Chemistry contact process

1. CONTACT PROCESS
IBRAHIM ALZIR

2. 1) TEMPERATURE
• 2SO2 (g) + O2 (g) ↔ 2SO3 (g) +E H = - 197 KJ
• The forward reaction is exothermic, so increasing temperature will shift
equilibrium to the left to use that increased energy in order to reduce the
disturbance. Reverse endothermic reaction is favored.
• Decreasing the temperature will shift the equilibrium to the right to replace
the lost heat in order to reduce the disturbance. Forward exothermic
reaction is favored.
• Optimum Temp to increase yield of sulfur trioxide is 450C

3. 2) REACTANT CONCENTRATION
• 2SO2 (g) + O2 (g) ↔ 2SO3 (g) H = - 197 KJ
• An increase in the concentration of the reactants will increase the rate of the
forward reaction, and hence equilibrium will shift to the right in order to decrease
the disturbance.
• A decrease in the concentration of the reactants will decrease the rate of the
forward reaction and hence equilibrium will shift to the left in favor of the reverse
reaction to decrease the disturbance.

4. 3) CATALYST
• 2SO2 (g) + O2 (g) ↔ 2SO3 (g) H = - 197 KJ
• Adding a catalyst (Vanadium (V) oxide, V2O5) will not be considered a
disturbance as it only lowers the activation energy of both reaction in the same
amount, and hence
• Equilibrium is reached in less time
• Rates of reactions increase equally
• The position of equilibrium is not altered
• Concentrations of neither reactants nor products are changed

5. 4) PRESSURE
• 2SO2 (g) + O2 (g) ↔ 2SO3 (g) +E H = - 197 KJ
• Increasing the pressure by decreasing the volume will increase the rate of the
forward reaction more as it has higher moles and thus more affected by changing
the pressure which increases the yield of sulfur trioxide as equilibrium shifts to
the right.
• Decreasing pressure by decreasing the volume will highly decrease the rate of the
forward reaction as it has higher moles which shifts equilibrium to the left
decreasing the yield of sulfur trioxide.
• An optimum pressure is a 2atm