Equilbrium Flashcards

(25 cards)

1
Q

IRREVRSIBLE Vs REVERSIBLE reactions

A

IRREVERSIBLE:
- Goes in one direction
- Goes to completion
- Shown by arrow pointing from reactants
to products
- The products do not convert back into the original reactants
- E.G decomposition of hydrogen peroxide

REVERSIBLE:
- The reactants are converted to the products at the same time as the products are converted to reactants
- The reversible sign shows the reaction is in EQUILBRIUM

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2
Q

DNAMIC EQUILBRIUM ~ Definition

A

DYNAMIC EQULIBRIUM:

  • the rate of the forward reaction is equal to the rate of the backward reaction
  • the concentrations of the reactants and products remain constant
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3
Q

DYNAMIC EQUILBRIUM ~ Explained

A

Occurs when a reversible reaction takes place in a CLOSED SYSTEM:

  • neither the products or reactants can enter or leave
  • the temperature, pressure or concentration of the reactants and products remains unchanged
  • The reactants are forming the products as FAST AS the products are forming the reactants
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4
Q

The POSITION of equilibrium

A
  • Indicates the EXTENT of the reaction
  • The PROPORTION of products to reactants in the equilibrium mixture
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5
Q

Le Chatelier’s Principle

A

According to le Chatelier’s principle:

’ When a system in equilibrium is subjected to an external change , the system readjusts itself to MINIMISE the effect of tat change’

The position of equilibrium shifts to CANCEL OUT any changes in:
- Temperature
- Pressure
- Concentration

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6
Q

Gas Pressure

A

Pressure = Force/Area

  • The greater number of MOLES of a reactant or product in an equilibrium mixture…
  • the greater the gas pressure it exerts
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7
Q

TEMPERATURE ~ effect on equilibrium reaction

A

INCREASE TEMPERATURE:
- Equilibrium moves in the ENDOTHERMIC
direction
- REDUCES the temperature of the system
- MINIMISES the change

DECREASE TEMPERATURE:
- Equilibrium moves in the EXOTHERMIC direction
- INCREASES the temperature of the system
- MINIMISES the change

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8
Q

CONCENTRATION ~ effect on equilibrium reaction

A

INCREASE CONCENTRATION OF REACTANTS:
- Equilibrium moves to the RIGHT
- REDUCES concentration of reactants
- MINIMISES the change

DECREASE CONCENTRATION OF REACTANTS:
- Equilibrium moves to the LEFT
- INCREASES concentration of reactants
- MINIMISES the change

INCREASE CONCENTRATION OF PRODUCTS:
- Equilibrium moves to the LEFT
- REDUCES concentration of products
- MINIMISES the change

DECREASE CONCENTRATION OF PRODUCTS:
- Equilibrium moves to the RIGHT
- INCREASES concentration of products
- MINIMISES the change

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9
Q

PRESSURE ~ effect on equilibrium reaction

A

INCREASE PRESSURE:
- Equilibrium moves to the side with FEWER
MOLES
- DECREASES the pressure of the system
- MINIMISES the change

DECREASE PRESSURE:
- Equilibrium moves to the side with MORE
MOLES
- INCREASES the pressure of the system
- MINIMISES the change

Only effects reactions that have BOTH:

  • Species in GASEOUS STATE
  • Different number of MOLES on either side
    of the reaction
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10
Q

CATALYST ~ effect on equilibrium reaction

A
  • Provides an alternative pathway with a lower activation energy
  • INCREASE THE RATE of BOTH the forward and backward reaction TO THE SAME EXTENT
  • The equilibrium is ATTAINED FASTER
  • NO EFFECT ON POSITION OF EQUILBRIUM~ remains the same
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11
Q

The HABER process & conditions we would EXPECT

A

N2(g) + 3H2(g) (reversible sign) 2NH3(g)

Enthalpy change = -92 KJmol-1

Reactants obtained from:
- Nitrogen ~ the fractional distillation of AIR
- Hydrogen ~ by reacting METHANE
and WATER

LOW TEMPERATURE:
- Reaction is exothermic
- Position of equilibrium to the right
- Favours production of ammonia

HIGH PRESSURE:
- 4 moles of reactants ——- 2 moles of reactants
- Position of equilibrium shifts to right as there are fewer moles of product
- Favours the production of ammonia

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12
Q

What are the COMPROMISE temperature and pressure conditions for the Haber Process?

A

350-500C:
- The forward reaction is EXOTHERMIC
- At a LOW temperature the equilibrium shifts towards the PRODUCTS
- Favours the production of ammonia
- HOWEVER temperature must be INCREASED to increase the RATE of reaction

100-200atm:
- A HIGH pressure shifts the position of equilibrium towards the PRODUCTS
- Favours the production of ammonia
- INCREASE the rate of reaction
- HOWEVER pressure must be LOWERED to:
- reduce cost
- Prevent danger for work force and
surrounding areas

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13
Q

The CATALYST used in the Haber Process and why?

A

IRON:

  • Used in the form of a SOLID GAUZE to maximise SURFACE AREA
  • INCREASES the rate of reaction by providing an alternative route with a lower activation energy
  • As a result, LOWER temperature can be used and costs are reduced
  • The LIFE and ACTIVITY of the catalyst is INCREASED when the operating temperature is LOWER.
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14
Q

What is the COMPRISED YIELD of ammonia from the Haber Process and WHY?

A

15%:

  • The removal of ammonia caused the position of equilibrium to shift to the RIGHT
  • This INCREASES the yield of ammonia produced
  • It takes MORE TIME for a higher conversion of ammonia to take place at the industrial conditions when the product is removed in a LARGER amount
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15
Q

The DRAWBACKS of using HIGH PRESSURES in industrial processes

A

VERY COSTLY TO:

-Compress gases to high pressures
- Produces thick pipes and reaction vessel to withstand high pressures

UNSAFE:

  • Risk of leakage , e.g toxic ammonia
  • Endangers the workforce & surrounding
    area
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16
Q

Any industrial process aims to achieve the:

A
  • Highest possible yield
  • In the shortest time (Fastest Rate)
  • At the lowest cost

This requires a COMPROMISE between temperature and pressure conditions and the use of a catalyst.

17
Q

How do we get VIRTUALLY 100% yield of ammonia from the Haber Process?

A
  • The unreacted gases are RECYCLED
  • Until eventually they are all converted to ammonia
18
Q

The Equilibrium Law

A
  • Gives the CINCENTRATION of reactants and products when a reversible reaction reaches EQUILIBRIUM
19
Q

The Equilibrium Constant (Kc)

A
  • Allows the CONCENTRATIONS of the reactants and products at equilibrium to be CALCULATED

aA + bB (reversible sign) cC + dD

Kc = [products] / [reactants]
= [C]^c[D]^d / [A]^a[B]^b

  • [] are the equilibrium concentrations in mol dm^-3
20
Q

UNITS for Equilibrium Constant

A
  • Write out units in equation

Examples:
mol dm^-3/ mol dm^-3 x mol dm^-3
- This would cancel out to give dm^-3 mol^-1
- Units with POSITIVE powers placed first

  • If there are no units , ‘NO UNITS’ must be written
21
Q

Homogenous Equilibria

A
  • ALL equilibrium species are in the SAME STATE or PHASE
  • The Kc expression will include the CONCENTRATIONS of ALL species in the equilibrium
22
Q

Heterogeneous Equilibria

A
  • The equilibrium species are in DIFFERENT STATES or PHASES
  • The Kc expression DOES NOT include the CONCENTRATIONS of SOLIDS or LIQUIDS.
  • This is because their concentration remain EFFECTIVLY CONSTANT
  • Only the concentrations of GASEOUS and AQUEOUS species are included
23
Q

Significance of a LARGE Kc value

A
  • > 100

Kc fraction:
- Top is LARGE
- Bottom is SMALL

Concentration of the products > concentration of reactants

POSITION OF EQUILIBRIUM:
RHS for the concentration of PRODUCTS to be HIGH

24
Q

Significance of a SMALL Kc value

A
  • <0.01

Kc Fraction:
- Top is SMALL
- Bottom is LARGE

Concentration of reactants > concentration of products

POSITION OF EQILIBRIUM:
LHS for the concentration of the reactants to be HIGH

25
Kc=1
Kc Fraction: The bottom and top are EQUAL Concentration of reactants = concentration of products POSITION OF EQUILIBRIUM: - Half way between the reactants and products