Question

The reactant concentration in a first-order reaction was 7.60 x 10-2 M after 35.0 s and 5.50 x 10-3 M after 85.0 s. What is the rate constant for this reaction? [Express or answer in units of s-1]

Answer 1

5.25*10^-2 s^-1

Step-by-step explanation:

For a first order reaction the rate of reaction just depends on the concentration of one specie [B] and it’s expressed as :

`-(d[B])/(dt)=k[B] - - -  -(d[B])/([B])=k*dt`

if we integrate between the initial concentration and the concentration at any time we get:

`\int\limits^B_B  \,-( d[B] )/([B])= \int\limits^t_t \, k*dt`

Solution:

`-(ln[B]-ln[B]_(o))=kt` (equation 1)

You can clear this equation to get a equation for [B] at any time but because we want to estimate k is easier to use this expression.

In equation 1 we don’t know the value of [B]o so we can’t clear directly to get the value of K, but we know the concentration at two different times. With this information, we can get a system with two mathematical unknowns and two equations that we can solve.

Equations:

(1)
`-(ln[B]_(1)-ln[B]_(o))=k*t_(1)`

(2)
`-(ln[B]_(2)-ln[B]_(o))=k*t_(2)`

With

`[B]_(1)= 7.60 *10^(-2) M, t_(1)=35s`

`[B]_(2)= 5.50*10^(-3) M, t_(2)=85s`

From (1)

`ln[B]_(o)=k*t_(1)+ln[B]_(1)`

Replacing this value for
`ln[B]_(o)` on (2) we get

`-ln[B]_(2)+( k*t_(1)+ln[B]_(1))=k*t_(2)`

Organizing

`-ln[B]_(2)+ ln[B]_(1)= k*t_(2)- k*t_(1)`

With k equals to

`k=(ln[B]_(1)- ln[B]_(2))/(t_(2)-t_(1))`

`k=(ln(7.60 *10^(-2))-ln(5.50*10^(-3)))/(85s-35s)=5.25*10^(-2)s^(-1)`

Answer 2

k = -0.0525 s⁻¹

Step-by-step explanation:

The equaiton for a first order reaction is stated below:

ln[A]=−kt+ln[A]₀.

[A] = 5.50 x 10⁻³ M

[A]₀ = 7.60 x 10⁻² M

t = 85.0 - 35.0 = 50.0 s

The rate constant is represented by k and can be calculated substituting the values given above:

k = (ln[A]₀ - ln[A])/t

k = (ln5.50 x 10⁻³ M - ln7.60 x 10⁻² M)/50.0s

k = -0.0525 s⁻¹