To obtain the rate of the reaction
3I-(aq) + H3AsO4(aq) + 2H+(aq) --> I3-(aq) + H3AsO3(aq) + H2O (l)
I got that the rate is
1/3 delta [I-]/ delta t = delta [H3AsO4]/ delta t
is this correct
Yes, you are correct! The rate expression for a reaction can be determined by using the stoichiometric coefficients of the reactants and products in the balanced chemical equation.
In this case, the balanced equation for the reaction is:
3I^-(aq) + H3AsO4(aq) + 2H^+(aq) → I3^-(aq) + H3AsO3(aq) + H2O(l)
The rate expression can be determined by considering the change in concentration of each reactant divided by the corresponding change in time.
So, for the reactant iodide ions (I^-), you correctly determined the rate expression as 1/3 * (change in [I^-] / change in time). This is because the stoichiometric coefficient of I^- in the balanced equation is 3, and so the change in concentration is divided by 3 to reflect the ratio of reactants in the reaction.
Similarly, for the reactant H3AsO4, the rate expression is (change in [H3AsO4] / change in time).
Your rate expression is correct and accurately represents the stoichiometry of the reaction.