Chemistry, 12.07.2019 12:00 drma1084

In a first-order decomposition reaction, 50.0% of a compound decomposes in 10.5 min. a. what is the rate constant of the reaction? b. how long does it take for 75.0% of the compound to decompose?

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Answer from: wig77777

In a decomposition reaction, the reactant breaks into two or more products. The rate of reaction is 0.066 per minute and the time required is 21 minutes.

What is the rate constant?

The rate constant relates the reaction rate with the molar concentration of the reactants and is called reaction rate coefficient (k).

In the reaction 50 % compound decomposed in 10.5 minutes, that will be its half-life $\rm (t^{\frac{1}{2}})$

Calculate the rate constant of the reaction by the formula:

$\begin{aligned}\rm k &= \dfrac{0.6932}{\rm t^{\frac{1}{2}}}\\\\&= \dfrac{0.6932}{10.5}\\\\&= 0.066 \rm \;min^{-1}\end{aligned}$

Hence the rate constant for the reaction will be 0.066 per minute.

Calculate the time taken for 75 % decomposition of the compound:

The rate of the first-order reaction is given as:

$\rm k = \dfrac{2.303}{t^{\frac{1}{2}}}log \dfrac{[A_{o}]}{[A_{t}]}$

Given,

Initial concentration = 100Final concentration = 25

Substituting values in the above equation:

$\begin{aligned}0.066 &= \dfrac{2.303}{\rm t^{\frac{1}{2}}}\rm log \dfrac{100}{25}\\\\&= \dfrac{2.303}{\rm t}(0.6020)\\\\\rm t &= 21 \rm \;minutes\end{aligned}$

Hence, for 75 % decomposition 21 minutes will be required.

Therefore, 0.066 per minute is the rate constant and 21 minutes is the time required for 75 % decomposition.

Learn more about rate constant and first-order reaction here:

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Answer from: hdolly81

In this reaction 50% of the compound decompose in 10.5 min thus, it is half life of the reaction and denoted by symbol $t_{1/2}$ .

(a) For first order reaction, rate constant and half life time are related to each other as follows:

$k=\frac{0.6932}{t_{1/2}}=\frac{0.6932}{10.5 min}=0.066 min^{-1}$

Thus, rate constant of the reaction is $0.066 min^{-1}$ .

(b) Rate equation for first order reaction is as follows:

$k=\frac{2.303}{t_{1/2}}log\frac{[A_{0}]}{[A_{t}]}$

now, 75% of the compound is decomposed, if initial concentration $[A_{0} ]$ is 100 then concentration at time t $[A_{t} ]$ will be 100-75=25.

Putting the values,

$0.066 min^{-1}=\frac{2.303}{t}log\frac{100}{25}=\frac{2.303}{t}(0.6020)$

On rearranging,

$t=\frac{2.303\times 0.6020}{0.066 min^{-1}}=21 min$

Thus, time required for 75% decomposition is 21 min.

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In a first-order decomposition reaction, 50.0% of a compound decomposes in 10.5 min. a. what is the...

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