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.
Question on Molarity and Solubility
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Re: Question on Molarity and Solubility
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.
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.
Re: Question on Molarity and Solubility
In my opinion, the best tactics to maximize C2H4Cl2 production are removing C2H4Cl2 as it forms and adding Cl2. Retro Bowl
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Re: Question on Molarity and Solubility
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.