A water bottling company is looking to achieve maximum production from the following reversible reaction. If the reaction has re
ached a point of equilibrium, what would be the best way to maximize production of water? 2 H2+O2⇔2 H2O Remove hydrogen and allow the reaction to reach equilibrium again. Stop of the process because no more water can be made after equilibrium is reached. Remove the already produced water and allow the reaction to reach equilibrium again. Remove oxygen and allow the reaction to reach equilibrium again.
Remove the already produced water and allow the reaction to reach equilibrium again.
Explanation:
<em>According to Le Chatelier principle, when a reaction is in equilibrium and one of the factors that influence the rate of reaction is altered, the equilibrium will shift so as to annul the effects of the change.</em>
If the product is continuously being removed from a reaction that is in equilibrium, more product will continued to be formed in another to annul the effect of reduction in the concentration of product.
Hence, in order to maximize production of water in the reaction, the product (water) needs to be removed and the reaction allowed to reach equilibrium again.
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