The rate constant for a certain reaction is kkk = 8.70×10−3 s−1s−1 . If the initial reactant concentration was 0.150 MM, what will the concentration be after 19.0 minutes?

7.40 x 10⁻⁶ M

Step-by-step explanation:

Our strategy  here is to use the definition for the rate which is

rate = - ΔA/Δt  = k[A]^n

where  k is the rate rate constant raised to the power of n, the reaction order respect to A and [A] is the concentration, M, of A.

Now if the rate is first order we can see that k will have units of s⁻¹ which is what we are told since the rate has units of M/s ( assuming the time is in seconds) . Therefore we know we have a first order reaction:

rate = k[A]

and its integrated rate law wil be

ln [A]/[A]₀ = - kt

where t is the time, [A] and [A]₀ are the concentrations after time t and inital concentration respectively. Thus, we have all the information required to solve this question:

ln( [A]/0.150 ) = - 8.70 x 10⁻³ s⁻¹ x (19.0 min x 60 s/min)

ln ( [A]/0.150 ) = - 9.92

Taking inverse log function to solve for [A]

[A] /0.150 = 4.9 x 10⁻⁵

[A] = 7.40 x 10⁻⁶ M

Note: One can always deduce the order of the reaction if we are given the units of the rate constant k.

For example for a second order reaction it is M⁻1s⁻1 if we are working in  units of second. For third order it is M⁻²s⁻¹, and hopefully you can see the pattern.

20days louise can read300 pg

step-by-step explanation:

1 day 15 pg 300: 15= 20

x day 300 pg