Question

Consider the reaction of peroxydisulfate ion (S2O2−8) with iodide ion (I−) in aqueous solution: S2O2−8(aq)+3I−(aq)→2SO2−4(aq)+I−3(aq). At a particular temperature the rate of disappearance of S2O2−8 varies with reactant concentrations in the following manner: Experiment S2O2−8(M) I−(M) Initial Rate (M/s) 1 0.018 0.036 2.6×10−6 2 0.027 0.036 3.9×10−6 3 0.036 0.054 7.8×10−6 4 0.050 0.072 1.4×10−5 What is the rate of disappearance of I− when [S2O2−8]= 1.8×10−2 M and [I−]= 5.0×10−2 M ?

Answer 1

Answer:

r = 3.61x$10^{-6}$ M/s

Explanation:

The rate of disappearance (r) is given by the multiplication of the concentrations of the reagents, each one raised of the coefficient of the reaction.

r = k.$[S2O2^{-8} ]^{x} x [I^{-} ]^{y}$

K is the constant of the reaction, and doesn't depends on the concentrations. First, let's find the coefficients x and y. Let's use the first and the second experiments, and lets divide 1º by 2º :

$\frac{r1}{r2} = \frac{0.018^{x} x0.036^{y} }{0.027^xx0.036^y}$

$\frac{2.6x10^{-6}}{3.9x10^{-6}} = (\frac{0.018}{0.027})^xx(\frac{0.036}{0.036})^y$

$0.67 = 0.67^x$

x = 1

Now, to find the coefficient y let's do the same for the experiments 1 and 3:

$\frac{r1}{r3} = \frac{0.018x0.036^y}{0.036x0.054^y}$

$\frac{2.6x10^{-6}}{7.8x10^{-6}} = (\frac{0.018}{0.036})x(\frac{0.036}{0.054})^y$

$0.33 = 0.5x 0.67^y$

$0.67 = 0.67^y$

y = 1

Now, we need to calculate the constant k in whatever experiment. Using the first :

$2.6x10^{-6} = kx0.018x0.036$$kx6.48x10^{-4} = 2.6x10^{-6}$

k = 4.01x10^{-3} M^{-1}s^{-1}[/tex]

Using the data given,

r = $4.01x10^{-3}x1.8x10^{-2}x5.0x10^{-2}$

r = 3.61x$10^{-6}$ M/s