Question

The reaction $2I^{-}\left(aq\right)+S_2{O}_8^{2-}\left(aq\right)\to I_2\left(aq\right)+2S{O}_4^{2-}\left(aq\right)$ was studied ${25}^{0}C$. The following results were obtained.
where, Rate $=-\frac{d\left[S_2{O}_8^{2-}\right]}{dt}$

[I${}^{-}$]	$\left[S_2{O}_8^{2-}\right]$	Inital rate (mol L${}^{-1}$ S${}^{-1}$
0.08	0.040	$12.5\times {10}^{-6}$
0.04	0.040	$6.25\times {10}^{-6}$
0.08	0.020	$5.56\times {10}^{-6}$
0.032	0.040	$4.35\times {10}^{-6}$
0.060	0.030	$6.41\times {10}^{-6}$

The value of the rate constant is:

• A. $1.8\times {10}^{-3}L$ $mo{l}^{-1}{s}^{-1}$
• B. $0.9\times {10}^{-3}L$ $mo{l}^{-1}{s}^{-1}$
• C. $3.7\times {10}^{-3}L$ $mo{l}^{-1}{s}^{-1}$
• D. $4.5\times {10}^{-3}L$ $mo{l}^{-1}{s}^{-1}$

Answer 1

Answer: (C)

$3.7\times {10}^{-3}L$ $mo{l}^{-1}{s}^{-1}$

From first and second experiment, when the concentration of iodide ion is doubled from 0.040 M to 0.080 M, the rate of the reaction is doubled. The concentration of $S_2{O}_8^{2-}$ remains same. Thus the reaction is first order with respect to iodide ion.
From first and third experiment, when the concentration of $S_2{O}_8^{2-}$ is doubled from 0.020 M to 0.040 M, the rate of the reaction increases by 2.248 times. The concentration of iodide ion remains same. Thus the order of the reaction with respect to $S_2{O}_8^{2-}$ is 1.1687.
The expression for the rate of the reaction is as shown below.
$r=k\left[I^{-}\right]\times \left[S_2{O}_8^{2-}\right]$
Substitute values for the first experiment.
$12.5\times {10}^{-6}=k\left(0.08M\right)\left(0.040M\right)$
$k=3.7\times {10}^{-3}L$.