2.6 chem

catalyst diagram

analysing reactions

1. The line is steep, indicating a fast rate of reaction. This is because at the start of a reaction, the concentration of reactant particles (example + example) are relatively high, resulting in an increased rate of effective collisions.

2. As the reaction continues, the amount of reactants decrease because some products are forming, so the concentration of reactant particles decreases. This results in a decreased rate of effective collisions, slowing down the rate of reaction and the line becomes less steep.

3. Eventually, the concentration of reactants would fall to zero as all reactants would be converted into the product. The line is flat, indicating the reaction is complete and the rate of reaction stops.

mass decreasing stuff maybe??

• state that the gas has been released

• the logic is that concentration decreases, mass decreases

Temperature affecting the rate of reaction

As the temperature increases, the kinetic energy of reactant particles also increases. This means that reactant particles move with greater speed, resulting in more frequent collisions between reactant particles, increasing the rate of successful collisions with sufficient energy to overcome the activation energy per unit time. Therefore, the rate of reaction will increase.

Concentration affecting rate of reaction

As the concentration of reactants increase, there will be a greater number of reactant particles per unit volume. This means that there are more frequent collisions between reactant particles, which increases the rate of successful collisions with sufficient energy to overcome the activation energy per unit time. Therefore, there will be a higher rate of reaction.

Surface area affecting the rate of reaction

An increase in surface area of the reactant will increase the rate of reaction. This is because more reactant particles are exposed and available to collide with particles in the other reactant per unit volume. This means that there are more frequent collisions between reactant particles, which increases the rate of successful collisions with sufficient energy to overcome the activation energy per unit time. Therefore, there will be a higher rate of reaction.

catalyst affecting rate of reaction

Catalysts speed up the rate of reaction by lowering the activation energy required by reactant particles for a chemical reaction to take place. This means that there will be more frequent collisions between reactant particles, which increases the number of successful collisions with sufficient energy to overcome the activation energy per unit time. Therefore, there will be a higher rate of reaction.

Pressure (gases)

More particles in the same volume = higher pressure = more collisions

When comparing the rate of reaction of two different compounds

• you must state what happens to the other type (even if you’ve mentioned it partially while explaining the first type)

• If asked about the total mass loss and they have the same concentration of reactants used:

  • Both reactions will eventually produce the same reduction in mass as the same mass of CO₂ is formed, because the same amounts of each reactant are used. 

Kc value

Kc > 1 = products are favoured at equilibrium (forwards direction)

Kc < 1 = reactants are favoured at equilibrium (reverse direction)

Kc = 1 neither is favoured

given Kc value, but have to calculate and explain why not at equilibrium

[calculations]

The Kc value is not at equilibrium as the value is higher/lower than it should be at equilibrium

As Kc is the ratio of products / reactants, to decrease/increase the Kc value, the concentration of reactants/products need to increase. To increase/decrease reactants/products, the reverse/forwards reaction would need to be favoured until the concentration of reactant/product falls/increases, and the system achieves equilibrium.

Concentration affecting equilibrium

As the concentration of reactants increases, the system will try to minimise this change and shift in the forwards direction. This means that at equilibrium, there will be products than reactants. This means that there will be a higher ratio of products / reactants, resulting in an increased Kc value.

Temperature rules for system

1. Increase in temp = endo. decrease in temp = exo

2. Increase in kc = products favoured, decrease in kc = reactants favoured

3. Forward reaction rules:

   1. kc increase + temp increase = forwards reaction is endo

   2. Kc increase + temp decrease = forwards reaction is exo

4. Reverse reaction rules:

   1. kc decrease + temp decrease = reverse reaction is exo

   2. kc decrease + temp increase = reverse reaction is endo

5. Positive △H = endo. Negative -△H = exo

Temperature affecting system (and you know Kc value increases)

When the temperature increases, the system will shift in the endothermic direction to minimise this change. Since the Kc value increases, the forward reaction is favoured, therefore there will be more products than reactants at equilibrium. As the kc value increases while the temp increases, the forward reaction is endothermic.

Temperature affecting system (and you know bond enthalpy)

Because the bond enthalpy for this reaction is __ mol^-1, the forward reaction is exothermic/endothermic and the reverse reaction is endothermic/endothermic. When the temperature increases/decreases, the system will shift in the endothermic/exothermic direction to absorb the added heat/release heat. This means that there will be more reactants than products (vice versa) at equilibrium. Because the ratio of products : reactants decreases/increases, the value of Kc decreases/increases as the temperature increases/decreases.

Catalyst affecting equilibrium

Adding a catalyst does not shift the reaction. Since there is no overall change to the relative amount of product or reactant present, there is no change in the equilibrium position and no change in Kc value. Catalysts speed up the rate of forward and reverse reactions equally.

Pressure affecting the rate of reaction

There are __ moles on the reactant side of the equation, and __ moles on the product side of the equation, therefore the __ side of the equation has more pressure. When the pressure is increased, the system will shift in the forwards/reverse direction to minimise this change. This means that at equilibrium, there are more products/reactants than products/reactants. This results in a higher/lower ratio of products/reactants, leading to an increased/decreased Kc value. The colour associated with the product/reactant will be more intense and the colour associated with the reactant/product will fade.

If same moles:

say everything up to when the pressure part

Since there is no difference in moles of gas particles on either side of the equation, increasing/decreasing the pressure would not affect equilibrium system and will remain unchanged.

Removing something as it forms

Removing the __ as it forms will result in the system responding to minimise the effect of this change. This means that it will favour the forwards/reverse direction, resulting in an increase in the amount of __ produced.

Strong acids

• completely dissociates in water

• Examples: HCl, HNO₃, H₂SO₄, HBr, HI

• Forward sign, h2o reactant

Conductivity of strong acids

Conductivity depends on the concentration of free-moving charged particles present in a solution. Strong acids completely dissociate, therefore there will be a higher concentration of free-moving charged particles, meaning strong acids are good conductors.

Weak acids

• partially dissociates in water

• Examples: HCN, HF, CH₃COOH (basically all COOH), NH₄⁺, H₂SO₃, H₃PO₄, HNO₂

• reversible sign

Weak acid conductivity

say conductivity part

Weak acids partially dissociate, therefore there will be a lower concentration of free-moving charged particles, meaning weak acids are poor conductors.

Strong bases

• completely dissociate

• Examples: NaOH, LiOH, Mg(OH)₂, Sr(OH)₂, Cs(OH)₂, Rb(OH)₂, KOH, Ba(OH)₂

• forward sign (h2o above)

Strong base conductivity

• same as strong acid

Weak base

• partially dissociate in water

• Examples: NH₃,CH₃NH₂, NH₄OH

• reversible sign

Weak base conductivity

• same as weak acid

Conjugate base/acid pairs

polyprotic

Example for polyprotic:

1. H₂SO₄ + H₂O → HSO₄^- + H₃O⁺  (reacts once, take the species to react again)

2. HSO₄ + H₂O ⇌  SO₄^2- + H₃O⁺ (reacts twice)

3. Overall: H₂SO₄ + 2H₂O → SO₄^- + 2H₃O⁺

• Basically just keep going until you get rid of H

comparing strong and weak acid/base

• you must state the dissociation and give equation

identifying nature of salts

1. dissociate salt in water (just split it up)

2. check active species

   1. Spectator species = group 1 and 2 cations, group 17 anions

3. react active species with water

4. check if it releases OH or H₃O⁺

Reasoning:

[active species] when reacted with water releases OH or H₃O⁺, therefore [salt] is acidic/basic in nature. talk about conjugate acid base pairs with active species if you want

kc value over 1

The value of K is significantly over 1, so there are far more products than reactants, which means the equilibrium favours the products.