Answer:
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.
