Question

a) The rate constant for a certain reaction is k = 8.20×10⁻³ s−1 . If the initial reactant concentration was 0.750 M, what will the concentration be after 10.0 minutes?

b) A zero-order reaction has a constant rate of 1.80×10⁻⁴M/s. If after 30.0 seconds the concentration has dropped to 9.00×10⁻²M, what was the initial concentration?

(Please explain how to do a problem like this, not just the answer. I appreciate any help. Thank you!)

Answer 1

a) It is a first order reaction. Because, rate constant has units of  s⁻¹ .

k =  8.20×10⁻³ s⁻¹ =0.0082  s⁻¹

   t = 10 min = 10 x 60 s = 600 s

[Ao] =  0.750 M

k = 1/t ln { [A]_o/[A]_t}

  [A]_t =   [A]_o e^-kt

     =  0.750 x e^{-(0.0082 x 600)}

   = 0.00547 M

  [A]_t = 0.00547 M

Therefore, concentration of reactant after 10.0 minutes = 0.00547 M

b) For a zero order reaction: [A]t = [A]o - kt ------------- Eq (1)

where [A]t is concentration at time t, [A]o is initial concentration and k is rate constant

Given that
t = 30.0 s, [A]t = 9 x 10^-2 M and k = 1.80×10⁻⁴ M/s

Substitute these values in Eq (1),

[A]t = [A]o - kt

9 x 10^-2 = [A]o - [1.80×10⁻⁴x 30.0]

[A]o = 9 x 10^-2 + [1.80×10⁻⁴x 30.0]

= 0.0954 M

[A]o = 0.0954 M

Therefore, initial concentration = 0.0954 M