Question

Molecular iodine, I2 (g), dissociates into iodine atoms at 545K with a first order rate constant of 0.344 1/s. If you start with 0.054 M I2 at this temperature how much will remain after 5.16 s?

Answer 1

Answer:

0.00915 M of $I_2$ remain after 5.16 seconds.

Explanation:

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,  

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration

Given that:

The rate constant, k = $0.344$ s⁻¹

Initial concentration $[A_0]$ = 0.054 M

Final concentration $[A_t]$ = ? M

Time = 5.16 s

Applying in the above equation, we get that:-

$[A_t]=0.054e^{-0.344\times 5.16}\ M=\frac{1\times \:0.054}{e^{1.77504}}\ M=0.00915\ M$

0.00915 M of $I_2$ remain after 5.16 seconds.