what is the rate law of S2O8^2- + 3I- ---> 2SO4^2- + I3-
idk
To determine the rate law of a chemical reaction, we need to analyze the dependence of the reaction rate on the concentrations of the reactants. The general rate law expression is given by:
Rate = k[A]^m[B]^n
Where:
- Rate represents the reaction rate
- k is the rate constant
- [A] and [B] represent the concentrations of the reactants A and B, respectively
- m and n are the reaction order with respect to reactants A and B, respectively.
In the given reaction: S2O8^2- + 3I^- ---> 2SO4^2- + I3-
To determine the rate law, we need to experimentally measure the reaction rate at different concentrations of the reactants and analyze the results. By comparing the rate when the concentrations of each reactant are changed individually, we can determine the reaction order for each reactant.
For example, if we keep [S2O8^2-] constant and vary [I^-], and observe that the rate doubles when [I^-] doubles, then the reaction is first-order with respect to I^- ([I^-]^1).
Similarly, if we keep [I^-] constant and vary [S2O8^2-], and observe that the rate quadruples when [S2O8^2-] doubles, then the reaction is second-order with respect to S2O8^2- ([S2O8^2-]^2).
Based on this analysis, the rate law for the reaction S2O8^2- + 3I^- ---> 2SO4^2- + I3- can be determined as follows:
Rate = k[S2O8^2-]^2[I^-]^1
Please note that the actual values of m and n, the reaction orders, can only be determined experimentally. The rate constant (k) can also be determined by conducting experiments at known concentrations of the reactants and measuring the corresponding reaction rates.