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
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 :
[tex]-\frac{d[B]}{dt}=k[B] - - - -\frac{d[B]}{[B]}=k*dt[/tex]
if we integrate between the initial concentration and the concentration at any time we get:
[tex]\int\limits^B_B \,-\frac{ d[B] }{[B]}= \int\limits^t_t \, k*dt[/tex]
Solution:
[tex]-(ln[B]-ln[B]_{o})=kt[/tex] (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) [tex] -(ln[B]_{1}-ln[B]_{o})=k*t_{1}[/tex]
(2) [tex] -(ln[B]_{2}-ln[B]_{o})=k*t_{2}[/tex]
With
[tex][B]_{1}= 7.60 *10^{-2} M, t_{1}=35s [/tex]
[tex][B]_{2}= 5.50*10^{-3} M, t_{2}=85s [/tex]
From (1)
[tex]ln[B]_{o}=k*t_{1}+ln[B]_{1}[/tex]
Replacing this value for [tex]ln[B]_{o}[/tex] on (2) we get
[tex]-ln[B]_{2}+( k*t_{1}+ln[B]_{1})=k*t_{2}[/tex]
Organizing
[tex] -ln[B]_{2}+ ln[B]_{1}= k*t_{2}- k*t_{1}[/tex]
With k equals to
[tex]k=\frac{ln[B]_{1}- ln[B]_{2}}{t_{2}-t_{1}}[/tex]
[tex]k=\frac{ln(7.60 *10^{-2})-ln(5.50*10^{-3})}{85s-35s}=5.25*10^{-2}s^{-1}[/tex]
