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

28.43 min

Step-by-step 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* 10^(-2)` min⁻¹

Initial concentration
`[A_0]` = 0.1 M

Final concentration
`[A_t]` =
`1.0* 10^(-2)` M

Time = ?

Applying in the above equation, we get that:-

`1.0* 10^(-2)=0.1e^{-8.1* 10^(-2)* t}`

`0.1e^{-8.1* \:10^(-2)t}=10^(-2)`

`e^{-8.1* \:10^(-2)t}=(1)/(10)`

`\ln \left(e^{-8.1* \:10^(-2)t}\right)=\ln \left((1)/(10)\right)`

`t=28.43\ min`