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What is the effect of increasing concentration of the product on the rate of the reaction between iodine and propanone?

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ago by (139k points)
The reaction between iodine and propanone is an example of a reversible reaction, which can be represented by the following equation: CH3 2CO + I2   CH3 2CI + HIIn this reaction, propanone  acetone  reacts with iodine to form iodopropane and hydrogen iodide. The rate of this reaction depends on the concentrations of the reactants and products, as well as the temperature and presence of a catalyst.According to Le Chatelier's principle, if a system at equilibrium is subjected to a change in concentration, temperature, or pressure, the system will adjust itself to counteract the change and restore a new equilibrium. In the context of this reaction, if the concentration of the product  iodopropane or hydrogen iodide  is increased, the system will shift to counteract this change by favoring the reverse reaction, which consumes the products and forms more reactants  propanone and iodine .As a result, the rate of the forward reaction  formation of iodopropane and hydrogen iodide  will decrease, while the rate of the reverse reaction  formation of propanone and iodine  will increase. This shift in reaction rates will continue until a new equilibrium is established, with a lower rate of the forward reaction compared to the initial state.In summary, increasing the concentration of the product in the reaction between iodine and propanone will cause the rate of the forward reaction to decrease and the rate of the reverse reaction to increase, ultimately leading to a new equilibrium with a lower rate of product formation.
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