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What is the effect of doubling the concentration of reactants on the rate of the reaction between potassium iodide and hydrochloric acid?

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ago by (139k points)
The reaction between potassium iodide  KI  and hydrochloric acid  HCl  can be represented by the following equation:2 KI  aq  + 2 HCl  aq   2 KCl  aq  + I2  aq  + H2  g To determine the effect of doubling the concentration of reactants on the rate of the reaction, we need to know the order of the reaction with respect to each reactant. The rate law for this reaction can be expressed as:Rate = k [KI]^m [HCl]^nwhere k is the rate constant, m and n are the orders of the reaction with respect to KI and HCl, respectively.Assuming the reaction is first order with respect to both KI and HCl  m = 1 and n = 1 , the rate law would be:Rate = k [KI] [HCl]If we double the concentration of both reactants, the new rate would be:New Rate = k [2KI] [2HCl] = 4k [KI] [HCl]Comparing the new rate to the initial rate, we can see that doubling the concentration of both reactants results in a four-fold increase in the rate of the reaction, assuming the reaction is first order with respect to both reactants.However, it is important to note that the actual order of the reaction with respect to each reactant must be determined experimentally. If the reaction is not first order with respect to both reactants, the effect of doubling the concentration of reactants on the rate of the reaction will be different.
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