particle with sufficient energy to overcome the EA 3. Increasing the surface area of the reactants if they are solids. Having more surface area of the solid exposed, increases the availability of the reactant for collisions 4. Using a catalyst. Catalysts are substances that increase the rate of a reaction by providing an alternative pathway for the reaction, which has a lower activation energy. This means that more particles will have enough energy to overcome the lowered activation energy barrier, reaction rate will be faster. 5. In other words, as less energy is required for the reaction to occur, a collision between reactants is more likely to be successful at forming products....