The reaction between selenous acid and the iodide ion in acid solutions is H_2 S

Question

The reaction between selenous acid and the iodide ion in acid solutions is H_2 SeO_3 (aq) + 6I^- (aq) + 4H^+ (aq) rightarrow Se(s) + 2I_3^- (aq) + 3 H_2 O(I). The data in the following table were measured at 0 degree C. Experiment [H_2 SeO_3]_0 (M) [H^+]_0 (M) [I^-]_0 (M) [I^_]_0 (M) Initial Rate [mol/(L middot s)] 1.00 times 10^-4 2.00 times 10^-2 3.00 times 10^-2 5.30 times 10^-7 2.00 times 10^-4 2.00 times 10^-4 2.00 times 10^-2 3.00 times 10^-2 1.06 times 10^-6 3.00 times 10^-4 4.00 times 10^-2 3.00 times 10^-2 6.36 times 10^-6 3.00 times 10^-4 8.00 times 10^-2 3.00 times 10^-2 2.54 times 10^-5 3.00 times 10^-4 8.00 times 10^-2 6.00 times 10^-2 2.04 times 10^-4 2.00 times 10^-4 2.00 times 10^-2 6.00 times 10^-2 8.48 times 10^-6 What is the reaction order with respect to I^- (aq)? Select one: a. 0 b. 1 c. 2 d. 3 e. 1/2

Explanation / Answer

Rate of the reaction in terms of concentration of reactants is given by:

rate = k [A]a [B]b.....,

where, k is rate constant

A , B .... are reactants and a , b... are orders with respect to reactants.

Considering data in question for expts. 4 and 5 we can take ratio of rates :

rate4 / rate5 = k([H2Se03]x[H+]y[I-]z)4 / k([H2Se03]x[H+]y[I-]z)5

Since k remains constant at a particular temp. so it will cancel in the ratio. Also H2Se03 and H+ will cancel because

[H2Se03]4 = [H2Se03]5 and [H+]4 = [H+]5

Therefore,

rate4 / rate5 = (3 x 10-6 / 6 x 10-6 )z = (2.54 x 10 -5 / 2.04 x 10-4)

Taking log on both sides, we get,

zlog (3 x 10-6 / 6 x 10-6 ) = log (2.54 x 10 -5 / 2.04 x 10-4)

solving we get z = 3.

Answer : d) 3

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