UNIT 1: SECTION 5-KINETICS, EQUILIBRIA AND REDOX REACTIONS Flashcards

(47 cards)

1
Q

A) Define “reaction rate”

B) Outline the simple formula for finding the rate of a chemical reaction

A

A)-change in concentration (amount) of reactant or product over time
B)-rate of reaction=amount of product formed/time

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2
Q

A) What are particles constantly doing liquids and gases?

B) Describe collision theory

A

A)-always moving and colliding with each other
B)-particles not react all the time only when conditions right
–>reaction wont happen unless:
1-they collide in right direction–>need to be facing each other in right way
2-they collide with a minimum amount of K. Energy (movement).

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3
Q

A) Define “activation energy” and briefly explain it

B) What does a high or low activation energy of a reaction mean?

A

A)-minimum amount of energy particles needs to react with
-the particles must have at least this much energy to break their bonds and start the reaction
B)-low activation energy reactions often happen easier
–>BUT reactions with high activation energies don’t as they particles need extra energy by heating them.

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4
Q

A) What does a “Maxwell-Boltzmann Distribution” curve show?

A

A)-different molecules in a gas have different kinetic energies

  • ->some don’t have much energy and move slowly
  • ->others have lots of K.E and whizz around a lot
  • ->but most molecules are in between
  • if you plot a graph of the n. of molecules in a gas with different kinetic energies you get a “Maxwell-Boltzmann Distribution” curve.
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5
Q

Describe the following parts of the Maxwell-Boltzmann curve:

A) Graph starts at (0, 0) 
B) Start of graph steepening
C) Graph continues to steepen
D) Peak of the curve
E) Average K. Energy (mean) of all molecules 
F) Start of shaded part
A

A)-this as no molecules have no energy
B)-some molecules are moving slowly
C)-most molecules moving at moderate speed–>their energies are somewhere in middle
D)-this represents most likely energy of any single molecule
E)-this may be to the right or left of the peak
F)-activation energy starts
–>area of this shows that some molecules have more than the activation energy
–>these are the only ones able to react.

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6
Q

A) What is the effect of increasing the temperature of a reaction?

B) Why as a result of a temperature increase would the shape of the Maxwell Distribution curve change and how?

A

A)-if increase the temperature the particles will on average have more K.E and will move faster
B)-so a greater proportion of molecules will have at least the activation energy and so be able to react
–>this changes the curve’s shape–>pushes over to right.

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7
Q

C) Also, how else does a temperature increase on the rate of a reaction?

D) Consequently, what does it mean when you get both of these 2 effects happening at once (increase reaction rate)?

A

C)-this is as at higher temp as molecules moving about faster they’ll collide more often
D)-more collisions and more energetic collisions means quite a small temperature increase in temperature can lead to quite a large increase in the rate of reaction.

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8
Q

A) Describe the effect of an increase in CONCENTRATION on the rate of reaction

A

A)-if increase conc of reactants in solution, the particles will on average be closer together

  • ->if they are closer together they would collide more often
  • ->if collisions more frequent they would have more chances to react
  • ->therefore increasing concentration increases reaction rate.
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9
Q

B) Explain the effect of an increase in PRESSURE on the rate of reaction

A

B)-if reaction involves gases increasing pressure works same way a increasing conc

  • ->it pushes all of gas particles closer together
  • ->making them more likely to collide
  • ->so collisions take place more frequently + reaction rate increases.
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10
Q

C) Define a “catalyst”

A

C)-a substance that increases reaction’s rate by providing alternative reaction pathway with lower activation energy
–>the catalyst is chemically unchanged end of reaction.

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11
Q

D) Why are catalysts useful?

E) How are catalysts v. specific?

A

D)-1: they don’t get used up in a reaction so only need tiny bit of it to catalyse lots of stuff
–>do take part in reaction but REMADE at end
2-a catalysts allow you to make same amount of product faster (+ often at lower temp too), they save lots of money in industrial processes
E)-very specific as to which reactions they catalyse
–>often only work on a single reaction.

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12
Q

A) How is the effect of a catalyst shown on a Maxwell-Boltzmann distribution curve?

A

A)-more particles have at least the activation energy in the catalysed reaction

  • ->lower activation means more particles have enough energy to react when they collide
  • ->does this via going a different route
  • ->so in certain amount of time more particles react
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13
Q

A) Outline and evaluate the method to follow to measure the rate of reaction by TIMING HOW LONG A PRECIPITATE TAKES TO FORM

A

A)1-can use this method when product a precipitate which clouds solution
2-watch a mark through a solution + time how long takes to be obscured
3-is same observer uses same mark each time, you can compare reaction’s rate as roughly same precipitate amount will have been formed when mark obscures
4-this methods subjective though–>different people may not agree exact moment mark disappears.

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14
Q

B) State the method and evaluate it to measure the rate of reaction by MEASURING A DECREASE IN MASS

A

B)1-When 1 + products is a gas you can measure rate of formation using mass balance
2-as gas given off–>mass of reaction mixture decreases
3-this method accurate + easy to do
–>but it does release gas into atmosphere so usually done in fume cupboard.

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15
Q

C) What method must you follow to measure the rate of reaction when MEASURING THE VOLUME OF GAS GIVEN OFF? Also evaluate this method

A

C)1-this involves using gas syringe to measure volume of gas being produced
2-can only use this method when 1 + products is a gas
3-gas syringes usually give volumes to nearest 0.1cm3, so this method is accurate.

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16
Q

A) Describe the reaction between Sodium Thiosulfate and Hydrochloric Acid and the rate of reaction may be measured

A

A)-sodium thiosulfate + HCl are both clear, colourless solutions

  • ->they react to form a yellow precipitate of sulfur
  • can use amount of time takes for precipitate formation as measure of reaction rate
  • ->this experiment often used to demonstrate effect of increasing temp on reaction’s rate.
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17
Q

B) Describe the method to follow to measure the rate of this reaction

A

B)1-measure fixed sodium thiosulfate volumes + HCl using measuring cylinder
2-use water bath to gently heat both solutions to desired temp before mixing them
3-mix solutions in conical flask
–>place flask over black cross that can be seen through solution
–>watch cross disappear through cloudy sulfur–>time it
4-can repeat reaction at different temperatures
–>depth of liquid must be kept same each time
–>concentrations of solutions must be kept same too

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18
Q

C) What results should you expect from this experiment?

A

C)-results should show that higher the temp

–>faster rate of reaction and so takes less time for mark to disappear.

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19
Q

A) What are “reversible reactions”?

B) Outline an example of a reversible reaction

A

A)-many chemical reactions reversible–>go both ways
–>2 way arrow represents this
B)-H2 (g) + I2 (g)–> 2HI (g) (reaction can go either way)
2HI
H2 (g) + I2 (g)

20
Q

C) Describe and explain “dynamic equilibrium”

A

C)1-as reactants in reaction get used up forward reaction slows AND as more product formed reverse reaction speeds up
2-after while forward reaction at exact same rate as backwards so amount of reactants + products not change no more
–>this is dynamic equilibrium because concentrations of reactants + products stay constant

21
Q

D) What is the link between dynamic equilibrium and “closed systems”?

A

D)-dynamic equilibrium only possible in closed system

–>this means nothing able to get in or out.

22
Q

A) What would happen if you change the concentration, pressure or temperature of a reversible reaction?

A

A)-you would be altering the position of equilibrium

–>mean would end up with different amounts of reactants and product at equilibrium.

23
Q

What would happen in each of the following reactions?:

A) H2 (g) + I2 (g)–> 2HI (g)
2HI (g)

A

A)-lots of H2 and I2 but not much HI in reaction
–>so you would get more reactants
B)-not much H2 and I2 but lots of HI
–>so more products

24
Q

C) Outline “Le Chatelier’s principle” and briefly explain it

A

C)-if a reaction at equilibrium is subjected to change in concentration, pressure or temperature, the position of of equilibrium will move to counteract the change

  • ->so if raise temp, the position of equilibrium will shift to try and cool things down
  • ->and if raise the pressure or concentration, the position of equilibrium will shift to try and reduce it again.
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Effect of CONCENTRATION on position of equilibrium: A) What would happen if you increase the concentration of a reactant? B) EXAMPLE: Outline the effect of increasing the concentration of SO2 or O2 the following reaction: 2SO2 (g) + O2 (g)--> 2SO3 (g)
A)-the equilibrium tries to get rid of extra reactant -->does so by making more product -->so equilibrium shifts to right B)-speed of forward reaction increases (to use up extra reactant) which moves equilibrium to right C)-the equilibrium tries to remove extra product -->this makes reverse reaction faster so equilibrium shifts to left D)-it has the opposite effect.
26
Effect of PRESSURE on position of equilibrium (only affects gases): A) Outline the effect of increasing the pressure of a reaction B) EXAMPLE: What would be the effect of increasing pressure on the following reaction?: 2SO2 (g) + O2 (g)--> 2SO3 (g)
A)-shifts position of equilibrium position to side of reaction with fewer gas molecules -->this reduces pressure B)-3 moles of gas on left BUT only 2 on right -->so increase in pressure shifts equilibrium to right C)-shifts equilibrium to side with more gas molecules -->this raises pressure again.
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Effects of TEMPERATURE on the position of equilibrium: A) What would be the effect of increasing the temperature of a reaction? B) EXAMPLE: State the effect of increasing the temperature of the following reaction: 2SO2 (g) + O2 (g)-->2SO3 (g) (Delta H= -197 kj/mol-1)
A)-means you would be adding more heat -->equilibrium shifts in endothermic (delta H) direction to absorb this heat B)-reactions exothermic in forward direction -->if you increase temp, the equilibrium shifts to left to absorb the extra heat C)-mean removing heat -equilibrium in exothermic (-delta H) direction to produce more heat to counteract drop in temp D)-exothermic and vice versa
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E) Does a catalysts affect the position of equilibrium in a reaction?
E)-no a catalyst has no effect - ->it CAN'T increase the yield - -but it does mean EQUILIBRIUM IS REACHED FASTER.
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A) What is the involvement of reversible reactions and conditions in industry?
A)-companies have to think about how much costs to run a reaction and how much money they can make from it -->means have to consider a few factors when choosing best conditions for a reaction
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EXAMPLE: A) Outline the reversible reaction for the production for the production of ethanol B) Under what conditions is ethanol produced?
A)-it may be produced via reversible exothermic reaction between ethene and steam -EQUATION: C2H4 (g) + H2O (g)--> C2H5OH (g)
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C) Explain why a lower temperature is used for the production of ethanol
C)-this is as it's an exothermic reaction so lower temp favours forward reaction -->means at lower temp more ethene + steam converted to ethanol-->get a better yield - but lower temp also means slower rate of reaction - ->no point in high yield of ethanol if take v. long time - ->so 300 degrees C is compromise between reasonable yield and a faster reaction.
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D) Why is 60-70 atmospheres a compromise pressure used in the production of ethanol-explain?
D)-high pressure shifts equilibrium to side with fewer molecules which favours forward reaction - ->this high pressure increases reaction rate too - ->using higher pressures than this are v. expensive so need v. strong containers + pipes to withstand pressure - ->therefore 60-70 atmospheres is a compromise-->it gives reasonable yield for lowest possible cost.
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A) When may the "equilibrium constant (Kc)" be calculated? B) Outline the general expression used to represent equilibrium constant
A)-when know molar concentrations of each substance at equilibrium you may work out Kc B) aA + bB--> dD + eE
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C) EXAMPLE: Hydrogen gas and iodine gas are mixed in a closed flask. Hydrogen iodide is formed H2 (g) + I2--> 2HI (g)
C)-just stick concentrations in to the expression for Kc: -Kc= (HI)^2/ (H2) (I2)= 0.08^2/0.10 *0.10=64 - to work out the units of Kc put units in the expression instead of numbers: - ->Kc units = (mol dm-3)^2/(mol dm-3)(mol dm-3) - ->the concentrations cancel, so NO UNITS for Kc - ->so answer for Kc=64.
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You may have to figure out some of the equilibrium concentrations before you can find Kc: A) EXAMPLE: 0.20 moles of phosphorus (V) chloride decomposes at 600 K in a vessel of 5.00 dm3. The equilibrium mixture is found to contain 0.080 moles of chlorine. Write the expression for Kc and calculate it's value, including units.
A) 1-calculate moles of PCl5 and PCl3 at equilibrium: - ->equations tells you 1 mole of PCl5 decomposes 1 mole of PCl3 and 1 mole of Cl2 formed - ->so if 0.08 chlorine moles produced at equilibrium then there will be 0.080 moles of PCl3 as well - ->0.08 PCl5 moles must of decomposed so there will be 0.12 moles left (0.20-0.080) 2-divide each n. of moles by flask volume to give molar concentrations: (PCl3)=(Cl2)=0.08/5.00=0.016 mol/dm-3 (PCl5)= 0.12/5.00=0.024 mol/dm-3. 3-put concentrations in expression for Kc and calculate it: Kc=(PCl3) (Cl2)/ (PCl5)=0.016 * 0.016/0.024=0.011 4-now find Kc units: Kc units= (mol/dm-3)(mol/dm-3)/(mol/dm-3)= mol/dm-3 5-ANSWER for Kc= 0.011 mol/dm-3
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Using Kc to find concentrations in an equilibrium mixtur: EXAMPLE: When ethanoic acid allowed to reach equilibrium with ethanol at 25 degrees C it was found equilibrium mixuture contained 2.0 mol/dm-3 ethanoic acid and 3.5 mol/dm-3 ethanol. The Kc of the equilibrium is 4.0 at 25 degrees C. What are the concentrations of the other components? CH3COOH (l) + C2H50H (l)--> CH3COOC2H5 (l) + H2O (l)
1-put all known values in to expression: Kc= (CH3COOC2H5)(H2O)/(CH3COOH)(C2H5OH) -->(CH3COOC2H5) (H20)/ 2.0 * 3.5= 4.0 2-rearranging this gives: (CH3COOC2H5) (H2O)= 4.0 *2.0 *3.5= 28 3-from equation you know (CH3COOC2H5)= (H20) so: -->(CH3COOC2H5)= (H20)= SQR 28= 5.3 mol/dm-3 4-the concentration of CH3COOC2H5 and H2O is 5.3 mol/dm-3
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A) Explain what would happen to the value of Kc if the temperature is changed and how exactly this temperature change will have an effect
A)-Kc value only for particular temperature -if change temp of system you will also change equilibrium concentrations of the products + reactants so Kc will change THEREFORE: -->if temp change means more product at equilibrium, Kc will rise -->if means there's less product at equilibrium, Kc will decreases
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B) EXAMPLE: The following reaction is exothermic in the forward direction and endothermic in the backwards direction: N2 (g) + 3H2 (g)--> 2NH3 (g) (delta H= -46.2kj/mol-1)
B)-if increase temp you favour endothermic reaction so less product formed - Kc= (NH3)^2/(N2)(H2)^3 - ->concentration of NH3 will be reduced and N2 and H2 concentrations will increase - ->therefore as temp increases (NH3) will decrease and (N2) + (H2) will increase so Kc will decrease.
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C) Outline the effect of changing the CONCENTRATION on the Kc value and adding a CATALYST to the reaction
C)-CONCENTRATION: changing the conc of a reactant OR product will not affect the value of Kc - CATALYST: it doesn't affect the Kc value either - ->they would speed up reaction in both directions by same amount so they just help the system to reach equilibrium faster.
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# Define the following terms: ``` A) Oxidation B) Reduction C) Redox D) Oxidising agnent E) Reducing agent ```
``` A)-loss of electrons B)-gain of electrons C)-reduction and oxidation occurings simultaneously D)-ACCEPTS electrons and is reduced E)-DONATES electrons and is oxidised. ```
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F) EXAMPLE: How is the formation of Sodium chloride from Sodium and Chlorine a redox reaction?: Na + 1/2Cl--> Na+ Cl- A) What is the "oxidation state" of an element?
F) 1e- lost from Na (it's oxidised) -1/2 Cl2 gains 1e- (its reduced) A)-it tells you total n. of electrons it has donated or accepted - ->E.G: in redox reaction above Na atom has donated 1 electron so it has +1 oxidation state - ->chlorine atom has accpeted 1 electron so has -1 oxidation state
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Rules to follow when working out oxidation state of an atom when it's in a compound, in an ion or just on it's own: 1) Uncombined elements... 2) Elements bonded to identical atoms... 3) Oxidation state of simple monatomic ion... 4) In compound ions... E.G: S04 2-
1)-uncombined elements like He or Ar have 0 oxidation state 2-elements bonded to identical atoms likr O2 and H2 also have 0 oxidation state 3)-oxidation state of simple monatomic ion like Na+ is same as it's charge 4)-the overall oxidation state is just charge on ion -E.G: SO4 2-: overall oxidation state= -2 --> O oxidation state = -2 (4*-2)= -8 -->so oxidation state of S= -2-(8)=+6
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``` 5- Sum of neutral compound oxidation state... E.G: Fe2O3 6-Peroxides E.G : H2O/H2O2 7-Combined hydrogens... -E.G: HF/H/NaH ```
5-sum of oxidation state for neutral compound is 0 -E.G: Fe2O3 overall oxidation state=0 -->oxidation state of O=-2 (total= 3*-2=-6) -->so oxidation state of Fe= (0-(-6))/2= +3 6-combined oxygen nearly always -2 except in peroxides where it's -1 -->and in fluorides (OF2) its +2 -->and O2F2 its +1 -->and O2 where its 0 -->E.G: H20 oxidation state of O= -2 but H2O2 oxidation state of O= -1 7-combined hydrogen is +1 except in metal hydrides where it's -1 (and H2 where its 0) -E.G: in HF oxidation state of H= +1 BUT in NaH oxidation state of H= -1.
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A) What do roman numerals show in relation to oxidation states? B) Example: Outline the oxidation states of the following chemicals -->iron (II) sulfate (FeSO4) AND iron (III) sulfate (Fe2(SO4)3)
A)-sometimes oxidation states not clear from formula of a compound -if see roman numerals in a chemical name its an oxidation number B)-iron in iron (II) had oxidation state of +2 BUT in iron (III) sulfate (Fe2(SO4)3) it's oxidation state is +3.
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A) Describe ionic "half-equations"
A)-they show oxidation or reduction - ->you show electrons that are being lost/gained in half-equations - ->E.G: half equation for Na- Na-->Na+ + e- - you can combine half-equations for different oxidising or reducing agents together to make full equations for redox reactions.
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A) Outline the formation of magnesium oxide including the half-equations
``` A)-magnesium burns in oxygen to form magnesium oxide 1-->Oxygen is reduced to O2- (O2 + 4e- -->2O2-) 2-Magnesium oxidised to Mg 2+: (Mg-->Mg2+ + 2e-) 3-->need both equations to contain same n. of electrons so double everything in 2nd equation (2Mg-->2Mg 2+ + 4e-) 4-->combining the half-equations: 2Mg + O2-->2MgPO ``` 5-(electrons not included in full equation but end up with 4 on each side: --> 2Mg + O2 + 4e- --> 2MgO + 4e-
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B) Similarly show the formation of aluminium chloride with the inclusion of the half-equations
B)-aluminium reacts with chlorine to form aluminium chloride 1-aluminium oxidised to Al 3+: (Al--> Al3+ + 3e-) 2-chlorine reduced to Cl- (Cl2 + 2e- -->2Cl-) 3-now make sure equations each contain same n. of electrons: Al--> Al3+ + 3e- -->x2 2Al--> 2Al3+ + 6e- Cl2 + 2e- --> 2Cl- -->x3 3Cl2 + 6e- -->6Cl- 4-combinig the half-equations makes: 2Al + 3Cl2 --> 2AlCl3.