5.1.2 How Far Flashcards

(33 cards)

1
Q

how do you find the units for Kc

A

1) substitute the units into the Kc expression
2) cancel out common units and show the final units in a single line

  • remember, all concentrations are given as mol dm-3
  • and you can have no units
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2
Q

what is homogeneous equilibrium

A

contains equilibrium species that all have the same state/phase

-Kc would contain ALL species

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3
Q

what is heterogeneous equilibrium

A
  • contains equilibrium species that have different states
  • for calculating Kc, you ONLY USE gases and aqueous species, as concentrations of solids and liquids are essentially constant, so are automatically incorporated into the total expression
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4
Q

what is the first step in calculating Kc, where the concentrations are not given, so you can’t just input them into the expression

A

MUST FIND THE AMOUNT OF MOLES AT EQUILIBRIUM:
using a RICE table: where you write the balanced equation given at the top
1) Reacting amount: the molar numbers in the balanced equation
2) Initial moles: given for reactants, and always 0 for products
3) Change in mole: gathered from the moles of product made, as will be subtracted from reactants in CORRECT MOLAR RATIO
4) Equilibrium moles: calculated at the end once you have applied the change in moles

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5
Q

what is the second step in calculating Kc, once the moles at equilibrium have been established

A
  • need to find the concentration in moldm-3
  • use the same volume as given in the question for all values
  • and divide each equilibrium moles by it
  • REMEMBER IT IS IN dm-3, NOT cm-3, SO CONVERT IF NECESSARY
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6
Q

what is the last step of Kc, once you have found the concentration of each species

A
  • write the Kc expression
  • and input in your concentrations
  • remembering to square values if in molar value 2
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7
Q

what is Kp

A

the equilibrium constant in terms of partial pressure

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8
Q

what is Kp used for

A

used for gases, where it is easier to measure partial pressure rather than concentration

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9
Q

what can be said about the relationship between Kc and Kp

A
  • concentration and pressure are directly proportional
  • so Kp and Kc have a direct relationship to eachother too
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10
Q

what can be said about gases in the same conditions of temperature and pressure

A

under the same conditions of temperature and pressure, the same volume of different gases will contain the same number of moles of gas molecules (via Avogadro’s Law)

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11
Q

what is mole fraction

A

the proportion of a gas by its volume to the total volume of gases in a gas mixture

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12
Q

how is mole fraction calculated

A

moles of A / total moles in gas mixture

  • if given as percentages, just do percent over 100
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13
Q

what is the symbol of mole fraction

A

x(A) = mole fraction of A

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14
Q

what is partial pressure

A

p = the contribution a gas makes towards the total pressure P
- sum of all p=P (sum of all partial pressures is the total pressure)

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15
Q

how do you find the partial pressure of a gas

A

p(A) = x(A) x P

  • the mole fraction of A times by the total pressure
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16
Q

how do you write out a Kp constant

A
  • same process of Kc (products over reactants, and use powers to show the moles), but [] are replaced with p()
17
Q

what is important to remember when writing out the Kp constant

A
  • ONLY includes gases, as only they have partial pressures
  • all other species MUST BE IGNORED
  • units individually of p() can be kPa, Pa, atm, as long as you stay constant throughout
  • the power is determined by balancing numbers
18
Q

how do you calculate Kp

A

1) need to find the mole fractions of each species (if moles not given at equilibrium, us ICE table) - via adding all the moles, and dividing each mole by this amount
2) find the partial pressures of each species - via multiplying the mole fraction by the total pressure
3) input the p() values into the Kp constant and determine units

  • REMEMBER ONLY GASES
19
Q

what are the basic 3 principles of Le Chatelier’s principle

A

1) as concentration increases, equilibrium position shifts to decrease this concentration
2) as pressure increases, equilibrium shifts towards the side with fewer molecules to decrease this pressure
3) as temperature increases, the equilibrium shifts in the endothermic direction, to decrease this temperature

20
Q

what do the equilibrium constants of K (c/p) indicate

A

the extent of a chemical equilibrium
- value of 1 showcases it is halfway
- high value 100 shows in favour of products
- low value 0.001 shows in favour of reactants

21
Q

when will the value of an equilibrium constant K change

A

ONLY WITH TEMPERATURE

  • so if the temperature stays constant, no other modification will cause it to change
22
Q

in an exothermic reaction, what happens to the equilibrium constant K when you increase temperature

A
  • K decreases with increasing temperature
  • and increasing the temperature decreases the equilibrium yield of products (equilibrium shifts to the left)
23
Q

what are the effects of Kp decreasing when temperature increases in an exothermic reaction

A
  • at equilibrium, there is a set Kp value
  • if temperature increases, Kp decreases
  • so the system is no longer at equilibrium, as the ratio of p(x)/p(y) is now greater than Kp
  • so the equilibrium partial pressures must now change to give new values of Kp
  • p(products) must decrease
  • and p(reactants) must decrease
  • so position of equilibrium shifts to the left
  • new equilibrium is reached, where p(x)/p(y) is now equal to Kp again
24
Q

what happens to K when you increase the temperature of an endothermic reaction

A
  • K increases
  • as temperature increases of endothermic reaction
  • so position of equilibrium shifts to the right
  • increasing the equilibrium yield of products
25
what are the effects of Kp increasing as temperature does in an endothermic reaction
- as temperature increases - so does Kp - so the system is no longer in equilibrium - and p(x)/p(y) is now less than Kp - so p() must now change to give the new equilibrium value of Kp - so p(products) must increase - and p(reactants) must decrease - meaning the position of equilibrium shifts to the right - new equilibrium is established again, where p(x)/p(y) is now equal to Kp
26
how would you talk about changing Kc values instead of Kp
- same layout and reasoning - just use concentration [] instead of talking about partial pressures
27
what happens to the value of K when you increase concentration
IT IS UNAFFECTED !!! - therefore the equilibrium has to shift because of this
28
what happens to the position of equilibrium when the concentration of reactants increases
- at normal equilibrium, you have a given value of K - if you increase the [] of reactants, the new ratio of [x]/[y] will be less than Kc at equilibrium (does not change) - so system is no longer at equilibrium - therefore, you must change concentrations to return back to the value of Kc at equilibrium - the [products] must increase - the [reactants] must decrease - so shifts to the right, and a new equilibrium is established where [x]/[y] returns back to the normal value of Kc
29
why does a shift in equilibrium occur when you change the concentrations of reactants/products at equilibrium
- as Kc does not change with change in concentration - so is able to control the relative concentrations of products and reactants at equilibrium
30
what effect does increasing pressure have on K
- UNAFFECTED BY IT !!!
31
where the pressure of reactants is greater than that of products, what happens when you double pressure
- doubling total pressure doubles the concentration and p() of both the reactants and products - so a new ratio of p(x)/p(y) is established [as the p(x) has a power due to having greater moles, so increases by a e.g. square of the value] - which is greater than value of Kp - the system is no longer in equilibrium - so we must change the p() values to return the ratio back to original Kp - p(products) decreases - and p(reactants) increases - new equilibrium reached where p(x)/p(y) is restored back to that of Kp - and equilibrium shifts to the left
32
why does increasing the pressure shift the equilibrium to the left (where products have more pressure than reactants)
- as via Le Chatelier's, it will shift to the side with fewer moles - and this shift is directed due to the value of Kp needing to be restored
33
what effect do catalysts have on K
UNAFFECTED !!! - only increases the rate of reaction, doesn't affect the position of equilibrium - as speeds up the rate of the forward and backward reaction by the same factor - so equilibrium is reached quicker, but the position of it is unchanged