The oxidation of arsenate ion by cerium(IV) ion in aqueous solution: AsO 3 3- +

Question

The oxidation of arsenate ion by cerium(IV) ion in aqueous solution: AsO33- + 2 Ce4+ + H2O ---> AsO43- + 2 Ce3+ + 2 H+ is first order in AsO33- and second order in Ce4+. In an experiment to determine the rate law, the rate constant was determined to be 0.70 M-2s-1. Using this value for the rate constant, calculate the rate of the reaction in mole per liter per seond (Ms-1) when [AsO33-] = 3.2 x 10-2 M and [Ce-4+] = 0.60 M the book says the answer is 8.1x10-3Ms-1    im not sure how to do the steps to get that answer thank you

Explanation / Answer

AsO33- + 2 Ce4+ + H2O ---> AsO43- + 2 Ce3+ + 2 H+

r = k [ AsO33-] [ Ce4+]2

k = 0.7 M-2s-1

[ AsO33-] = 3.2 x 10^-2 M

[ Ce4+] = 0.6 M

r = 0.7 * 3.2 x 10^-2 * ( 0.6 ) ^2

r = 8.1 x 10^-3 Ms-1

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