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
To determine the correctness of the rate expression for the given reaction, we need to consider the stoichiometry of the reaction.
The balanced equation of the reaction is:
3I-(aq) + H3AsO4(aq) + 2H+(aq) -> I3-(aq) + H3AsO3(aq) + H2O(l)
From the balanced equation, we can see that the coefficient of I- is 3 and the coefficient of H3AsO4 is 1.
To write the rate expression, we consider the change in concentration of reactants or products over time. The rate of the reaction can be expressed as the change in concentration of any reactant or product divided by the change in time.
In this case, you correctly identified that the rate expression can be written as:
1/3 * (Δ[I-] / Δt) = (Δ[H3AsO4] / Δt)
This rate expression is correct for this reaction because the coefficient of I- is 3. Therefore, dividing Δ[I-] by 3 to account for the stoichiometry gives you the correct rate expression.
Additionally, since the coefficient of H3AsO4 is already 1, there is no need to divide the rate of change of its concentration by any factor for stoichiometric consistency.
Hence, your given rate expression is correct for the provided reaction.