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 and a greater proportion of collisions having sufficient energy to overcome the activation energy. This increases the number of successful collisions per unit time. Therefore, the rate of reaction increases.

Concentration affecting rate of reaction

As the concentration of reactants increases, there will be a greater number of reactant particles per unit volume. This means that there are more frequent collisions between reactant particles, increasing the number of collisions with sufficient energy to overcome the activation energy and therefore increasing the number of successful collisions 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. This means that there are more frequent collisions between reactant particles, increasing the number of successful collisions 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 for the [reaction]/chemical reaction to occur. This is done by providing an alternative pathway. This means that a greater proportion of collisions between __ particles have sufficient energy to overcome the activation energy. This increases the number of successful collisions 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. 

  • YOU MUST STATE WHETHER THE MASS OF THINGS ARE THE SAME/SAME MASS IS USED

  • The mass declines over time in both reactions as the reactants are used up forming products. The added line starts and ends at the same mass of Al2O3 as the original line, because all of the Al2O3 has successfully reacted into products at the end of both reactions. However, as the added higher concentration line indicates, the end point is reached sooner due to the increased rate of reaction.

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 system is not at equilibrium because the calculated ratio of products to reactants is greater/lower than the Kc value. This means that there are too many/less products relative to reactants. As Kc is the ratio of products to reactants at equilibrium, the fowards/reverse reaction is favoured until the concentration of products increases/decreases and the concentration of the reactants increases/decreases, re-establishing equilibrium.

Concentration affecting equilibrium

• reactant increase = follow hcl template

• reactant decrease = follow agno3 template

• product increase = follow hcl but switch reactants to product, opp for product decrease

Temperature rules for system

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

2. Increase in kc = forward favoured, decrease in kc = reverse 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

• use kc value to determine if temp increase or decrease

• use

asking if forward/reverse reaction is endo/exo (tells you temp and change in Kc)

Increase temp increase kc

Increasing the temperature increases the Kc value. As Kc is the ratio of products to reactants, the ratio of products to reactants must increase. Therefore, the forward reaction is favoured when the temperature is increased. Since increasing temperature favours the endothermic direction, the forward reaction must be endothermic.

Increase temp decreases Kc

Increasing the temperature decreases the Kc value. As Kc is the ratio of products to reactants, the ratio of products to reactants must decrease. Therefore, the reverse reaction is favoured when the temperature is increased. Since increasing temperature favours the endothermic direction, the reverse reaction must be endothermic. Therefore, the forward reaction is exothermic.

1. kc decrease + temp decrease = reverse reaction is exo (switch words for template)

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

asking whether the temp increased/decreased (gives you endo/exo and kc value)

Kc increase, forward reaction endo = temp increase

Since Kc is the ratio of products to reactants, an increase in Kc means the forward reaction is favoured. As the forward reaction is endothermic (because wtv is identified as the endothermic direction), the temperature must have increased because increasing temperature favours the endothermic direction. This shifts the equilibrium in the forward direction, absorbing the added heat and re-establishing equilibrium. Therefore, the increase in Kc indicates that the temperature has increased.

Kc increase, forward reaction exo = temp decrease

Since Kc is the ratio of products to reactants, an increase in Kc means the forward reaction is favoured. As the forward reaction is exothermic (because wtv is identified as the exothermic direction), the temperature must have decreased because decreasing temperature favours the exothermic direction. This shifts the equilibrium in the forward direction, replacing the heat removed from the system and re-establishing equilibrium. Therefore, the increase in Kc indicates that the temperature has decreased.

Kc decrease, forward reaction endo (reverse reaction is exo) = temp decreased

Since Kc is the ratio of products to reactants, a decrease in Kc means the reverse reaction is favoured. As the reverse reaction is exothermic (or wtv they give), the temperature must have decreased because decreasing temperature favours the exothermic direction, which is the reverse reaction. This shifts the equilibrium in the reverse direction, replacing the heat removed from the system and re-establishing equilibrium. Therefore, the decrease in Kc indicates that the temperature has decreased.

Kc decrease, forward reaction exo (reverse reaction is endo) = temp increased

Since Kc is the ratio of products to reactants, a decrease in Kc means the reverse reaction is favoured. As the reverse reaction is endothermic (or wtv they give), the temperature must have increased because increasing temperature favours the endothermic direction, which is the reverse reaction. This shifts the equilibrium in the reverse direction, absorbing the added heat from the system and re-establishing equilibrium. Therefore, the decrease in Kc indicates that the temperature has increased.

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 of gas particles on the reactant side of the equation, and __ moles of gas particles on the product side of the equation. Therefore the [reactant/product] side of the equation has more pressure. When the pressure is increased, the equilibrium shifts in the forwards/reverse direction to minimise this change, favouring the side with fewer gas moles. This means that at equilibrium, there are more products/reactants and fewer products/reactants. The colour associated with the product/reactant becomes more intense, while the colour associated with the reactant/product fades.

• pressure decrease = favours side with more moles

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 has no affect on the position of equilibrium. Therefore, the equilibrium system remains unchanged and there is no change in colour.

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

• look for a unit and charge

• if it loses H⁺ it gains a conjugate base, if not acid

• basically whatever a species loses, the other side of the equation = their conjugate something

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.

[Cu (H₂O)₆]²⁺(aq) + 4Cl^-(aq) ⇄ [CuCl₄]^2-(aq) + 6H₂O(l) Kc = 1.1 × 10⁴

Colour of equilibrium mixture is green. Explain, using equilibrium principles, the effect on the relative concentration of the coloured species and therefore the overall colour of the solution if:
- concentrated HCl is added
- AgNO₃(s) is added. Ag⁺ ions react with the Cl^- ions to form AgCl (s).

Conc HCl:

• If HCl is added, the concentration of Cl^- ions increases. As Cl^- is a reactant, the equilibrium shifts in the forwards direction to minimise this change, removing the added cl- ions (by converting more reactants to products), re-establishing equilibrium. As a result, the concentration of [CuCl₄]^2- increases and the concentration of [Cu (H₂O)₆]²⁺ decreases. Therefore, the colour of the equilibrium mixture will be more yellow OR more green because [CuCl₄]^2- is yellow and [Cu (H₂O)₆]²⁺ is blue (yellowish green is fine).

AgNO₃(s):

• If AgNO₃ is added, the concentration of Cl^- ions decreases because the Ag⁺ ions react with the Cl^- ions, a reactant, removing them from the mixture. This causes the equilibrium to shift in the reverse direction to minimise this change, replacing the Cl^- ions that were removed and re-establishing equilibrium. As a result, the concentration of [CuCl₄]^2- decreases and the concentration of [Cu (H₂O)₆]²⁺ increases. Therefore, the colour of the equilibrium mixture will be more blue OR less green because [CuCl₄]^2- is yellow and [Cu (H₂O)₆]²⁺ is blue (bluish green is fine).

BASICALLY, identify what increases, move to other direction, AND ALWAYS MENTION THE INCREASE OR DECREASE OF THE ACTUAL REACTANT ITSELF, finally the colour change

when they give you the bond enthalpy + temperature change (find kc)

Because ΔH is positive/negative, the forward reaction is endothermic/exothermic. Increasing/decreasing the temperature favours the endothermic/exothermic direction. This shifts the equilibrium towards the products/reactants, increasing/decreasing the ratio of products to reactants. Therefore, the value of Kc increases/decreases.

• it’s always the forward reaction!

Describe the chemical observations made when the calcium carbonate shells react with acid in the ocean.

The shells will slowly lose mass (become thinner), and bubbles will form as the carbonate reacts to form carbon dioxide.

• basically fizzing observed, mass loss