M5-S6: Further Practical Investigations of Le Chatelier's Principle

 

Formation of hydrogen iodide

• Hydrogen gas and iodine gas form an equilibrium mixture to produce hydrogen iodide (also a gas).

$$H_{2(g)} + I_{2(g)} \rightleftharpoons 2HI_{(g)}$$

 

• While hydrogen gas and hydrogen iodide are both colourless, iodine gas has a strong dark blue/purple colour.

 

Concentration/partial pressure

• Addition of hydrogen gas into a closed reaction chamber (closed system) will increase the rate of forward reaction. This causes reactants to be consumed to produce more hydrogen iodide. The solution will become less purple as the reaction proceeds.

 

• Removal of hydrogen gas favours the reverse reaction. By Le Chatelier’s principle. This causes [I₂] to increase gradually. The equilibrium mixture will turn more purple.

 

• Addition of iodine gas instantaneously increases its partial pressure (concentration), so the purple colour will initially become more intense. By Le Chatelier’s principle, reactants will be consumed to produce more hydrogen iodide. The solution will gradually become less purple.

 

Pressure/volume

• Changes in pressure and volume affects the partial pressure (concentration) of reactants and products equally because the chemical equation has equal number of moles of gas on both sides. This means the colour of the solution will change with vessel’s volume but not the equilibrium position.

 

Temperature

• The forward reaction is endothermic.
• Increase in temperature favours the forward reaction (endothermic), which causes [HI] to increase. The reaction becomes less purple.
• Decrease in temperature favours the reverse reaction (exothermic), which causes [I₂] to increase. The reaction becomes more purple.

 

Practice Question

1. Using Le Chatelier’s principle, explain the changes that will occur when hydrogen iodide is removed from the system.

 

 

 

 

2. Using Le Chatelier’s principle, explain the changes that will occur when the pressure on the system is increased.