Question

The half-life of a reaction of compound A to give compounds D and E is 8.50 min when the initial concentration of A is 0.150 mol/L. How long will it take for the concentration to drop to 0.0300 mol/L if the reaction is (a) first order with respect to A or (b) second order with respect to A?

Short answer

Half-life periods may be defined as the time period at which half concentration of reactant converts into the product.

1. The half-life period of the reaction for the first order is \(8.5\,min\) having rate constant \(0.082L/mole/min\). The time taken by the reactant of concentration from \(0.150mol/L\) to \(0.0300mole/L\) is \(19.62\,minutes\).
2. The half-life period of the reaction for the second order is \(8.5min\) having rate constant \(0.0088L/mole/min\). The time taken by the reactant of concentration \(0.150mol/L\) to \(0.0300mole/L\) is \(3034\,minutes\).

Step-by-step solution

Reaction Rate

Reaction involved the effective collision of two reactants to produce the desired products. Reactions can be natural which are occurring in surrounding environment whereas it can be artificially done in the laboratory to form a desired required product.

The reaction rate can be defined as the speed of reaction to produce the products. The reaction rate can be slow, fast or moderate. The reaction can take a lesser than millisecond to produce products or it can take years to produce a desired product.

Half-life period can be defined as the time period at which half concentration of the reactants gets converted into product.

Explanation

The half-life period of the first-order decomposition of the A is:

\(Half - life{\bf{ }}period{\bf{ }} = {\bf{ }}\frac{{\ln {\bf{ }}\left( 2 \right)}}{k}\)

\(K{\bf{ }} = {\bf{ }}{\rm{Rate Constant}}\)

\(\begin{align}Half - life{\rm{ }}period{\bf{ }} &= {\bf{ }}\frac{{0.693}}{k}\\8.5\min {\bf{ }} &= {\bf{ }}\frac{{0.693}}{k}\\{\bf{ }}k &= {\bf{ }}\frac{{0.693}}{{8.5\min }}\\k &= 0.082L/mole/\min \end{align}\)

Now, the rate constant, \(k{\bf{ }} = {\bf{ }}0.082{\bf{ }}L/mole/\min \).

The initial concentration of the reactant \(\left( A \right)^\circ = 0.300{\bf{ }}mole/L\)

The final concentration of the reactant \(A = 0.0300mol/L\)

Let the time taken by the reactant reaching from concentration \(0.150mol/L\) to \(0.0300mole/L\) be ‘t.’

\(\begin{align}\ln (A) &= \ln (A0) - kt\\\ln (0.0300) &= \ln (0.150) - 0.082 \times t\\ - 3.506 &= - 1.897 - 0.082 \times t\\ - 3.506 + 1.897 &= - 0.082 \times t\end{align}\)

\(\begin{align} - 0.082 \times t &= - 1.609\\t = \frac{{1.609}}{{0.082}}\\t &= 19.622\,\min utes\end{align}\)

The time taken by the reactant of concentration \(0.150mol/L\) to \(0.0300mole/L\) is \(19.62\,minutes\).

As the first-order reaction does not depend upon the concentration of the reactant. That is why, the half-life period does not change.

The half-life period

\(Half - life{\bf{ }}period{\bf{ }} = {\bf{ }}\frac{{\left( A \right)^\circ }}{{2k}}\)

Where \(\left( A \right)^\circ = \) initial concentration of reactant

\(\begin{align}Half - life{\rm{ }}period{\bf{ }} &= {\bf{ }}\frac{{0.150mole/L}}{{2 \times k}}\\8.50\min {\bf{ }} &= {\bf{ }}\frac{{0.150mole/L}}{{2k}}\\k{\bf{ }} &= {\bf{ }}\frac{{0.150mole/L}}{{2 \times 8.50\min }}\\k &= 0.0088L/mole/\min \end{align}\)

Rate constant, \(K{\bf{ }} = {\bf{ }}0.0088{\bf{ }}L/mole/\min \)

Initial concentration of reactant, \(\left( A \right)^\circ = 0.300{\bf{ }}mole/L\)

The final concentration of the reactant \(A = 0.0300mol/L\)

Let the time taken by the reactant reaching from concentration \(0.150mol/L\) to \(0.0300mole/L\) be ‘t.’

\(\begin{align}\frac{1}{{(A)}} - \frac{1}{{(A0)}} &= kt\\\frac{1}{{0.0300}} - \frac{1}{{0.150}} &= 0.0088 \times t\\33.33 - 6.67 &= 0.0088 \times t\\26.7 &= 0.0088 \times t\end{align}\)

\(\begin{align}t &= \frac{{26.7}}{{0.0088}}\\t &= 3034\min utes\end{align}\)

Therefore, the time taken by the reactant of concentration \(0.150mol/L\) to \(0.0300mole/L\) is \(3034\,minutes\).