For the reaction: $$N_2 + 3H_2\rightarrow 2NH_3$$, If $$\cfrac{d[NH_3]}{dt}= 2 \times 10^{-4} mol\ L^{-1}s^{-1}$$, the value of $$\cfrac{-d[H_2]}{dt}$$ would be:

$$1 \times 10^{-4} mol \; L^{-1}s^{-1}$$

$$4 \times 10^{-4} mol\; L^{-1}s^{-1}$$

$$6 \times 10^{-4} mol\; L^{-1}s^{-1}$$

MCQ Questions for Class 12 Engineering Chemistry Chemical Kinetics Quiz 9

What is the activation energy for a reaction if the rate doubles when the temperature is raised from

$20^{o}C$ to

$35^{o}C$ ?

$(R=8.134\ J\ mol^{-1}K^{-1})$

0%
$15.1\ kJ\ mol^{-1}$
0%
$342\ kJ\ mol^{-1}$
0%
$269\ kJ\ mol^{-1}$
0%
$34.7\ kJ\ mol^{-1}$

Explanation

Given :

$T_1 = 20 + 273 = 293K$

$T_2 = 35 + 273 = 308K$

$log \dfrac{{k}_{2}}{{k}_{1}} = \dfrac{{E}_{a}}{2.303 \times R} \times (\dfrac{1}{{T}_{1}} - \dfrac{1}{{T}_{2}})$

Substituting the values, we get

$log\space 2 = \dfrac{{E}_{a}}{2.303 \times 8.314} \times (\dfrac{1}{293} - \dfrac{1}{308})$

${E}_{a}$ =

$34.7\space kJ\space {mol}^{-1}$

Therefore, the option is D.

K for a zero order reaction is

$2\times10^{-2}\,mol\,L^{-1}sec^{-1}$ . If the concentration of the reactant after 25 sec is 0.5 M, the initial concentration must have been:

0%
0.5 M
0%
1.25 M
0%
12.5 M
0%
1.0 M

For a first order process

$A\rightarrow B$ rate constant

${k}_{1}=0.693\ {min}^{-1}$ and another first order process

$C\rightarrow D$ ,

${K}_{2}=x$

${min}^{-1}$ . If

$99.9$ % of

$C\rightarrow D$ requires time same as

$50$ % of reaction

$A\rightarrow B$ , value of

$x$ ? (in

${min}^{-1}$ )

0%
$0.0693$
0%
$6.93$
0%
$23.03$
0%
$13.86$

Explanation

The solution is as follows:

1. Time required for

$50$ % completion for reaction

$1$ is:

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

Given

$k=0.639\ min^{-1}$

$t_{1/2}=\dfrac{0.693}{0.693}=1\ min$

2. For reaction

$2$ , it takes

$1$ min to get

$99.99$ % of reaction to be completed. Thus, it takes

$0.5$ min to get

$50$ % reaction.

Thus,

$t_{1/2}=\dfrac{0.693}{K_2}$

$K_2=\dfrac{0.693}{t_{1/2}}$

$K_2=\dfrac{0.693}{0.5}$

Option D is correct.

$K_2=13.86\ min^{-1}$

The temperature coefficient of a reaction is

$2$ . When the temperature is increased from

$30^{\circ}C$ to

$90^{\circ}C$ , the rate of reaction is increased by

0%
$150$ times
0%
$410$ times
0%
$72$ times
0%
$64$ times

Explanation

For every

$10^{\circ}C$ rise in temperature the rate of reaction doubles.

$\therefore$ From

$30^{\circ}C$ to

$90^{\circ}C$ there 6 times

$10^oC$ rise in temperature.

So, rate increase by

$2^6=64$ times.

A reaction

$A+B\leftrightarrow C+D$ follows the mechanism:

$A+B\leftrightarrow AB$

$AB+C\leftrightarrow D$

In which first step remains essentially in equilibrium. If

$\Delta H$ is the enthalpy change for the first reaction the activation energy for the second reaction, the activation energy of the overall reaction will be given?

0%
$E_{0}$
0%
$E_{0}-\Delta H$
0%
$E_{0}+\Delta H$
0%
$E_{0}+2\Delta H$

If reaction A and B are given with Same temperature and same concentration but rate of

$A$ is double than

$B$ . Pre exponential factor is same for both the reaction then difference in activation energy

$E_{A} - E_{B}$ is?

0%
$-RT\ ln2$
0%
$RT\ ln2$
0%
$2RT$
0%
$\dfrac {RT}{2}$

Explanation

$\dfrac {r_{A}}{r_{B}} = \dfrac {A_{1}e^{\dfrac {-E_{A}}{RT}}}{A_{2}e^{\dfrac {-E_{B}}{RT}}}$

$\dfrac {2}{1} = \dfrac {e^{\dfrac {-E_{A}}{RT}}}{e^{\dfrac {-E_{B}}{RT}}}$

$ln2 = E_{B} - E_{A} / RT$

$E_{B} - E_{A} = RT\ ln2$

$E_{A} - E_{B} = -RT\ ln2$ .

Among the following, the maximum covalent character is shown by the compound

0%
$MgCl_{2}$
0%

$FeCl_{2}$
0%
$SnCl_{2}$
0%
$AlCl_{3}$

Raw milk sours in 4 hours at

$27^\circ C,$ but in 40 hours in refrigerator at

$7^\circ C.$ What is activation energy for souring of milk:

0%
402.1 J/mol
0%
30.2 J/mol
0%
8.04 KJ/mol
0%
80.4 KJ/mol

Explanation

$K\alpha\dfrac{1}{t}$

$Log\dfrac{K_2}{K_1}=log\dfrac{t_1}{t_2}=\dfrac{E_a}{R}(\dfrac{T_2-T_1}{T_I\times T_2})$

$Log\dfrac{40}{4}=\dfrac{E_a}{8.314}(\dfrac{20}{300\times 280})$

So solving

$E_a=30.2J/mol$

Two substances

$A$ and

$B$ are initially present as

$\left[ { A }_{ 0 } \right] =8\left[ { B }_{ 0 } \right]$ and

${ t }_{ 1/2 }$ for the first-order decomposition of

$A$ and

$B$ are

$10$ and

$20$ min, respectively. If they start decomposing at the same time, after how much time, the concentration of both of them would be same?

0%
$20min$
0%
$40min$
0%
$60min$
0%
$200min$

The decomposition of

${ NH }_{ 3 }$ on nitrogen surface follows zero-order kinetics. The half-life is

$315s$ for an initial pressure of

$70mm$ of

${NH}_{3}$ . If the initial pressure had been

$150mm$ , what would be the half-life?

0%
$315s$
0%
$472.5s$
0%
$675s$
0%
$630s$

For the reaction :

$NH_2COONH_4(s) \leftrightharpoons 2NH_3(g) +CO_2(g) , K_p = 3.2 \times 10^{-5}atm^3$
the total pressure of the gaseous products when sufficient amount of reactant is allowed to achieve equilibrium , is:

0%
$0.02$ atm
0%
$0.04$ atm
0%
$0.06$ atm
0%
$0.095$ atm

$k$ for a zero order reaction is

$2\times 10^{-2}\ mol\ L^{-1}\ s^{-1}$ . If the concentration of the reactant after

$25\ s$ is

$0.5\ M$ , the initial concentration must have been:

0%
$0.5\ M$
0%
$1.25\ M$
0%
$12.5\ M$
0%
$1.0\ M$

Explanation

$K=2\times 10^{-2}mole/lit-sec$

$t=25\ sec$ . concentration

$=0.5\ M$

$=Ao-Kt$

$0.5=Ao-2\times 10^{-2}\times 25\Rightarrow Ao=1\ M$

${ SO }_{ 2 }{ Cl }_{ 2 }\rightarrow { SO }_{ 2 }+{ Cl }_{ 2 }$ is a first-order gaseous reaction with

$K=2.5\times { 10 }^{ -5 }{ s }^{ -1 }$ at

${320}^{o}C$ . The percentage of

${ SO }_{ 2 }{ Cl }_{ 2 }$ decomposed on heating for

$100min$ is:

$\left( \ln { 1.16 } =0.15 \right)$

0%
$86.2$
0%
$15.0$
0%
$85.0$
0%
$13.8$

A zero-order reaction is one

0%
in which reactants do not react
0%
in which one of the reactants is in large excess
0%
whose rate does not change with time
0%
whose rate increases with time

Explanation

In a zero-order reaction, the rate is independent of the concentration of the reactants. The rate of change of concentration would be constant.

Therefore,

$Rate=k=constant$

Hence, the correct option is (C).

A solution of

${ N }_{ 2 }{ O }_{ 5 }$ in

$C{Cl}_{4}$ yields by decomposition at

${45}^{o}C$ ,

$4.8ml$ of

${O}_{2}$ ,

$20min$ after the start of the experiment and

$9.6ml$ of

${O}_{2}$ after a very long time. The decomposition obeys first-order kinetics. What volume of

${O}_{2}$ would have evolved,

$40min$ after the start?

0%
$7.2ml$
0%
$2.4ml$
0%
$9.6ml$
0%
$6.0ml$

Nitric oxide, NO, and bromine vapour react together according to the following equation.

$2NO(g)+Br_2(g)\rightarrow 2NOBr(g)$

$\Delta H^o=-23$ kJ

$mol^{-1}$
The reaction has an activation energy of

$+5.4$ kJ

$mol^{-1}$ .
What is the correct reaction pathway diagram for this reaction?

Explanation

${2 NO}_{\left( g \right)} + {{Br}_{2}}_{\left( g \right)} \longrightarrow {2 NOBr}_{\left( g \right)}$

$\Delta{{H}^{°}} = \Delta{{H}_{P}} - \Delta{{H}_{R}}$

$\Delta{{H}^{°}} = - 23 \; {KJ}/{mol} = -ve$

We also know that activation enrgy always has a positive value and is always at a higher energy level than the reactants or products.

Hence for the above reaction, the correct pathway diagram is

$\left( D \right)$ .

Which of the following statement is/are incorrect?

0%
When $\Delta t$ is infinitesimally small, the average rate equals the instantaneous rate
0%
Activation energy for the forward reaction equals activation energy for the reverse reaction in a catalysed reaction
0%
For a reversible reaction, an increase in temperature, increase the rate for both forward and backward reaction
0%
Larger the initial reactant concentration for zero-order reaction, shorter is the half-life

For what type of the following reactions is the law of mass action, never obeyed?

0%
Zero order
0%
First order
0%
Second order
0%
Third order

$10\%$ of a reactant decomposes in

$1$ hour ,

$20\%$ in

$2$ hours and

$30\%$ in

$3$ hours. The order of the reaction is

0%
$0$
0%
$1$
0%
$2$
0%
$3$

In a first order reaction of the type

$A(g)\rightarrow 2B(g)$ , the initial and final pressures are

$p_{1}$ and

$p$ respectively. The rate constant can be expressed by

0%
$k=\dfrac{1}{t}\ln \dfrac{p_{1}}{2p_{1}-p}$
0%
$k=\dfrac{1}{t}\ln =dfrac{p_{1}}{p_{1}-p}$
0%
$k=\dfrac{1}{t}\ln \dfrac{p_{1}}{p-p_{1}}$
0%
$k=\dfrac{1}{t}\ln \dfrac{p_{1}}{p}$

The half-life of

$_6C^{14}$ if its K or

$\lambda$ is

$2.31 \times 10^{-4}$ is:

0%
$2 \times 10^2$ yrs
0%
$3 \times 10^3$ yrs
0%
$3.5 \times 10^4$ yrs
0%
$4 \times 10^3$ yrs

Explanation

$t_{1/2} = \dfrac{0.693}{k} = \dfrac{0.693}{2.31 \times 10^{-4}}$

$= 0.3 \times 10^4 \ yrs$

$= 3.0 \times 10^3 \ yrs$

The rate constant for two parallel reactions were found to be

$1.0\times 10^{-2}dm^{3}$

$mol^{-1}s^{-1}$ and

$3.0\times 10^{-2}dm^{3}$

$mol^{-1}s^{-1}$ . If the corresponding energies of activation of the parallel reactions are 60.0 KJ

$mol^{-1}$ and 70.0 KJ

$mol^{-1}$ respectively, then what is the apparent overall energy of activation?

0%
130.0 KJ $mol^{-1}$
0%
65.0 KJ $mol^{-1}$
0%
67.5 KJ $mol^{-1}$
0%
100.0 KJ $mol^{-1}$

Explanation

Overall energy of activation =

$\dfrac{k_{1}E_{1}+k_{2}E_{2}}{k_{1}+k_{2}}$

$\implies \dfrac{60 \times 1 \times 10^{-2}+70\times 3\times 10^{-2}}{4\times10^{-2}}$

$\implies \dfrac{270}{4}=67.5 \: KJ mol^{-1}$

Consider an endothermic reaction

$X \rightarrow Y$ with the activation energies

$E_{b}$ and

$E_{f}$ for the backward and forward reactions respectively. In general:

0%
$E_{b}> E_{f}$
0%
$E_{b}< E_{f}$
0%
$E_{b}=E_{f}$
0%
Any of the above

Explanation

Here,

$E_f>E_b$

Option B is correct.

In a zero-order reaction:

0%
The rate constant has the unit mol $L^{-1}s^{-1}$
0%
The rate is independent of the concentration of the reactants.
0%
The half-life depends on the concentration of the reactants.
0%
The rate is independent of the temperature of the reaction.

Explanation

For zero order,

Rate

$=K[A]^0 \Rightarrow rate =K$

Zero-order is independent of the concentration and unit of rate constant is

$mole\ L^{-1}sec^{-1}$

$t_{^1/_2} =\cfrac{A_0}{2 K}$

From the given equation it is clear that for a zero-order reaction the half-life period is dependent on concentration.

Statement - I : Every collision of reactant molecule is not successful.
Statement - II: Every collision of reactant molecule with proper orientation is successful one.

0%
Both statements are true Statement - II is correct explanation of Statement - I
0%
Both statements are true but Statement -II is not correct explanation of Statement - I
0%
Statement - I is true but Statement - II is false
0%
Statement - I is false but Statement - II is true

During the hydrogenation of vegetable oil at

$25^{0}C$ , the pressure of

$H_{2}$ reduces from 2 atmospheres to 1.2 atmospheres in 50 minutes. The rate of reaction in terms of molarity per second is:

0%
$1.09\times 10^{-6}$
0%
$1.09\times 10^{-5}$
0%
$1.09\times 10^{-7}$
0%
$1.09\times 10^{5}$

Explanation

For hydrogenation reaction the rate of reaction is written as

$r = -\dfrac{d[H_2]}{dt}$ .

Assuming the hydrogen an ideal gas

$P = CRT$

Therefore

$C = \dfrac{P}{RT}$

Hence

$\Delta C = \dfrac{\Delta P}{RT}$

$\Delta P = -0.8$ .

So,

$\Delta C = \dfrac{-0.8}{0.0821\times 298}M$

So,

$r = -\dfrac{ \dfrac{-0.8}{0.0821\times 298}}{50\times60} = 1.09\times10^{-5}M/s$

$r = 1.09\times10^{-5}M/s$

Option B is correct.

For the first order gaseous reaction,

$x(g)\rightarrow 2y(g)+z(g)$ , the initial pressure,

$P_{x}=90$ mm Hg. The pressure after 10 minutes is 180 mm Hg. The rate constant of the reaction is:

0%
$2\times 10^{-3}sec^{-1}$
0%
$2\times 10^{3}sec^{-1}$
0%
$1.15\times 10^{-3}sec^{-1}$
0%
$1.15\times 10^{3}sec^{-1}$

Explanation

The inital pressure of x,

$P_x$ is 90 mm Hg.

Total pressure after 10 minute is 180 mm Hg. Let X be the change in pressure of x.

Total pressure will be

$(P_x-X)+2X+X=P_x+2X=90+2X=180$ or

$X=45 mm.$

The expression for the rate constant is

$k=\dfrac {2.303} {t}log \dfrac {a} {a-X}=\dfrac {2.303} {10 \times 60}log \dfrac {90} {90-45}=1.15\times 10^{-3}sec^{-1}.$

The gas phase decomposition of dimethyl ether follows first order kinetics:

$CH_{3}-O-CH_{3}(g)\rightarrow CH_{4}(g)+H_{2}(g)+CO(g)$

The reaction is carried out in a constant volume container at

$50^0 C$ and has a half life of 14.5 minutes. Initially, only dimethyl ether is present at a pressure of 0.40 atm. What is the total pressure of the system after 12 minutes? (Assume the ideal gas behaviour.)

0%
0.946 atm
0%
0.785 atm
0%
0.777 atm
0%
0.749 atm

Explanation

$t_{\frac{1}{2}}=\dfrac{0.693}{k}\Rightarrow k=\dfrac{0.693}{14.5}=0.0478$

$P_t=P_o[e^{kt}]$

$P_t = 0.225$

$CH_3OCH_3\rightarrow CH_4+H_2+CO$

$\begin{matrix}f=0 &\ .40 &0\ \ \ \ &0\ \ \ &0 \\f=t \ &.40-x &x &x &x \end{matrix}$

$0.40 - x = 0.225$

$x = 0.40 - 0.225 = 0.17 46$ atm
total pressure

$= 0.40 + 2x$

$= 0.40 + 0.349$

$=0.749$ atm

If the initial pressure of

$CH_{3}CHO(g)$ is 80 mm and the total pressure at the end of 20 min is 120 mm.

$CH_{3}CHO(g)\rightarrow CH_{4}(g)+CO(g)$

What is the half-life of the first-order reaction?

0%
80 min
0%
120 min
0%
20 min
0%
40 min

Explanation

Let

$p^o$ be the initial pressure.

$p^o=80.$

At time t, the total pressure is

$\mathrm{p}^{o}+\mathrm{p}=120.$

$\mathrm{p}=120-80=40.$

The expression for the rate constant is

$\mathrm{K}=\cfrac{1}{\mathrm{t}}\ln\cfrac{\mathrm{p}^{o}}{\mathrm{p}^{o}-\mathrm{p}}$

$=\cfrac{1}{20}\ln\cfrac{80}{80-40}=\cfrac{1}{20}\ln 2.$

The expression for the half life period is

$\displaystyle \mathrm{t}_{1/2}=\frac{\ln 2}{\mathrm{K}}.$

Hence,

$\displaystyle \mathrm{t}_{1/2}=\frac{\ln 2}{\ln 2}\times 20=20\min.$

Statement - I : If in a zero order reaction, the concentration of the reactant is doubled, the half-life period is also doubled.
Statement - II: For a zero order reaction, the rate of reaction is independent of initial concentration.

0%
Both statements are true. Statement - II is correct explanation of Statement - I
0%
Both statements are true but Statement -II is not correct explanation of Statement - I
0%
Statement - I is true but Statement - II is false
0%
Statement - I is false but Statement - II is true

Explanation

For a zero-order reaction,

$t_{1/2}=\dfrac{[A]_0}{2k}.$
Hence, if in a zero order reaction, the concentration of the reactant is doubled, the half-life period is also doubled.

The rate of a zero-order reaction is independent of the concentration of the reaction. The reaction proceeds at constant rate throughout.

Thus, both statements are true but Statement -II is not correct explanation of Statement - I.

$75\%$ of a first order reaction is completed in

$32$ minutes.

$50\%$ of the reaction will be completed in:

0%
$24$ mins
0%
$16$ mins
0%
$18$ mins
0%
$23$ mins

Explanation

For a first-order reaction, we have:

$t_{75\%}=2t_{50\%}$

Given:

$t_{75\%}=32$ min

Substituting the values in the above expression, we get-

$32$ min

$=2\ t_{50\%}$

$t_{50\%}=16$ mins

Hence,

$50\%$ of the reaction would have been completed in

$16$ mins.

Which of the following graphs are correct for a zero - order reaction?

Explanation

For a zero-order reaction,

$t_{1/2}=\frac{[A]_0}{2k}$ . Thus, as the initial concentration increases, the half-life increases. Hence, graph A is correct.

Also,

$[A_{0}] - [A] = kt$ . Hence, with an increase in time, x increases and a-x decreases. Graph B and C are also correct.

For the first-order reaction, the half-life period is independent of the concentration. Hence, graph D is for the first-order reaction.

In a first order reaction, the concentration of the reactant, decreases from 0.8 M to 0.4 M in 15 minutes. The time taken for the concentration to change from 0.1 M to 0.025 M is:

0%
30 min
0%
15 min
0%
7.5 min
0%
60 min

The total pressure after 200 seconds, if the initial pressure is

$0.1$ atm is _______ .

0%
$0.154$ atm
0%
$0.248$ atm
0%
$0.174$ atm
0%
$0.114$ atm

Explanation

After 200 sec,

$log \dfrac{P_{o}}{P}=\dfrac{kt}{2.303}=\dfrac{23.03\times 10^{-3}\times 200}{2.303}=200\times 10^{-2}=2.$

Hence,

$\dfrac{P_{o}}{P}=100.$

$P=P_{o}\times 0.01.$

The pressure of dinitrogen pentoxide is

$P_{N_{2}O_{5}}=(P_{o}-2x)=P_{o}\times 0.01$

Hence,

$x=\dfrac{1}{2}P_{o}(1-0.01)=0.495P_{o}=0.0495.$

The total pressure is,

$(P_{o}+3x)=P_{o}+3\times0.0495=0.1 +0.1485 = 0.2485\;atm.$

The time for half-life period of a certain reaction

$\mathrm{A}\rightarrow$ products is 1 hour. When the initial concentration of the reactant

$A$ is 2.0 mol L

$^{-1}$ , how much time does it take for its concentration to come from 0.50 to 0.25 mol L

$^{-1}$ if it is a zero-order reaction?

0%
4 h
0%
0.5 h
0%
0.25 h
0%
1 h

Explanation

For a zero order reaction

$\displaystyle \mathrm{k}=\frac{\mathrm{x}}{\mathrm{t}} \rightarrow(1)$
Where

$\mathrm{x}=$ amount decomposed

For a zero order reaction,

$\displaystyle \mathrm{k}=\frac{[\mathrm{A}]_{0}}{2\mathrm{t}_{\frac{1}{2}}}$

Since

$[\mathrm{A}_{0}]=2\mathrm{M}, \mathrm{t}_{1/2}=1$ hr;

$\mathrm{k}=1$

From equation (1),

$\displaystyle \mathrm{t}=\frac{0.25}{1}=0.25$ hr

For a first order reaction, (A)

$\rightarrow$ product, the concentration of A changes from 0.1 M to 0.025 M in 40 minutes. The rate of reaction when the concentration of A is 0.01 M, is:

0%
$1.73\displaystyle \times 10^{-5}Mmin^{-1}$
0%
$3.47\displaystyle \times 10^{-4}Mmin^{-1}$
0%
$3.47\displaystyle \times 10^{-5}Mmin^{-1}$
0%
$1.73\displaystyle \times 10^{-4}Mmin^{-1}$

Explanation

For first order reaction,

$\displaystyle k= \frac{2.303}{40} log\frac{0.1}{0.025}$

$= \dfrac{2.303}{40}log\ 4$

$= \dfrac{2.303}{20}log\ 2$

$= \dfrac {2.303}{20} \times 0.301 = 0.0347$

$\displaystyle k= 0.0347$

$Rate=k[A]$

$\displaystyle Rate= 0.0347 \times 10^{-2}=3.47 \times 10^{-4} M min^{-1}$

Which of the following statement(s) is/are true for a zero order reaction?

0%
$t_{\frac{1}{2}}$ for a zero order reaction is proportional to a, the initial concentration.
0%
The rate constant is equal to the rate of the reaction at all concentrations.
0%
$t_{\frac{1}{2}}$ is related to initial concentration of the reactant as shown in the graph.
0%
The unit of rate constant is mole time $^{-1}$ .

Explanation

For $zero^{th}$ order reaction
$t_{1/2} =\frac{a}{2K}$

The graph between $t_{1/2}$ and 'a' is shown in the figure.

The unit of rate constant is $molL^{-1} time^{-1}.$

$A$ follows the first-order reaction.

$(A)$

$\longrightarrow$ product.

The concentration of

$A$ changes from 0.1

$\mathrm{M}$ to 0.025

$\mathrm{M}$ in 40 minutes. Find the rate of reaction of

$A$ when concentration of

$A$ is 0.01

$\mathrm{M}$ ?

0%
$3.47\displaystyle \times 10^{-4}\mathrm{M}$ $\min^{-1}$
0%
$3.47\displaystyle \times 10^{-5}\mathrm{M}$ $\min^{-1}$
0%
$1.73\displaystyle \times 10^{-4}\mathrm{M}$ $\min^{-1}$
0%
$1.73\displaystyle \times 10^{-3}\mathrm{M}$ $\min^{-1}$

Explanation

As given, concentration changes from

$0.01\ M$ to

$0.025\ M$ in

$40$ minutes.

As we know,

$t_{1/2}$ is the time when concentration becomes half of the original.

$\Rightarrow 2\mathrm{t}_{1/2}=40\min$

$\mathrm{t}_{1/2}=20\min$

$K = \dfrac{0.693}{t_{\frac{1}{2}}}$

$k = \dfrac{0.693}{20}$

For first order reaction,

$\displaystyle \mathrm{r}=\mathrm{K}[\mathrm{A}]=\frac{0.693}{20}\times 0.01=3.47\times 10^{-4}\ M\ min^{-1}$

The time required for the decomposition of

$N_{2}O_{5}$ , so that the total pressure becomes 0.15 atm is ___________.

(Given

$log\ 1.8=0.255$ )

0%
25.5 sec
0%
35.5 sec
0%
45.5 sec
0%
55.5 sec

Explanation

The reaction for the decomposition of nitrogen pentoxide is as shown below:

$2N_{2}O_{5}\rightarrow 4NO_{2}+O_{2}$

Initial pressures of

$N_{2}O_{5}, NO_2$ and

$O_2$ are

$P_{o},\;0$ and

$0$ respectively. At equilibrium, the pressures of

$N_{2}O_{5},\;NO_{2}$ and

$O_2$ are

$(P_{o}-2x),\; 4x$ and

$x$ respectively.

Total pressure is

$(P_{o}-2x)+4x+x=(P_{o}+3x)$

But,

$P_{o}=0.09\;atm.$

Hence,

$P_{total}=0.15\;atm=0.09+3x \implies x=0.02\;atm.$

And

$P_{N_{2}O_{5}}=(P_{o}-2x)=(0.09-0.04)=0.05\;atm.$

The integrated rate law for the first-order reaction is

$t =\dfrac{2.303}{k}log\dfrac{a}{(a-x)}=\dfrac{2.303}{23.03\times 10^{-3}}log (\dfrac{0.09}{0.05})=\dfrac{1}{10^{-2}}\times 0.255=25.5\;sec.$

$\mathrm{l}\mathrm{n}$ a first order reaction the concentration of reactant decreases from 800

$\mathrm{m}\mathrm{o}l/\mathrm{d}\mathrm{m}^{3}$ to 50

$\mathrm{m}\mathrm{o}l/\mathrm{d}\mathrm{m}^{3}$ is

$2\times 10^{4}$ sec. The rate constant of reaction in

$\sec^{-1}$ is:

0%
$2\times 10^{4}$
0%
$3.45\times 10^{-5}$
0%
$1.386\times 10^{-4}$
0%
$2\times 10^{-4}$

Explanation

Assuming decay is happening in the above example, we can say it will take 4 half-lives in

$2\times{ 10 }^{ 4 }$ sec.

So, each half-life will take

$5\times { 10 }^{ 3 }$ sec.

Now using,

$t_{ 1/2 }=\dfrac { 0.693 }{ k }$ and putting values we get

$k=\dfrac { 0.693 }{ 5\times{ 10 }^{ 3 } }$

$k=1.386\times{ 10 }^{ -4 }$

Option C is correct.

Select the correct statements out of I, II and III for zero order reaction.
I: Quantity of the product formed is directly proportional to time.
II: Larger the initial concentration of the reactant, greater the half-life period.
III: If 50% reaction takes place in 100 minutes, 75% reaction take place in 150 minutes.

0%
I only
0%
I and II only
0%
II and III only
0%
I, II and III

Explanation

We know for zeroth order Rxn
For (I)

$x=kt$
For (II)

$t_{1/2} \propto [A_0]$ [Initial concentration]
For (III)

$\displaystyle A_0- \frac {A_0}{2} = k \times 100 .... (i)$

$\displaystyle A_0 - \frac{A_0}{4} = k \times t .... (ii)$

$\displaystyle \implies \frac {\frac{A_0}{2}}{(\frac{3A_0}{4})} = \frac{k \times 100}{k \times t}$

$\displaystyle t= \frac{100 \times 3}{2} = 150$ minutes

For a reaction

$P \rightarrow Q$ , the half-life of the reaction was 3h, when the initial concentration of P was 0.5M. As the concentration of P was increased to 1.0M, half life changes to 6h. The order of reaction with respect to P is:

0%
zero
0%
one
0%
two
0%
three

Explanation

We know the relation between half time and concentration i.e.

$t_{\frac{1} {2}} \propto \displaystyle \frac {1}{a^{(n-1)}}$ where

$n$ is the order of reaction

$t_{\frac{1} {2}} = k \displaystyle \frac {1}{a^{(n-1)}}$

$(t_{\frac{1} {2}})_1 \times a_1^{(n-1)}=(t_{\frac{1} {2}})_2 \times a_2^{(n-1)}$

$3\times(0.5)^{n-1}=6(1)^{n-1}$

$(0.5)^{n-1}=2$

$2^{1-n}=2^1$

So comparing both the sides, we get

$1-n=1$

$n=0$

Hence, it is a zero-order reaction.

For the

$1^{st}$ order reaction,

$A(g) \rightarrow 2B(g) + C(s),$

$t_{1/2}= 24$ min. The reaction is carried out by taking a certain mass of 'A' enclosed in a vessel in which it exerts a pressure of 400 mm Hg. The pressure of the reaction mixture after the expiry of 48 min will be:

0%
700 mm
0%
600 mm
0%
500 mm
0%
1000 mm

Explanation

$A(g) \longrightarrow 2B(g)+ C(s)$

$P_{0}$

$0$

$(P_{0} - P)$

$2P$

$t_{1/2} = 24\ min; \quad k=\dfrac{0.693}{t_{1/2}}$

$k = \dfrac{2.303}{t} log \dfrac{P_{0}}{P_{0}-P}$

$\dfrac{0.693}{24}= \dfrac{2.303}{48} log \dfrac{400}{400-P}$

$4 = \dfrac{400}{400-P}$

$400-P = 100$

$P = 300$

$\therefore$ The pressureof the reaction mixture after expiry time

$=P_{0} + P = 400+300=700.$ mm

The gaseous decomposition reaction:

$A(g)\rightarrow 2B(g) + C(g)$ , is observed to first order over the excess of liquid water at

$25^oC$ . It is found that after 10 minutes, the total pressure of the system is 188 torr and after a very long time it is 388 torr. Calculate the rate constant of the reaction in

$hr^{-1}$ . The vapour pressure of

$H_2O$ at

$25^oC$ is 28 torr.

$[ln 2 = 0.7,\: ln 3 = 1.1, \: ln 10 = 2.3]$

0%
$0.02$
0%
$1.2$
0%
$0.2$
0%
$0.12$

Explanation

$A(g)\rightarrow 2B (g) + C(g)$

Let the initial pressure be

$P_0$ 0 0
After 10 mins,

$(P_0 - x)$

$2x$

$x$
After long time

$(t\rightarrow \infty )$ 0

$2P_{0}$

$P_{0}$

Now,

$(P_0 - x) + 2x + x +$ vapour pressure of

$H_2O = 188$ .

$P_0 + 2x = 160$ and

$3P_0 + 28 = 388$

So,

$P_0=120$ and

$x =20 \: torr$ .

$k=\cfrac{1}{t}ln\left ( \cfrac{P_{0}}{P_{0}-x} \right )=0.02\: min^{-1}=1.2 \: hr^{-1}.$

Given

$X \rightarrow$ product (Taking

$1^{st}$ order reaction)

conc of $X$ (mol/lit.)	$0.01$	$0.0025$
Time (min.)	$0$	$40$

Half life period of this reaction is :

0%
$0\ min$
0%
$20\ min$
0%
$40\ min$
0%
$\sqrt{20}\ min$

Explanation

Initial concentration is 0.01 M. The concentration after 40 min is 0.0025 M.

Ratio of concentrations

$=\frac {0.0025} {0.01}=\frac {1} {4}.$

Thus in 40 minutes, the concentration is reduced to one fourth. This corresponds to two half-life periods.

Thus the half-life period of the reaction is 20 min.

For the reaction:

$N_2 + 3H_2\rightarrow 2NH_3$ ,

$\cfrac{d[NH_3]}{dt}= 2 \times 10^{-4} mol\ L^{-1}s^{-1}$ , the value of

$\cfrac{-d[H_2]}{dt}$ would be:

0%
$1 \times 10^{-4} mol \; L^{-1}s^{-1}$
0%
$3 \times 10^{-4} mol\; L^{-1}s^{-1}$
0%
$4 \times 10^{-4} mol\; L^{-1}s^{-1}$
0%
$6 \times 10^{-4} mol\; L^{-1}s^{-1}$

Explanation

${ N }_{ 2 }+{ 3H }_{ 2 }\longrightarrow { 2NH }_{ 3 }$

Applying rate law.

$\dfrac { -1 }{ 3 } \dfrac { d\left[ { H }_{ 2 } \right] }{ dt } =\dfrac { 1 }{ 2 } \dfrac { d\left[ { NH }_{ 3 } \right] }{ dt }$

Now, given that

$\dfrac { d\left[ { NH }_{ 3 } \right] }{ dt } =2\times { 10 }^{ -4 }mol\ { L }^{ -1 }{ S }^{ -1 }$

Using above relation,

$\dfrac { -d\left[ { H }_{ 2 } \right] }{ dt } =\dfrac { 3 }{ 2 } \dfrac { d\left[ { NH }_{ 3 } \right] }{ dt }$

$\dfrac { -d\left[ { H }_{ 2 } \right] }{ dt } =\dfrac { 3 }{ 2 } \times 2\times { 10 }^{ -4 }\ mol\ { L }^{ -1 }{ S }^{ -1 }$

$\therefore$

$\dfrac { -d\left[ { H }_{ 2 } \right] }{ dt } =3\times { 10 }^{ -4 }\ mol\ { L }^{ -1 }{ S }^{ -1 }$

Hence, option B is correct.

The conversion of vinyl allyl ether to pent-4-enol follows first-order kinetics. The following plot is obtained for such a reaction. Determine rate constant for the reaction.

0%
$4.6 \times 10^{-2}s^{-1}$
0%
$1.2 \times 10^{-2}s^{-1}$
0%
$2.3 \times 10^{-2}s^{-1}$
0%
$8.4 \times 10^{-2}s^{-1}$

Explanation

$Slope = \frac{k}{-2.303}$

$\implies k = Slope \times (-2.303)$

$\implies k = \frac{-6}{10 \times 60} \times (-2.3)$

$\implies k= 2.3 \times 10^{-2} s^{-1}$

It takes 32 minutes to complete 99% of a first order reaction from start. Calculate the time required (in a minute) to complete 99.9% of the reaction from the start?

0%
50
0%
48
0%
55
0%
46

Explanation

It takes 32 minutes to complete 99% of a first order reaction.

$K=\frac{1}{32}ln\frac{a}{0.01a}=\frac{ln 100}{32}$

$t=\frac{32}{ln 100}ln\frac{a}{0.01a}$

$t=48 min$

Half-life is independent of the concentration of the reactant. After

$10$ minutes, the volume of

$N_{2}$ gas is

$10\ L$ and after complete reaction, it is

$50\ L$ . Hence, the rate constant is:

0%
$(2.303 /10)\ log\ 5$ $min^{-1}$
0%
$(2.303 /10)\ log\ 1.25\ min^{-1}$
0%
$(2.303 /10)\ log\ 2\ min^{-1}$
0%
$(2.303 /10)\ log\ 4\ min^{-1}$

$A(g) \rightarrow 2B(g)+C(g)$ is observed to be a first order reaction. On starting with pure

$A$ , it is found that, at the end of 10 min, the total pressure of the system is

$176\ mm$ of

$Hg$ and after a long time, it is

$270\ mm$ of

$Hg$ . Which of the following is /are correct for the given data?

0%
The initial pressure $A$ is $90\ mm\ Hg$
0%
The partial pressure of $A$ after 10 min is $47\ mm\ Hg$
0%
The rate constant of the reaction is $0.0649 /min$
0%
None of the above

Explanation

Let the initial pressure of

$A$ will be a

$mm\ Hg$ .

After a long time, the pressures of

$A, B$ and

$C$ will be

$0, 2a$ and

$a$ atm respectively. Thus,

$3a=270\ or\ a\ = 90\ mm \ Hg$ .

Hence, the initial pressure

$A$ is

$90$

$mm\ Hg$ .

Option

$A$ is correct.

After 10 min, the pressures of

$A, B$ and

$C$ will be

$a-x, 2x$ and

$x$ , respectively. Total pressure is

$176\ mm\ Hg$ .

$a-x+2x+x=176\ or\ x=43\ mm\ Hg$

$a-x=90-43=47\ mmHg$
Hence, the partial pressure of

$A$ after 10 min is

$47\ mm\ Hg$ .

Thus, the option

$B$ is correct.

$k=\dfrac {2.303}{t}log \dfrac {a}{a-x}=\dfrac {2.303}{10}log \dfrac {90}{47}=0.0649 /min$

Hence, the rate constant of the reaction is 0.0649 /min.

Thus, option

$C$ is correct.

CBSE Questions for Class 12 Engineering Chemistry Chemical Kinetics Quiz 9 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

CBSE Questions for Class 12 Engineering Chemistry Chemical Kinetics Quiz 9 - MCQExams.com

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Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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