Chemistry Forum

This is the chemical equation: C₂H₄(g)+Cl₂(g)⇌C₂H₄Cl₂(g)

If you were a chemist trying to maximize the amount of C₂H₄Cl₂ produced, which tactic(s) might you try? Assume that the reaction mixture reaches equilibrium.

Check all that apply.

removing C₂H₄Cl₂ from the reaction mixture as it forms

adding Cl₂

lowering the reaction temperature

increasing the reaction volume

I honestly have no idea how C₂H₄CL₂ can be maximized. I would think lowering the temperature, and maybe even increasing the reaction volume. Please help me understand what the question means and why or why not an option would make sense. This is not an exam question, just a homework question from Pearson Textbook.

Read the principles of Le Chatelier.

Here you make from two reactants one product. So it means decrease Volume or increase pressure will give more product .Figure out is the reaction exothermic or endothermic. Means consume or release heat. If it consumes rise temperature, in other case decrease temperature.
Then adding more reactants and removing more product also drive it to a maximum.

In my opinion, the best tactics to maximize C₂H₄Cl₂ production are removing C₂H₄Cl₂ as it forms and adding Cl₂. Retro Bowl

According to Le Chatelier's Principle, if C₂H₄Cl₂ is removed from the reaction mixture, the equilibrium will shift to produce more C₂H₄Cl₂ to counteract the loss. Therefore, removing C₂H₄Cl₂ will favor the forward reaction and increase the production of C₂H₄Cl₂.

According to Le Chatelier's principle, when a stress is applied to a system at equilibrium, the system will shift in a direction that relieves the stress.

To maximize the production of C₂H₄Cl₂ in the given reaction, several tactics can be employed. Firstly, removing C₂H₄Cl₂ from the reaction mixture as it forms would shift the equilibrium towards the product side, following Le Chatelier's Principle. Additionally, adding more Cl₂, a reactant, would increase the concentration of reactants, thus driving the reaction towards the product side. Lowering the reaction temperature would favor the exothermic reaction, which is the forward reaction in this case, leading to more C₂H₄Cl₂ formation. However, increasing the reaction volume would not significantly impact the equilibrium position since both reactants and products have the same number of moles of gas. Therefore, the most effective strategies would involve removing the product and increasing the concentration of the reactant, Cl₂, while also lowering the temperature to promote the desired reaction.