Question

The decomposition of \rm XY is second order in \rm XY and has a rate constant of6.96Ã10â3M^{-1} \cdot s^{-1} at a certain temperature.

a) What is the half-life for this reaction at an initialconcentration of 0.100 {\rm M}?
b) How long will it take for the concentration of \rm XY to decrease to 12.5% of its initial concentration whenthe initial concentration is 0.100 {\rm M}?
c) How long will it take for the concentration of \rm XY to decrease to 12.5% of its initial concentration whenthe initial concentration is 0.200 {\rm M}?
d) If the initial concentration of \rm XY is 0.160 M, how long will it take for the concentration to decreaseto 6.20Ã10â2M?
Express your answer using twosignificant figures.
e) If the initial concentration of{\rm{SO}}_2 {\rm{Cl}}_2 is 0.050 {\rm M}, what is the concentration of \rm XY after 55.0 s?
Express your answer using twosignificant figures.
f) If the initial concentration of\rm XY is 0.050 {\rm M}, what is the concentration of \rm XY after 500 s?
Express your answer using twosignificant figures.

Answer 1

Step-by-step explanation:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

Half life for second order kinetics is given by:

`t_(1/2)=(1)/(k* a_0)`

k = rate constant =?

`a_0` = initial concentration

a = concentration left after time t

Integrated rate law for second order kinetics is given by:

`(1)/(a)=kt+(1)/(a_0)`

a) Initial concentration of XY =
`a_o=0.100 M`

Rate constant of the reaction = k =
`6.96* 10^(-3) M^(-1) s^(-1)`

Half life of the reaction is:

`t_(1/2)=(1)/(6.96* 10^(-3) M^(-1) s^(-1)* 0.100 M)`

`=1,436.78 s`

1,436.78 seconds is the half-life for this reaction.

b) Initial concentration of XY = 0.100 M

Final concentration after time t = 12.5% of 0.100 M = 0.0125 M

`(1)/(0.0125 M)=6.96* 10^(-3) M^(-1) s^(-1)* t+(1)/(0.100M)`

Solving for t;

t = 10,057.47 seconds

In 10,057.47 seconds the concentration of XY will become 12.5% of its initial concentration.

c) Initial concentration of XY = 0.200 M

Final concentration after time t = 12.5% of 0.200 M = 0.025 M

`(1)/(0.025 M)=6.96* 10^(-3) M^(-1) s^(-1)* t+(1)/(0.200M)`

Solving for t;

t = 5,028.73 seconds

In 5,028.73 seconds the concentration of XY will become 12.5% of its initial concentration.

d) Initial concentration of XY = 0.160 M

Final concentration after time t =
`6.20* 10^(-2) M`

`(1)/(6.20* 10^(-2) M)=6.96* 10^(-3) M^(-1) s^(-1)* t+(1)/(0.200M)`

Solving for t;

t = 1,419.40 seconds

In 1,419.40 seconds the concentration of XY will become
`6.20* 10^(-2) M`.

e) Initial concentration of
`SO_2Cl_2= 0.050 M`

Final concentration after time t = x

t = 55.0 s

`(1)/(x)=6.96* 10^(-3) M^(-1) s^(-1)* 55.0 s+(1)/(0.050 M)`

Solving for x;

x = 0.04906 M

The concentration after 55.0 seconds is 0.04906 M.

f) Initial concentration of XY= 0.050 M

Final concentration after time t = x

t = 500 s

`(1)/(x)=6.96* 10^(-3) M^(-1) s^(-1)* 500 s+(1)/(0.050 M)`

Solving for x;

x = 0.04259 M

The concentration after 500 seconds is 0.0.04259 M.