Question

It is found that a gas undergoes a first-order decomposition reaction. If the rate constant for this reaction is 8.1 x 10-2 /min, how long will it take for the concentration of the gas to change from an initial concentration of .1M to 1.0 x 10-2 M?

Answer 1

Answer:

28.43 min

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 = $8.1\times 10^{-2}$ min⁻¹

Initial concentration $[A_0]$ = 0.1 M

Final concentration $[A_t]$ = $1.0\times 10^{-2}$ M

Time = ?

Applying in the above equation, we get that:-

$1.0\times 10^{-2}=0.1e^{-8.1\times 10^{-2}\times t}$

$0.1e^{-8.1\times \:10^{-2}t}=10^{-2}$

$e^{-8.1\times \:10^{-2}t}=\frac{1}{10}$

$\ln \left(e^{-8.1\times \:10^{-2}t}\right)=\ln \left(\frac{1}{10}\right)$

$t=28.43\ min$