Question

The decomposition of dimethylether at 504 degree Celsius is first order with a half-life of 1570 seconds. What fraction of an initial amount of dimethylether remains after 4710 seconds?

Answer 1

Answer:

Given: It is a first order reaction in which half-life of the reaction is 1570seconds.

To find: The amount of dimethylether remains after 4710 seconds.

Solution:

The half life of the first order reaction is represented by the expression given below:

$t_{1/2} =\frac{0.693}{k}$

Here, $t_{1/2}$ is the half-life and $k$ is the rate constant.

Substituting the given values in the above expression as shown below:

$1570 s =\frac{0.693}{k}$

$k=\frac{0.693}{1570} =0.00044 s^{-1}$

Hence, the rate constant of this reaction is $0.00044 s^{-1}$.

The expression for rate constant of first-order reaction is shown below:

$k=\frac{2.303}{t} log\frac{[A_{0} ]}{[A_{t}] }$

Here, $t$ is the time, $A_{o}$ is the initial concentration of substance and $A_{t}$ is the amount of substance at time

Substitue the given values in the above expression as mentioned below:

$0.00044 =\frac{2.303}{4710} log\frac{[A_{0} ]}{[A_{t}] }$

$0.899=log\frac{[A_{0} ]}{[A_{t}] }$

$7.925=\frac{[A_{0} ]}{[A_{t}] }$

Thus, the correct answer is 7.925.