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

Answer:

5.25*10^-2 s^-1

Explanation:

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

$-\frac{d[B]}{dt}=k[B] - - - -\frac{d[B]}{[B]}=k*dt$

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

$\int\limits^B_B \,-\frac{ d[B] }{[B]}= \int\limits^t_t \, k*dt$

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

$k=\frac{ln[B]_{1}- ln[B]_{2}}{t_{2}-t_{1}}$

$k=\frac{ln(7.60 *10^{-2})-ln(5.50*10^{-3})}{85s-35s}=5.25*10^{-2}s^{-1}$

Answer 2

Answer:

k = -0.0525 s⁻¹

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