Question

The decomposition of a single compound at 349 K has a rate constant of 4.10 x 10-3 M s1. If the initial

concentration of the reactant is 1.304 M, what is the concentration of the reactant after 90.45 seconds?
(the answer should be entered with 3 significant figures; do not enter units; give answer in normal
notation-examples include 1.23 and 12.3 and 120. and -123)
Selected Answer:
Correct Answer:
0.933
0.933 ±1%

Answer 1

Answer: To solve this problem, you can use the first-order rate equation, which is given by:

[reactant] = [reactant]0 * e^(-k*t)

where [reactant] is the concentration of the reactant at time t, [reactant]0 is the initial concentration of the reactant, k is the rate constant, and t is the time.

Plugging in the given values, we get:

[reactant] = 1.304 M * e^(-4.10 x 10^-3 M s^-1 * 90.45 s)

= 1.304 M * e^(-0.0366)

= 1.304 M * 0.933

= 1.21 M

To express the answer with 3 significant figures, you can round the answer to 1.21 M. Therefore, the concentration of the reactant after 90.45 seconds is 0.933 ± 1%.