Question

The second order decomposition of HI has a rate constant of 1.80*10^-3/MS

A. How much HI remains after 72.5 s if the initial concentration of HI is 4.78m

B. How long does it take to reduce the concentration by 25% if the initial concentration is 4.78m

C. What is the half life at 4.78m initial concentration

Answer 1

A.

The concentration-time equation for a second order reaction is

1/[A]_t = kt + 1/[A]_o

Here k = 1.80 x 10^-3 M^-1S^-1

[A]_o = 4.78 M

t = 72.5 s

1/[A]_t = (1.80 x 10^-3 M^-1S^-1) (72.5 s) + (1 / 4.78 M)

1/[A]_t = 0.13 M^-1 + 0.21 M^-1

1/[A]_t = 0.34 M^-1

[A]_t = 1 / 0.34 M^-1

[A]_t = 2.94 M

B.

Here k = 1.80 x 10^-3 M^-1S^-1

[A]_o = 4.78 M

The concentration by 25%.

So, [A]_t = 25% of 4.78 M

= (25/100) x 4.78 M

= 1.195 M

t = ?

The concentration-time equation for a second order reaction is

1/[A]_t = kt + 1/[A]_o

1/ (1.195 M) = (1.80 x 10^-3 M^-1S^-1) (t) + (1 / 4.78 M)

0.84 M^-1 = (1.80 x 10^-3 M^-1S^-1) (t) + 0.21 M^-1

(1.80 x 10^-3 M^-1S^-1) (t) = 0.84 M^-1 - 0.21 M^-1

(1.80 x 10^-3 M^-1S^-1) (t) = 0.63 M^-1

t = 0.63 M^-1 / (1.80 x 10^-3 M^-1S^-1)

t = 350 s

C.

The half-life of a second order reaction is expressed as

t_1/2 = 1 / (k [A]₀)

t_1/2 = half life = ?

k = 1.80 x 10^-3 M^-1s^-1

[A]₀ = 4.78 M

So,

t_1/2 = 1 / [ (1.80 x 10^-3 M^-1s^-1) (4.78 M) ]

t_1/2 = 1 / (0.0086 s^-1)

t_1/2 = 116 s