Question

The rate constant for a certain reaction is k = 4.70×10−3 s−1 . If the initial reactant concentration was 0.700 M, what will the concentration be after 4.00 minutes? Express your answer with the appropriate units.

Answer 1

Answer:

Therefore the concentration of the reactant after 4.00 minutes will be 0.686M.

Explanation:

The unit of k is s⁻¹.

The order of the reaction = first order.

First order reaction: A first order reaction is  a reaction in which the rate of reaction depends only the value of the concentration of the reactant.

$-\frac{d[A]}{dt} =kt$

[A] = the concentration of the reactant at time t

k= rate constant

t= time

Here k= 4.70×10⁻³ s⁻¹

t= 4.00

[A₀] = initial concentration of reactant = 0.700 M

$-\frac{d[A]}{dt} =kt$

$\Rightarrow -\frac{d[A]}{[A]}=kdt$

Integrating both sides

$\Rightarrow\int -\frac{d[A]}{[A]}=\int kdt$

⇒ -ln[A] = kt +c

When t=0 , [A] =[A₀]

-ln[A₀]  = k.0 + c

⇒c= -ln[A₀]  

Therefore

-ln[A] = kt - ln[A₀]

Putting the value of k, [A₀] and t

- ln[A] =4.70×10⁻³×4 -ln (0.70)

⇒-ln[A]=  0.375

⇒[A] = 0.686

Therefore the concentration of the reactant after 4.00 minutes will be 0.686M.