The data below were determined for the reaction S_2O_8^2- + 3I^- (aq) rightarrow

Question

The data below were determined for the reaction S_2O_8^2- + 3I^- (aq) rightarrow 2SO^2-_4 + I^-_3 Write the rate law for this reaction Calculate the rate constant for this reaction What is the overall reaction order At 25 degree C, the rate constant for the first-order decomposition of a pesticide solution is 6.40 times 10^-3 min^-1. If the starting concentration of pesticide is 0.0314 M. what concentration will remain alter 62.0 min at 25 degree C? At 25 degree C, the second-order reaction: NOCl(g) rightarrow NO(g) + 1/2 Cl_2 (g) is 50% complete after 5.82 hours when the initial concentration of NOC1 is 4.46 mol/L. How long will it take for the reaction to be 75% complete? Which of the following reactions will occur spontaneously as written? Sn^4+ (aq) + Fe^3+ (aq) rightarrow Sn^2+ (aq) + Fe^2+ (aq) 3Fe(s) + 2Cr^3+ (aq) rightarrow 2Cr(s) + 3Fe^2+ (aq) ss Sn^4+ (aq) + Fe^2+ (aq) rightarrow Sn^2+ (aq) + Fe(s) 3Sn^++ (aq) + 2Cr(s) rightarrow 2Cr^3+ (aq) + 3Sn^2+ (aq) 3Fe^2+ (aq) rightarrow Fe(s) + 2Fe^3+ (aq)

Explanation / Answer

Solution:- (10) If we take a look at exprement (1) and (2) then concentration of [I-] is constant and that of [S2O8-2] is doubled, same time the rate is also doubled it means the reaction is first order with respect to [S2O8-2].

Now if we compare experement (2) to (3) then concentration of [S2O8-2] is constant where as that of [I-] is halved, same time the rate is also halved. It again indicates a first order reaction.

So, the reaction is first order with respect to each of the reactant.

(A) the rate law would be..   rate = k [S2O8-2] [I-]

(B) To calculate the rate constant we could use the values from any of the experiment and plug in them to the rate law equation...

Let's say first experiment.

1.4 x 10-5 M.s-1 = k (0.038M) (0.060M)

k = 1.4 x 10-5 M.s-1/(0.038M) (0.060M)

k = 0.102 M-1.s-1

so, the rate constant is 0.102 M-1.s-1

(c) Over all reaction order is the sum of order of reactions with respect to each reactant. So, the over all order is 1+1 = 2 (second order reaction)

(11) First order equation is...

ln(A) = -kt + ln(A0)

where A0 is initial amount or concentration, A is final amount, k is rate constant and t is time.

Let's plug in the given values...

ln(A) = - 6.40 x 10-3 x (62.0) + ln(0.0314)

ln(A) = -0.3968 - 3.46 = -3.8568

taking anti ln to both sides...

A = e-3.8568

A = 0.0211 M

So, the final concentration would be 0.0211 M.

(12) 50% complete after 5.82 hours that means the half life is 5.82 hours.

initial concentration, (A0) = 4.46 mol/L or M

how long will it take for the reaction to be 75% completed.

If reaction is 75% completed then it means 75% of initial concentration is used up and 25% would be remaining.

So, 75% of 4.46 M would be = 4.46 M x 75/100 = 3.345 M

remaining concentration, (A) = 4.46 M - 3.345 M = 1.115 M

we know that, k = 0.693/half life = 0.693/5.82 hours = 0.119 hour-1

let's plug in the values in first order equation...

ln(1.115) = - 0.119 hour-1(t) + ln(4.46)

0.109 = - 0.119 hour-1(t) +1.495

0.109 - 1.495 = - 0.119 hour-1(t)

-1.386 = - 0.119 hour-1(t)

t = 1.386/0.119 hour-1

t = 11.65 hours

So, the 75% reaction will be completed in 11.65 hours.

(13) For redox reactions we could use the standard redox reaction potentials from the table to predict if the equation is spontaneous or not.

The reaction is spontaneous if cell potential is positive and non spontaneous if cell potential is negative.

In equation (A) the reduction is taking place for both the species so it does not look the right one.

half reactions for

equation (C) is also not the right one since the reduction is taking place for both the species.

The half equations for reaction (B) are...

3Fe(s) -------> 3Fe2+(aq) + 6e-       E0 = 0.036 V       (Oxidation)

2Cr3+(aq) + 6e- ------> 2Cr(s)       E0 = -0.73 V         (Reduction)

cell potential = -0.73V + 0.036V = -0.694V

cell potential is negative so it is not the spontaneous reaction.

For equation (D) the half reaction are...

2Cr(s) -------> 2Cr3+(aq) + 6e-           E0 = 0.73V

3Sn4+(aq) + 6e- ----> 3Sn2+(aq)   E0 = 0.15V

cell potential = 0.15V + 0.73V = 0.88V

cell potential is positive so the reaction is spontaneous.

For reaction (E) the half equations are..

2Fe2+(aq) ------------> 2Fe3+(aq) + 2e-          E0 = -0.77V

Fe2+(aq) + 2e-   --------> Fe(s)                      E0 = -0.44V

cell potential = -0.44V +(-.77V) = -1.21V

the cell potential is negative so it is not a spontaneous cell.

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