Question

The rate constant for a certain reaction is k = 5.40×10−3 s−1 . If the initial reactant concentration was 0.100 M, what will the concentration be after 17.0 minutes?

Answer 1

Answer : The concentration after 17.0 minutes will be, $4.05\times 10^{-4}M$

Explanation :

The expression for first order reaction is:

$[C_t]=[C_o]e^{-kt}$

where,

$[C_t]$ = concentration at time 't'  (final) = ?

$[C_o]$ = concentration at time '0' (initial) = 0.100 M

k = rate constant = $5.40\times 10^{-3}s^{-1}$

t = time = 17.0 min = 1020 s (1 min = 60 s)

Now put all the given values in the above expression, we get:

$[C_t]=(0.100)\times e^{-(5.40\times 10^{-3})\times (1020)}$

$[C_t]=4.05\times 10^{-4}M$

Thus, the concentration after 17.0 minutes will be, $4.05\times 10^{-4}M$

Answer 2

The concentration after 17 minutes is 4.1 * 10^-4 M.

First order reaction

From the unit of the rate constant ( s−1 ) we know that this is a first order reaction hence;

ln[A] = ln[A]o - kt

[A] = final concentration

[A]o = initial concentration

k = rate constant

t = time

[A] = e^ln[A]o - kt

[A] = e^[ln( 0.100) - (5.40×10−3 * 17 * 60)]

[A] = 4.1 * 10^-4 M

Learn more about first order reaction: https://brainly.com/question/867804