The rate and extent of chemical change

Collision Theory 

Particles need to collide with sufficient energy (activation energy) in the right orientation to react successfully. 

more frequent successful collisions  = faster rate of reaction 

so factors that increase the rate of reaction increase the frequency of successful collisions, usually by increasing the overall amount of particle collisions. 

Factors affecting rate of reaction  

Surface area: increasing surface area increases the surface area to volume ratio (size of solid decreased but volume is constant) which increases the rate of reaction; 

• More surface area exposed for collisions to occur, thus they happen more frequently, thus amount of overall collisions increases, which means more successful collisions 

Concentration: increasing the concentration of a solution increases the rate of reaction; 

• More reacting particles present in the same volume so collisions occur more frequently, thus amount of overall collisions increases, which means more successful collisions 

Temperature: increasing temperature increases rate of reaction; 

• Temperature is a measure of a substance’s average kinetic energy, so a higher temperature means a higher kinetic energy which means more frequent collisions because particles moving faster. thus amount of overall collisions increases, which means more successful collisions 

Pressure: increasing pressure increases rate of reaction; 

• Smaller volume for particles to travel through, so more likely to collide with each other. More frequent collisions = amount of overall collisions increases, which means more successful collisions 

Catalysts 

Speed up the rate of reaction by lowering activation energy – increases proportion of particles that have enough energy to react  

provide a different pathway with a lower activation energy 

not used up in the reaction 

to summarise:  

Required Practical 1: disappearing cross (temperature) 

 Method: 

1. Measure 10cm³ of sodium thiosulfate solution in a measuring cylinder 

2. Pour into a conical flask 

3. Place on a laminated image of a cross 

4. Collect 10cm³ of HCl from a water bath – each should have a different temperature 

5. Pour the HCl into the conical flask 

6. Time how long it takes for the cross to disappear – the solution becomes opaque due to the formation of a sulfur precipitate.  

7. Once it does, pour the solution into a waste bucket in a fume hood 

8. Scrub the conical flash thoroughly  

9. Repeat steps 1 – 8 for each temperature of HCl 

Variables:  

Dependent: time taken for cross to disappear 

Independent: temperature of hydrochloric acid 

Control: the cross, sodium thiosulfate concentration, temperature, volume, pressure  

Consider the impact of increased temperature on reactions. You may see slightly altered versions of this in the exam in terms of what factor is being tested – they’ll always look for the disappearing cross though.  

Calculating rates of reaction + graphs 

 You may be asked to sketch/identify the effects of changing factors on the rate of reaction graphically. Some key points: 

• Halving/doubling the concentration of a solution halves/doubles the amount of product 

• Changes in temperature, pressure or surface area affect rate of reaction but not the amount of product. Increases will mean a steeper gradient 

to calculate the average rate of a reaction, use the equation: 

rate = product made OR  reactant used / time. remember your units (mins for minutes as m is metres)

to find the r.o.r at a specific point, you need to draw a tangent that touches the graph at the specified time and draw a right angles triangle. The gradient is the rate of reaction – bigger the triangle the more accurate. 

to explain this graph:

1. The reactants have not yet been used up, so concentration of them is high, thus more collisions occur and the graph has a steep gradient

2. Fewer reactant particles now available as the reaction has progressed so fewer total + successful collisions

3. No more product being produced as time continues, so reaction is over. Reactants have been used up and graph has no gradient

Required Practical 2: surface area to volume

You may be asked to draw this diagram. 2 crucial things to remember that could lose marks:

1. draw the markings on the gas syringe

2. the delivery tube should not be in the solution, else the gas won’t be collected

Method

1. set up the apparatus as shown above. Ensure the gas syringe is pushed all the way in, to 0.

2. Weigh approximately 2g of large calcium carbonate chips using a weighing boat and mass balance. Note the exact mass used.

3. Transfer the calcium carbonate to the conical flask

4. Use a measuring cylinder to measure 25ml of HCl

5. Transfer the HCl to the conical flash, starting the stopwatch and replacing the bung as soon as you do so

6. Measure the volume of gas produces every 15 seconds for 3 minutes

7. Use a sieve to prevent calcium carbonate chips from blocking the drain. Clean out the conical flask.

8. Repeat steps 1-7 for smaller calcium carbonate chips.

Variables:

Dependent: volume of carbon dioxide produced

Independent: surface area of calcium carbonate chips

Control: volume of HCl, mass of calcium carbonate chips, concentration of HCl

Consider the impact of increased surface area to volume ratio on reactions. Remember to discuss that the volume of both sizes of chips was the same in your explanation.

The graph for the results will produce a line with a steep gradient for the small chips, and one with a gentler gradient for the larger chips.

Reversible reactions

In some reactions, the products can react to reproduce the reactants. This is a reversible reaction, represented by a ⇌ instead of a →

dynamic equilibrium has been reached when the rate of the forward reaction = the rate of the backwards reaction.

• the rate of the forward reaction will initially be much greater, but will decrease as the reactants are used up. In turn, the rate of the reverse reaction increases and the amount of the products will build up until the rates are the same.

• occurs in a closed system, where no reactants enter and no products leave

• the concentration of the reactants and products is constant in this case – not equal.

Le Chatelier’s principle: states that if a system at dynamic equilibrium gets distributed by changing conditions, equilibrium shifts to counteract the change.

Temperature:

If forward reaction is exothermic

If forward reaction is endothermic

• Temp increase will make the system shift to the endothermic side (the left/right) to decrease the temperature. Heat energy is absorbed, temperature decreases, restoring equilibrium

• Temp decrease will make the system shift to the exothermic side (the left/right)  to increase the temperature. Heat energy is emitted, temperature increases, restoring equilibrium

Pressure:

• If pressure is increased, equilibrium will shift to the side with fewer moles* (moles being the big number in front of a compound). As a result, pressure decreases, restoring equilibrium

• If pressure is decreased, equilibrium will shift to the side with more moles*. As a result, pressure decreases, restoring equilibrium

*give the number of moles on both sides of the equation.

Concentration:

• if the concentration of the reactants is increased (by adding reactants or removing products) the reaction will shift to the right, producing more product to remove the excess of reactants, restoring equilibrium

• if the concentration of the reactants is decreased the reaction will shift to the left, producing more of the reactants to increase their concentration, restoring equilibrium

These principles will often be related to yield of products. Yield will decrease if equilibrium shifts to the left, and increases if it shifts to the right.

The Haber Process

N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

used to produce ammonia for plant fertilisers

nitrogen is extracted from the air

hydrogen is produced by reacting methane and steam

Ideal conditions:

• high pressure

• low temperatures due to forward reaction being exothermic

• high concentration of products

all because they increase yield.

however this is unachievable, because the rate of reaction at low temperatures is far too slow to be efficient (takes ages to reach equilibrium) and the cost of building equipment that can withstand high pressures is too high. instead, the reaction takes place in compromised conditions:

• an iron catalyst speeds up the reaction, but doesn’t increase yield!!

• 450 degrees C

• pressure of 200 atmospheres

in terms of the actual apparatus:

1. nitrogen and hydrogen are pumped into the machine

2. mixture is compressed at pressure of 200 am and heated to 450 degrees C

3. mixture passed into reaction vessel where iron catalyst helps it along

4. mixture is cooled, ammonia condenses into a liquid and is removed

5. unreacted nitrogen and hydrogen are recycled, returning to the compressor before being passed back into the reaction vessel.