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What is the effect of doubling the concentration of reactants on the reaction rate of the iodine clock reaction between potassium iodide (KI) and hydrogen peroxide (H2O2)?

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The iodine clock reaction between potassium iodide  KI  and hydrogen peroxide  H2O2  is a classic example of a chemical reaction that demonstrates the effect of concentration on reaction rate. The overall reaction can be represented as:2 KI  aq  + H2O2  aq   2 KI  aq  + 2 H2O  l  + I2  aq The reaction rate depends on the concentrations of the reactants, KI and H2O2. According to the rate law, the reaction rate is proportional to the concentration of the reactants raised to their respective orders in the reaction:Rate = k [KI]^m [H2O2]^nwhere k is the rate constant, m and n are the orders of the reaction with respect to KI and H2O2, respectively.In the iodine clock reaction, the reaction is generally found to be first-order with respect to both KI and H2O2:Rate = k [KI] [H2O2]Now, let's consider the effect of doubling the concentration of both reactants. If we double the concentration of KI and H2O2, the new rate will be:New Rate = k  2[ KI]   2[ H2O2] Since the reaction is first-order with respect to both reactants, the new rate will be:New Rate = k  2 ^1  2 ^1 [KI] [H2O2] = 4 k [KI] [H2O2]Comparing the new rate to the original rate, we can see that doubling the concentration of both reactants results in a four-fold increase in the reaction rate. This is because the reaction rate is directly proportional to the product of the concentrations of the reactants, and in this case, both reactants have a first-order dependence.
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