The oxidation of iodide ion by peroxydisulfate ion is described by the equation:

Question

The oxidation of iodide ion by peroxydisulfate ion is described by the equation: 3 I^- (aq) + S_2O_3^2- (aq) rightarrow I_3^- (aq) + 2 SO_4^2- (aq) Over a certain interval of time, it is found that the average rate of change in the concentration of sulfate ion is 3.0 middot 10^-3 M middot s^-1. i. The average rate of change in the concentration of iodide ion over the same interval of time is _____ M middot s^-1 ii. The average rate of change in the concentration of S_2O_3^2- over the same interval of time is _____ M middot s^-1 iii. The average rate of change in the concentration of I_3^- over the same interval of time is _____ M middot s^-1 For the reaction 2 NO(g) + O_2(g) rightarrow 2 NO_2(g) if is found that the rate of the reaction is given by: Rate = 2 Delta [NO_2]/Delta t = k[NO]^2[O_2] i. The overall reaction order is _____. ii. If the concentrations of both NO and O_2 are doubled, the reaction rate _____ (increases/decreases) by a factor of a. 1 b. 2 c. 3 d. 4 e. 5 f. 6 g. 8 h. None of the above; the reaction rate changes by a factor of _____. iii. The units of the rate constant, k, are _____.

Explanation / Answer

Q1.

[SO42] = +3*10^-3 M/s

i)

find the rate of Iodine ion.. I-

thn, the ratio is 3:2

so 3/2 is the factor

[I-] = 3/2*[SO42] = 3/2*(3*10^-3) M/s = -4.5*10^-3 M/s

ii)

Similarly

1 mol of S2O3-2 reacts per 2 mol of SO4-2 forms so

[S2O3-2] = 1/2*[SO4-2] = 1/2*3*10^-4 = -1.5*10^-4 M/s

iii

finally, the rate for I3-

once again, 1:2

[I3-] = 1/2*[SO4-2] = 1/2*3*10^-4 =+ 1.5*10^-4 M/s

Q2.

the overal rate is

[NO2]^2 [O2] ^1 = a + b = 2+1 =3

third order

ii

if we doulbe NO --> 2^2 = 4

if we doulbe O2 --> 2^1 = 2

4*2 = 8

the overall factor will be 8

iii

the units for third order:

k = 1/(M^2 -s )

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