He second-order decomposition of hi has a rate constant of 1.80 x 10-3 m-1s-1. how much hi remains after 27.3 s if the initial concentration of hi is 4.78 m? a. 4.55 m b. 0.258 m c. 3.87 m d. 2.20 m e. 2.39 m

The amount of HI that remains after 27.3 s is 3.87 M.

The formula for second order reaction is;

1/[A]t = 1/[A]0 + kt

where

 [A]t = concentration of A at time t

 [A]0 = initial concentration of A  

 k = the rate constant

From the information provided;

[A]t = ?

[A]0 =  4.78 M

k = 1.80 x 10-3 m-1s-1

t = 27.3 s

Substituting values;

1/[A]t = 1/4.78  + (1.80 x 10-3 * 27.3)

1/[A]t =   0.20921  + 0.04914

[A]t = 3.87 M

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c. 3.87 M. After 27.3 s, the concentration of the remaining HI is 3.87 M

Whenever a question asks you, "How long to reach a certain concentration?", you must use the appropriate integrated rate law expression.

The integrated rate law for a second-order reaction is

1/[A]t = 1/[A]0 + kt

where

• [A]t = the concentration of A at time t

• [A]0 = the concentration of A at time 0 (i.e., at the beginning of the reaction)

•k = the rate constant for the reaction

1/[A]t = 1/(4.78 M) + 1.80 × 10^(-3) M·s^(-1) x 27.3 s

= 0.2092 M^(-1) + 0.04914 M^(-1) = 0.2583 M^(-1)

[A]t = 1/[0.2583 M^(-1)] = 3.87 M

yes

explanation:

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