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Equilibria (Chapter 7)
Note: This is a final version of the notes for chapter 7. There will not be any further version of notes for this, and this will be used for the final exam as preparation material Reversible Reaction: a reaction in which products can react to form the original reactants Features of Equilibrium:
- It is dynamic
- Rate of forward reaction is equal to rate of backward reaction
- Concentration of reactants and products are the same
- Closed System Note: 1. Dynamic means continuously reacting 2. Rate of breakdown of products = Rate of formation of products 3. Concentration is constant only when point of equilibrium is reached 4. Reaching equilibrium is why we donโt get our products 5. Equilibrium only occurs as gas or liquid Examples of Reversible Reactions: N 2 + 3H 2 โ 2 NH 3 2 Na + Cl 2 โ 2 NaCl [100% yield irreversible] N 2 + H 2 โ 2 NH 3 [10-15% yield reversible] Note: โ Everytime 2 mol of NH 3 breaks, we get 1 mol of N 2 and 3 mol of H 2 โ Everytime 1 mol of N 2 and 3 mol of H 2 react, we get 2 mol of NH 3 H 2 I 2 HI ratio 1 1 2 initial 5 5 0 change -4.32 -4.32 +8. equilibrium moles 0.68 0.68 8.
Le Chatelierโs Principle: if one or more factors that affect equilibrium is changed, the position of equilibrium shifts in the direction that opposes the change Factors affecting Equilibrium:
- Concentration
- Pressure
- Temperature
- Catalysts Concentration: โ Increasing the concentration of reactants shifts the equilibrium to the right โ Increasing the concentration of products shifts the equilibrium to the left Pressure: โ Increasing the pressure shifts the equilibrium to the side with fewer moles โ Decreasing the pressure shifts the equilibrium to the side with more moles Temperature: โ Increasing the temperature shifts the equilibrium in the endothermic reaction path โ Decreasing the temperature shifts the equilibrium in the exothermic reaction path Catalysts: โ A catalyst is a substance that increases the rate of a chemical reaction โ Catalysts speed up time taken to reach equilibrium, but have no effect on position โ Catalysts increase rate of forward and backwards reaction
Equilibrium Constant of Concentration Kc: N 2 + 3 H 2 โ 2 NH 3 Molar Ratio of Reactants and Products 1 3 2 Initial Moles 3 6 0 Change -0.5 -1.5 + Equilibrium Moles [ Initial Moles - Change ] 2.5 4.5 1 Equilibrium Concentration [ equilibrium moles / volume ]
mol dm-
mol dm-
mol dm- A mixture containing 0.5 mol of CO 2 , 0.5 mol of H 2 , 0.2 mol of CO, 0.2 mol of H 2 O was placed in a 1 dm^3 flask and allowed to come to equilibrium at 1200 K Calculate the amount in moles, of each substance present in the equilibrium mixture at 1200 K. [ CO 2 + H 2 โ CO + H 2 O ] CO 2 H 2 โ CO H 2 O initial moles 0.5 0.5 0.2 0. change -x -x +x +x equilibrium moles 0.5 - x 0.5 - x 0.2 + x 0.2 + x
A mixture of 0.02 mol of hydrogen and 0.02 mol of iodine was placed in 1 dm^3 flask and allowed to come to equilibrium at 650 K. Calculate the amount, in moles, of each substance present in the equilibrium mixture at 650K. H 2 I 2 โ 2HI initial moles 0.02 0.02 0 change -x -x +2x equilibrium moles 0.02 - x 0.02 - x 2x equilibrium concentration 0.02 - x 0.02 - x 2x A mixture containing 0.4 of CO, 0.4 mol of H 2 O, 0.2 mol of CO 2 and 0.2 mol of H 2 was placed in a 1 dm^3 flask and allowed to come to equilibrium at 1100 K CO (g) + H 2 O (g) โ CO 2 (g) + H 2 (g) Kc = 6.4 x 10-1^ at 1100 K CO H 2 O CO 2 H 2 initial moles 0.4 0.4 0.2 0. change -x -x +x +x equilibrium moles 0.4-x 0.4-x 0.2+x 0.2+x
Equilibrium expressions in terms of partial pressures are written in a similar way to equilibrium expressions in terms of concentrations:
- p โ partial pressure
- reactants / products written as subscripts after p
- number of moles of reactants / products is written as power after p
- no square brackets
- equilibrium constant K Nitrogen dioxide exists in equilibrium with dinitrogen tetroxide N 2 O 4 2 NO 2 (g) โ N 2 O 4 (g) 2 mol of dinitrogen tetroxide was sealed in a container at 350 K. After equilibrium had been established the total pressure was 104 kPa and the mixture of gases contained 1.84 mol of dinitrogen tetroxide NO 2 N 2 O 4 molar ratio 2 1 initial moles 0 2 change +0.32 -0. equilibrium moles
total moles 2. mole fraction 0.
total pressure
partial pressure x 140,
2.16 x 140,
Give the equation for the equilibrium constant Kp for this equilibrium a. Kp = ๐ [๐2๐4] ๐^2 [๐๐2]
Calculate the number of moles of NO 2 present at equilibrium and hence the mole fraction of each gas present at equilibrium a. 0.
Calculate the total number of moles of gas present at equilibrium and hence the mole fraction of each gas present at equilibrium a. total = 2.16 = 0.15 = 0. 0. 2. 1. 2.
Calculate the partial pressure of each gas present at equilibrium a. x 140,000 = 20, 741 Pa 0. 2.16 b.^ x 140,000 = 119,259 Pa 1. 2.
Calculate the value of Kp at 350 K
a. Kp = = 2.77 x 10-4^ =
119, (20,741) 2 ๐๐ ๐๐ 2 1 ๐๐ 2 mol of sulfur dioxide and 2 mol of oxygen were put in a flask and left to reach equilibrium. At equilibrium, the pressure in the flask was 2 x 10^5 Pa and the mixture contained 1.8 mol of sulfur trioxide 2 SO 2 (g) + O 2 (g) โ 2 SO 3 (g) SO 2 O 2 SO 3 molar ratio 2 1 2 initial moles 2 2 0 change -1.8 -0.9 +1. equilibrium moles
total moles 3. mole fraction 0.
total pressure 200, partial pressure 12,903 70,967 116,
Acids and Bases Acids โ Proton Donor / H+^ Donor Base โ Proton Acceptor / H+^ Acceptor Strong / Weak Acids and Bases: โ A strong acid / base is completely ionised in solution โ A weak acid / base is partially ionised in solution โ A concentrated acid/base has a large number of moles per unit volume of the acid/base โ A dilute acid/base has a small number of moles per unit volume of the acid/base Acid and Base: Arrhenius Theory: only applicable for aqueous solutions โ Acid โ acids dissociate to give H+ ions โ Bases โ alkalis dissociate to give OH-^ ions Bronsted-Lowry Definition: โ Acid โ Proton Donor / H+^ donor โ Bases โ Proton Acceptor / H+ acceptor
Bronsted-Lowry Acid: for an acid to behave as a proton donor, a base must be present to accept protons from it. Bronsted-Lowry Base: for a base to behave as a proton acceptor, an acid must be presented to donate protons to it. Conjugate Acid: it is the species that results when a base accepts a proton Conjugate Base: is the species that results when an acid donates a proton Example #1:
Indicators and Acid-Base Titrations An acid-base indicator changes color over a range of pH values.
- Many indicators are considered as weak acids in which the acid (Hln) and its conjugate base (ln-) have different colors Hln โ H+^ + ln- Un-ionised indicator color A Conjugate base color B Note: indicators usually change color over a pH range of between 1 and pH units. In the middle of the range, there is a recognisable end-point where the indicator has a color in between the two extremes of the color
Strong Acids with Strong Bases The diagram below shows how the pH changes when 0.1 mol dm-3^ of NaOH ( strong base ) is titrated with 0.1 mol dm-3^ of HCl ( strong acid ), in the presence of a bromothymol blue indicator. This graph shows that: A sharp fall in the graph line between pH 10.5 and
โ In that region, tiny additions of H+^ ions results in change of pH A midpoint of the steep slope at pH 7 โ This midpoint indicates the point where H+ ions of the acid have reacted with OH-^ ions in the alkali Bromothymol blue indicator changes from blue to yellow in the range of 7.6 to 7.0 where the slope is the steepest Strong Acids with Weak Bases The diagram below shows how the pH changes when 0.1 mol dm-3^ of aqueous NH 3 ( weak base ) is titrated with 0.1 mol dm-3^ of HNO 3 ( strong acid ) This graph shows that: โ The sharp fall in the graph line between pH 7.5 and pH 3. โ The midpoint of the steep slope is about pH 5 Because there is a sharp change in pH over the region of 3.5 to 7.5, we can use methyl red as an indicator for the titration
- This is because methyl red changes color between pH 4.2 and pH 6.3, values which correspond to regions of sharpest pH change
Tiny Notes
- Kc is only affected by temperature
- exothermic: Kc and temperature are inversely proportional
- endothermic: Kc and temperature are directly proportional
- Kp and Kc can only be changed by the temperature
- Kp is inversely proportional to temperature
- acids can act as bases, bases can also act as acids, stop falling for the name
- elements in the solid state can't be used for Kc
- when reactants/products are on the same side of the equation, they have equal partial pressures
- dynamic equilibrium: it is when the rate of forward reaction is equal to the rate of backward reaction
- large scale use of hydrogen in chemical industry is manufacture of Ammonia
- definition of weak base:
- a base is a proton acceptor
- a weak base doesn't fully ionize
- NH3 + H2O --> NH4+ + OH-
- converting SO3 + H2SO4:
- by dissolving in concentrated H2SO
- and then diluted with water
- Le Chatelier's Principle: If the conditions of a system in equilibrium is changed, the position of equilibrium moves so as to reduce that change
- SO3 is not directly dissolved in water because reaction would be too exothermic
- SO2 is used as an additive because it is a preservative / anti-oxidant
