Q: Show that in a first order reaction, time required for completion of 99.9% is 10 times of half-life (\${t}_{1/2}\$) of the reaction.

For first order reaction,\$t = \cfrac{2.303}{k} \log{\cfrac{a}{a-x}}\$Given that:- \$x = 99.9 \; a\$ for \${t}_{99.9 \%}\$\$x = 0.5 \; a\$ for \${t}_{50 \%}\$\$\therefore \; \cfrac{{t}_{99.9 \%}}{{t}_{50 \%}} = \cfrac{\cfrac{2.303}{k} \log{\cfrac{a}{a - 0.999a}}}{\cfrac{2.303}{k} \log{\cfrac{a}{a - 0.5a}}}\$\$\Rightarrow \; \cfrac{{t}_{99.9 \%}}{{t}_{50 \%}} = \cfrac{\log{\cfrac{1}{0.001}}}{\log{\cfrac{1}{0.5}}}\$\$\Rightarrow \; \cfrac{{t}_{99.9 \%}}{{t}_{50 \%}} = \cfrac{- \log{0.001}}{\log{2}}\$\$\Rightarrow \; \cfrac{{t}_{99.9 \%}}{{t}_{50 \%}} = \cfrac{- \left( -3 \right)}{0.3010}\$\$\Rightarrow \; {t}_{99.9 \%} = 10 \times {t}_{50 \%}\$Hence proved.

Q: A reaction is first order in \$A\$ and second order in \$B\$ :(i) Write differential rate equation.(ii) How is the rate affected when the concentration of \$B\$ is tripled?(iii) How is the rate affected when the concentration of both \$A\$ and \$B\$ is doubled?

It is given that a reaction is first order in \$A\$ and second order in \$B\$.i) The differential rate equation is as follows: Rate \$\displaystyle = k[A][B]^2\$(ii) On increasing the concentration of \$B\$ three times, rate becomes \$9\$ times.\$\displaystyle (d[B])^2 = 3^2 = 9\$(iii) When the concentrations of both \$A\$ and \$B\$ are doubled, rate becomes \$8\$ times.\$\displaystyle d[A] (d[B])^2 = 2 \times 2^2 = 8\$

Q: The rate constant for a first order reaction is \$60\ s^{-1}\$. How much time will it take to reduce the initial concentration of the reactant to its \$1/16^{th}\$ value?

Given that the rate constant for a first order reaction is \$60\ s^{-1}\$.Let \$a\$ M be the initial concentration. Final concentration will be \$\displaystyle \frac {a}{16}\$ M.Rate constant, \$\displaystyle k = 60 \: /s\$\$\displaystyle t = \frac {2.303}{k} log\left( \frac {[A]_0}{[A]}\right)\$\$\displaystyle t = \frac {2.303}{60} log \left(\frac {a}{\dfrac {a}{16}}\right)\$\$\displaystyle t = 4.6 \times 10^{-2}\$ seconds

Q: A first order reaction takes \$40\$ min for \$30\$% decomposition. Calculate \$t_{1/2}\$.

Let \$a\$ M be the initial concentration. After \$40\$ minutes, the concentration is \$\displaystyle a - \dfrac {30a}{100} = 0.70a\$.\$\displaystyle k = \dfrac {2.303}{t}log\dfrac {[A]_o}{[A]} \$\$\displaystyle k =\dfrac {2.303}{40} log\dfrac {a}{0.70a}\$\$\displaystyle k = 8.92 \times 10^{-3} \: /min\$The half life period, \$\displaystyle t_{1/2} = \dfrac {0.693}{k} =\dfrac {0.693}{8.92 \times 10^{-3} \: /min} = 77.7\$ min.

Q: For a first order reaction \$A \rightarrow B\$, the reaction rate at reactant concentration of 0.01 M is found to be \$2.0 \times10^{-5} mol L^{-1}s^{-1}\$. The half-life period of the reaction is:

For first order reaction,\$A \longrightarrow B\$Rate \$= k\times [A]\$Given, rate \$=2.0\times 10^{-5}mol \,L^{-1}s^{-1}\$[A] = Conc. of A = 0.01 M So, \$2.0\times 10^{-5}=k\times 0.01\$\$k=\frac{2.0\times 10^{-5}}{0.01}s^{-1}\$ \$ =2.0\times 10^{-3}s^{-1}\$For first order reaction,\$t_{1/2}=\frac{0.693}{k}=\frac{0.693}{2.0\times 10^{-3}}\$\$=346.5=347s\$

Q: What are the units of rate constant for the first order reaction?

Let \$R\$ be the rate of reaction.For first order reaction,\$R = K \; {\left[ A \right]}^{1}\$\$\Rightarrow K = R \; {\left[ A \right]}^{-1}\$Whereas, \$K\$ and \$\left[ A \right]\$ are rate constant and initial concentration of reactant respectively.Therefore,Unit of rate constant \$= {\left( mol \; {L}^{-1} \right)}^{1 - n} {sec}^{-1}\$For first order reaction,\$n = 1\$Unit of rate constant \$= {sec}^{-1}\$Hence the unit of rate constant for first order reaction is \${sec}^{-1}\$.

Q: The unit of rate constant for zero order reaction is:

\$Rate=\dfrac{dC}{dt} = k[R_0]^o = k\$ or \$k=\dfrac{dC}{dt} = \dfrac{conc.}{time}\$ \$=\dfrac{mol\,L^{-1}}{s} = mol\,L^{-1}s^{-1}\$.

Question 1

Show that in a first order reaction, time required for completion of 99.9% is 10 times of half-life (${t}_{1/2}$) of the reaction.

Accepted Answer

For first order reaction,$t = \cfrac{2.303}{k} \log{\cfrac{a}{a-x}}$Given that:- $x = 99.9 \; a$ for ${t}_{99.9 \%}$$x = 0.5 \; a$ for ${t}_{50 \%}$$\therefore \; \cfrac{{t}_{99.9 \%}}{{t}_{50 \%}} = \cfrac{\cfrac{2.303}{k} \log{\cfrac{a}{a - 0.999a}}}{\cfrac{2.303}{k} \log{\cfrac{a}{a - 0.5a}}}$$\Rightarrow \; \cfrac{{t}_{99.9 \%}}{{t}_{50 \%}} = \cfrac{\log{\cfrac{1}{0.001}}}{\log{\cfrac{1}{0.5}}}$$\Rightarrow \; \cfrac{{t}_{99.9 \%}}{{t}_{50 \%}} = \cfrac{- \log{0.001}}{\log{2}}$$\Rightarrow \; \cfrac{{t}_{99.9 \%}}{{t}_{50 \%}} = \cfrac{- \left( -3 \right)}{0.3010}$$\Rightarrow \; {t}_{99.9 \%} = 10 \times {t}_{50 \%}$Hence proved.

Question 2

A reaction is first order in $A$ and second order in $B$ :(i) Write differential rate equation.(ii) How is the rate affected when the concentration of $B$ is tripled?(iii) How is the rate affected when the concentration of both $A$ and $B$ is doubled?

Accepted Answer

It is given that a reaction is first order in $A$ and second order in $B$.i) The differential rate equation is as follows: Rate $\displaystyle = k[A][B]^2$(ii) On increasing the concentration of $B$ three times, rate becomes $9$ times.$\displaystyle (d[B])^2 = 3^2 = 9$(iii) When the concentrations of both $A$ and $B$ are doubled, rate becomes $8$ times.$\displaystyle d[A] (d[B])^2 = 2 \times 2^2 = 8$

Question 3

The rate constant of the reaction $A\rightarrow B$ is $0.6\times 10^{-3} \ moleL^{-1}s^{-1}$. If the concentration of $A$ is $5 \ M$, then concentration of $B$ after $20 \ minutes$ is:

Accepted Answer

The rate constant has unit mole per second which indicates zero order reaction.x = ktThe concentration of $B$ after 20 minutes is $\displaystyle 20 \times 60 \times 0.6\times 10^{-3} = 0.72 \: M$

Question 4

The rate constant for a first order reaction is $60\ s^{-1}$. How much time will it take to reduce the initial concentration of the reactant to its $1/16^{th}$ value?

Accepted Answer

Given that the rate constant for a first order reaction is $60\ s^{-1}$.Let $a$ M be the initial concentration. Final concentration will be $\displaystyle \frac {a}{16}$ M.Rate constant, $\displaystyle k = 60 \: /s$$\displaystyle t = \frac {2.303}{k} log\left( \frac {[A]_0}{[A]}\right)$$\displaystyle t = \frac {2.303}{60} log \left(\frac {a}{\dfrac {a}{16}}\right)$$\displaystyle t = 4.6 \times 10^{-2}$ seconds

Question 5

A first order reaction takes $40$ min for $30$% decomposition. Calculate $t_{1/2}$.

Accepted Answer

Let $a$ M be the initial concentration. After $40$ minutes, the concentration is $\displaystyle a - \dfrac {30a}{100} = 0.70a$.$\displaystyle k = \dfrac {2.303}{t}log\dfrac {[A]_o}{[A]} $$\displaystyle k =\dfrac {2.303}{40} log\dfrac {a}{0.70a}$$\displaystyle k = 8.92 \times 10^{-3} \: /min$The half life period, $\displaystyle t_{1/2} = \dfrac {0.693}{k} =\dfrac {0.693}{8.92 \times 10^{-3} \: /min} = 77.7$ min.

Question 6

What is the unit for the rate constant of a second order reaction?

Accepted Answer

The unit for the rate constant of a second order reaction is $\displaystyle \displaystyle L { mol }^{ -1 }{ s }^{ -1 } $. For a second order reaction, $\displaystyle rate = k [A]^2 $ $\displaystyle mol { L }^{ -1 }{ s }^{ -1 }  = k (mol { L }^{ -1 })^2$ $\displaystyle k = L { mol }^{ -1 }{ s }^{ -1 }  $

Question 7

For a first order reaction $A \rightarrow B$, the reaction rate at reactant concentration of 0.01 M is found to be $2.0 \times10^{-5} mol L^{-1}s^{-1}$. The half-life period of the reaction is:

Accepted Answer

For first order reaction,$A \longrightarrow B$Rate $= k\times [A]$Given, rate $=2.0\times 10^{-5}mol \,L^{-1}s^{-1}$[A] = Conc. of A = 0.01 M So, $2.0\times 10^{-5}=k\times 0.01$$k=\frac{2.0\times 10^{-5}}{0.01}s^{-1}$ $ =2.0\times 10^{-3}s^{-1}$For first order reaction,$t_{1/2}=\frac{0.693}{k}=\frac{0.693}{2.0\times 10^{-3}}$$=346.5=347s$

Question 8

What are the units of rate constant for the first order reaction?

Accepted Answer

Let $R$ be the rate of reaction.For first order reaction,$R = K \; {\left[ A \right]}^{1}$$\Rightarrow K = R \; {\left[ A \right]}^{-1}$Whereas, $K$ and $\left[ A \right]$ are rate constant and initial concentration of reactant respectively.Therefore,Unit of rate constant $= {\left( mol \; {L}^{-1} \right)}^{1 - n} {sec}^{-1}$For first order reaction,$n = 1$Unit of rate constant $= {sec}^{-1}$Hence the unit of rate constant for first order reaction is ${sec}^{-1}$.

Question 9

The unit of rate constant for zero order reaction is:

Accepted Answer

$Rate=\dfrac{dC}{dt} = k[R_0]^o = k$ or $k=\dfrac{dC}{dt} = \dfrac{conc.}{time}$                  $=\dfrac{mol\,L^{-1}}{s} = mol\,L^{-1}s^{-1}$.

Question 10

Write four differences between molecularity and order of reaction.

Accepted Answer

MolecularityOrder  It represents the number of the reactant molecules taking part in the elementary reaction. It represents the sum of the exponents to which the concentration term in the rate law are raised. It is theoretical property and indicates the number of reactant molecules involved in each act leading to the reaction.It is an experimental property and indicates the dependence of observed reaction rate on the concentration of the reactants.  It is always an integer and can never be fraction or zero.It may be an integer, fraction or zero.  It is the property of only elementary reaction and has no meaning for complex reactions.It is property of both elementary and complex reactions.