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Reversible Reactions and Equilibrium

  • In a reversible reaction, you can change the direection of the reaction by changing the reaction conditions.
  • The symbol ⇄ is used to indicate that a reaction is reversible.
  • A reversible reaction in a closed system is at equilibrium when:
    • the rate of the forward reaction is equal to the rate of the reverse reaction
    • the concentrations of reactants and products are no longer changing

Changing conditions to change direction

  • The general reversible reaction for hydrated/anhydrous compounds is:
    • hydrated salt ⇄ anhydrous salt + water

Effect of heat on hydrated compounds

  • The forward reaction, hydrated → anhydrous + water, is endothermic because it requires heat to break the bonds between the salt and water moleucules.
    • When heat is applied, the equilibrium shifts to the right (forward reaction)
    • The hydrated compount loses its water of crystallization, forming the anhydrous compound.
    • This is essentially thermal decomposition.
  • Example: Blue hydrated copper(II) sulfate, $CuSO_{4}·5H_{2}O$, is heated, then it turns white as it becomes $CuSO_{4}$, releasing water vapor.
    • Equation: $CuSO_{4}·5H_{2}O → CuSO_{4} + 5H_{2}O$

Addition of water to anhydrous compounds

  • The reverse reaction, anhydrous + water → hydrated, is exothermic because energy is released when water bonds to the salt.
    • When water is added to the anhydrous salt, the equilibrium shifts to the left (reverse direction).
    • The anhydrous compound absorbs the water, reforming the hydrated compound.
    • This is the reverse of dehydration.
  • Example: White anhydrous $CuSO_{4} has water added → it turns blue again as $CuSO_{4}·5H_{2}O$ reforms, and the mixture gets warm (exothermic).
    • Equation: $CuSO_{4} + 5H_{2}O → CuSO_{4}·5H_{2}O$

Shifting the equilibrium

  • When a reversible reaction is in equilibrium and you make a change, the system acts to oppose the change, and restore equilibrium. A new equilibrium mixture forms.

Change in temperature

  • A high temperature favors the endothermic direction.
    • TIP: If ΔH is given, it is endo. if it is +Δ.
  • A low temperature favors the exothermic direction.
    • TIP: If ΔH is given, it is exo. if it is -Δ.

Change in pressure (GASES ONLY)

  • A high pressure shifts to the side with fewer moles of gas.
  • A low pressure shifts to the side with more moles of gas.
  • There is no effect is the gas moles are equal on both sides.

Change in concentration

  • A high concentration of a reactant shifts the equilibrium to use up a reactant.
  • A low concentration of a reactant shifts the equilibrium to replace it.

Using a catalyst

  • There is NO CHANGE in position of equilibrium. Rate of both forward and reverse reactions increase equally.
    • Remember that a catalyst speeds up the rate of a reaction without it being used up. It does not affect the equilibrium.

The Haber process

  • The equation for the Haber process:
    • $N_{2}(g) + 3H_{2}(g) ⇄ 2NH_{3}(g)$
    • nitrogen + hydrogen ⇄ ammonia

How ammonia is made

  1. Nitrogen is obtained by:
    • reacting natural gas, methane with steam
    • cracking hydrocarbons from petroleum. The two gases are mixed and scrubbed to remove impurities.
  2. The mixture is compressed. More gas is pumped in until pressure reaches 200 atm.
  3. The compressed gas flows into the converter, a round tank with beds of iron at 450°C, where the key reaction occurs. Iron is the catalyst.
  4. The mixture is cooled until the ammonia condenses to a liquid. The nitrogen and hydrogen are recycled to the converter for another chance to react. Steps 3 and 4 are continually repeated.
  5. The ammonia is run into tanks and stored as a liquid under pressure.
  • The typical conditions in the Haber process:
    • 450°C to give a faster rate
    • 200 atm / 20,000 kPa is safer and saves money
    • an iron catalyst to speed up the reaction but does not affect yield.

The contact process

  • The equation for the Contact process:
    • $2SO_{2}(g) + O_{2}(g) ⇄ 2SO_{3}(g)$
    • sulfur dioxide + oxygen ⇄ sulfur trioxide
  1. Sulfur is burned in air.
  2. The equation to make sulfur dioxide is $S(s) + O_{2}(g) → SO_{2}(g)$. The sulfur dioxide is mixed with more air.
    • Sulfur dioxide is obtained by burning sulfur or roasting sulfur ores and oxygen (air).
  3. It is then passed over four beds of catalyst, pellets of vanadium(V) oxide at 450°C.
  4. The sulfur trioxide gas is dissolved in concentrated sulfuric acid. Any unreacted sulfur dioxide is recycled.
    • Concentrated sulfuric acid, $H_{2}SO_{4}$:
      • $H_{2}O(l) + SO_{3}(g) → H_{2}SO_{4}$
  5. Oleum, a thick fuming liquid, is mixed carefully with water.
  • The typical conditions in the Contact process:
    • 450°C for the catalyst since it is inactive below 400°C
    • 2 atm / 200 kPa is low for acceptable yield
    • vanadium(V) oxide catalyst to give a faster rate.