Question

A substance decomposes with a rate constant of 9.05 × 10⁻⁴ s⁻¹. How long does it take for 28.0% of the substance to decompose?

Answer 1

Answer:

$ln(\frac{A_0}{A_t})=k_t\\ln(\frac{x}{1-0.15x})=(9.05x10^-^4)t\\ln(\frac{1}{0.85})=(9.05x10^-^4)t\\t=180 seconds$

Explanation:

Use the integrated rate law to solve for the time it takes for concentration A to decrease to the given amount. The units s⁻¹ indicate that this is a first-order reaction.

Answer 2

Answer:

The time required to decompose the substance is 360 seconds.

Explanation:

The unit of the given rate constant "k" is "s⁻¹" which shows that it is a first-order reaction.

And for the first-order reaction from chemical kinetics we have,

$k=\frac{2.303}{t}log \frac{R_o}{R}$        (1)

Where,

k=rate constant for the reaction

R₀=initial concentration of the reactant

R=remaning concentration of the reactant after time t

t=time taken for the change to occur

From the question we have,

k=9.05×10⁻⁴s⁻¹

R₀=100g

R=100-28=72g

By substituting the required values in equation (1) we get;

$t=\frac{2.303}{9.05\times 10^-^4}log \frac{100}{72}$

$t=\frac{2.303}{9.05\times 10^-^4}(log 100-log 72)$

$t=\frac{2.303}{9.05\times 10^-^4}\times 0.143$

$t=0.036\times 10^4sec$

$t=360 sec$

Hence, the time required to decompose the substance is 360 seconds.