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(a q). 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 ?

r = 3.61x 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.

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 :

x = 1

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

y = 1

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

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

Using the data given,

r =

r = 3.61x M/s

n2(g) + 3h2-> 2nh3(g)   this is the balanced equation

note the mole ratio between n2, h2 and nh3.   it is 1 : 3 : 2   this will be important.

moles n2 present = 28.0 g n2 x 1 mole n2/28 g = 1 mole n2 present

moles h2 present = 25.0 g h2 x 1 mole h2/2 g = 12.5 moles h2 present

based on mole ratio, n2 is limiting in this situation because there is more than enough h2 but not enough n2.

moles nh3 that can be produced = 1 mole n2 x 2 moles nh3/mole n2 = 2 moles nh3 can be produced

grams of nh3 that can be produced = 2 moles nh3 x 17 g/mole = 34 grams of nh3 can be produced

note:   the key to this problem is recognizing that n2 is limiting, and therefore limits how much nh3 can be produced.