Find the values of $$A, B$$ and $$C$$ in the following table for the reaction $$X + Y\rightarrow Z$$. The reaction is of first order w.r.t $$X$$ and zero w.r.t. $$Y$$. Exp. $$[X] (mol\ L^{-1})$$ $$[Y] (mol\ L^{-1})$$ Initial rate $$(mol\ L^{-1}s^{-1})$$ $$1.$$ $$0.1$$ $$0.1$$ $$2\times 10^{-2}$$ $$2.$$ $$A$$ $$0.2$$ $$4\times 10^{-2}$$ $$3,$$ $$0.4$$ $$0.4$$ $$B$$ $$4.$$ $$C$$ $$0.2$$ $$2\times 10^{-2}$$

$$A = 0.2\ mol\ L^{-1}, B = 8\times 10^{-2} mol\ L^{-1}s^{-1}, C = 0.1\ mol\ L^{-1}$$

$$A = 0.4\ mol\ L^{-1}, B = 4\times 10^{-2} mol\ L^{-1}s^{-1}, C = 0.2\ mol\ L^{-1}$$

$$A = 0.2\ mol\ L^{-1}, B = 2\times 10^{-2} mol\ L^{-1}s^{-1}, C = 0.4\ mol\ L^{-1}$$

$$A = 0.4\ mol\ L^{-1}, B = 2\times 10^{-2} mol\ L^{-1}s^{-1}, C = 0.4\ mol\ L^{-1}$$

The following data were obtained during the first order thermal composition of $$SO_{2}Cl_{2}$$ at a constant volume. $$SO_{2}Cl_{2(g)} \rightarrow SO_{2(g)} + Cl_{2(g)}$$ Experiment $$Time/ s^{-1}$$ Total pressure (atm) $$1$$ $$2$$ $$0$$ $$100$$ $$0.5$$ $$0.6$$ What is the rate of reaction when total pressure is $$0.65\ atm$$?

$$2.235\times 10^{-3}\ atm\ s^{-1}$$

$$1.55\times 10^{-4}\ atm\ s^{-1}$$

MCQ Questions for Class 12 Engineering Chemistry Chemical Kinetics Quiz 12

In the presence of an acid, the initial concentration of cane sugar was reduced from

$0.20$ to

$0.10$ molar in

$5$ hours and from

$0.2$ to

$0.05$ molar in

$10$ hours. The reaction is of:

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

Which of the following are the correct representations of a zero order reaction, where A represents the reactant?

0%
a, b, c
0%
a, b, d
0%
b, c, d
0%
a, c, d

Explanation

The graphs a, b, d are the correct representations of a zero-order reaction, where A represents the reactant.

For a zero order reaction, rate

$\displaystyle =k[A]^0=k=$ constant. Hence, the graph of rate vs

$[A]$ is a straight line parallel to [A] axis.

Also,

$\displaystyle [A]_t = -kt +[A]_0$

The plot of

$\displaystyle [A]_t$ vs time (t) is a straight line with a negative slope.

Slope

$\displaystyle = -k$

The half life period

$\displaystyle t_{1/2} = \dfrac {[A]_0}{2k}$

The graph of

$\displaystyle t_{1/2}$ with

$\displaystyle [A]_0$ is a straight line with positive slope. The slope is

$\displaystyle \dfrac {1}{2k}$

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

$20^{\circ}C$ to

$35^{\circ}C$ ?

0%
$521\ kJ\ mol^{-1}$
0%
$34.7\ kJ\ mol^{-1}$
0%
$15.1\ kJ\ mol^{-1}$
0%
$342\ kJ\ mol^{-1}$

Explanation

$K_2=2K_1, T_2=35+273=308$ $K, T_1=20+273=293$ $K$

$\ln \cfrac {K_2}{K_1}=\cfrac {E_a}{R}\left [\cfrac {1}{T_1}-\cfrac {1}{T_2}\right]$

$\therefore \ln 2=\cfrac {E_a}{8.314}\left[\cfrac {1}{293}-\cfrac {1}{308}\right]$

$\therefore \cfrac {8.314\times \ln 2}{(3.41-3.24)\times 10^{-3}}=E_a$

$\therefore E_a\approx 34.7$ $kJ/mol$

Match the List-I and List-II:

List-I	List-II
(A) Rate constant has the same unit as the rate of reaction.	(i) One
(B) Reactions having apparent molecularity more than three reaction	(ii) Zero order
(C) Reactions having molecularity two but order of reaction is one.	(iii) Complex reaction
(D) For a reaction, $A \rightarrow B$ , the rate of reaction doubles as the ocular reaction concentration of A is doubled.	(iv) Pseudo unimolecular reaction

0%
(A) - ii (B) - iv (C) - iii (D) - i
0%
(A) - ii (B) - iii (C) - iv (D) - i
0%
(A) - iii (B) - ii (C) - iv (D) - i
0%
(A) - ii (B) - iv (C) - i (D) - iii

Explanation

(A) Rate constant has the same unit as the rate of reaction for zero order reactions.
(B) Reactions having apparent molecularity more than three reaction for pseudo uni-molecular reactions.
(C) Reactions having molecularity two but order of reaction is one are Complex reaction.
(D) For a reaction,

$A \rightarrow B$ , the rate of reaction doubles as the ocular reaction concentration of A is doubled in first order reactions.
So the correct options are (A) - ii (B) - iv (C) - iii (D) - i

Hence option A is correct.

At a certain temperature, the first order rate constant

$k_{1}$ is found to be smaller than the second order rate constant

$k_{2}$ . If

$E_{a}(1)$ of the first order reaction is greater than

$E_{a}(2)$ of the second order reaction, then as temperature is raised:

0%
$k_{2}$ will increase faster than $k_{1}$
0%
$k_{1}$ will increase faster than $k_{2}$ but will always remain less than $k_{2}$
0%
$k_{1}$ will increase faster than $k_{2}$ and become equal to $k_{2}$
0%
$k_{1}$ will increase faster than $k_{2}$ and become greater than $k_{2}$

A substance

$A$ undergoes decomposition in solution following the first order kinetics. Flask

$I$ contains 1 litre of 1 M solution of

$A$ and flask

$II$ contains

$100$

$ml$ of 0.6 M solution. After 8 hours, if the concentration of

$A$ in flask

$I$ became 0.25 M, what will be the time required in hours for the concentration of

$A$ in flask

$II$ to become 0.3 M?

0%
0.4
0%
2.4
0%
4.0
0%
Unpredictable as rate constant is not given

Explanation

Initial concentration of

$A$ =

$1M$

Concentration after

$2$ hours=

$0.25M$

Since, the reaction follows first order kinetics; so, time taken to have concentration

$0.5M=4$ hours

This is half life

$t_{1/2}$ of the reaction.

$\Rightarrow$

$t_{1/2}=4$ hours

Now in Flask II

Initial concentration of

$A=0.6M$

Final concentration of

$A=0.3M$

Since, the concentration is reduced to half of its initial value so time required for it to happen will be

$t_{1/2}$ i.e.

$4$ hours.

A certain reaction rate increases

$1000$ folds in the presence of a cataylst at

$27^{\circ}C$ . The activation energy of the original pathway is

$98\ kJ/mol$ . What is the activation energy of the new pathway?

0%
$80.77\ kJ$
0%
$56.38\ kJ$
0%
$24.67\ kJ$
0%
$90.34\ kJ$

Explanation

Given that rate increase by 1000 tines

$\Rightarrow \dfrac{K_2}{K_1}=1000$

$\log \dfrac {K_{2}}{K_{1}} = \log 1000=3$

$3= \dfrac {E_{1} - E_{2}}{2.303RT}$

$3= \dfrac {98000 - E_{2}}{2.303\times 8.314 \times 300}$

$E_{2}(catalyst) = 80.77\ kJ$ .

It takes

$1\ h$ for a first order reaction to go to

$50$ % completion. The total time required for the same reaction to reach

$87.5$ % completion will be:

0%
$1.75\ h$
0%
$6.00\ h$
0%
$3.50\ h$
0%
$3.00\ h$

Explanation

For first order reaction we have:-

$K=\cfrac {1}{t} ln \left(\cfrac {a}{a-x}\right)$

$- (i)$

where,

$K$ = rate constant

$t$ =time

$a$ = initial concentration of the reactant

$x$ = amount of reactant reacted in time

$t$

Case I:-

$t=1$ hour &

$x=0.5a$

$(a-x)=0.5a$

Putting the values in

$(i)$ :-

$K=\cfrac {1}{1} ln \left(\cfrac {1}{0.5}\right)= ln 2$

$- (ii)$

Case II:-

$t=?$ &

$x=0.875a$

$(a-x)=0.125a$

Putting the value in

$(ii)$ :-

$K=\cfrac {1}{t} ln \left(\cfrac {1}{0.125}\right)=\cfrac {1}{t} ln (8)$

$- (iii)$

From,

$(ii)$ and

$(iii)$

$\Rightarrow ln2=\cfrac {1}{t} ln 8$

$\Rightarrow t=\cfrac {ln8}{ln2}=3$ hours

$[\because ln(2^3)=3ln2]$

Calculate the rate constant for decomposition of ammonium nitrate from the following data;

$NH_4NO_2\rightarrow N_2+2H_2O$

Time (minutes)	10	25	$\infty$
Vol.of $N_2$ ml	6.25	13.65	35.05

Given that the reaction follows first order kinectics.

0%
$0.0197 min^{-1}$
0%
$1.97 min^{-1}$
0%
$0.197 min^{-1}$
0%
None of these

Explanation

The decomposition of ammonium nitrate is given by

$NH_4NO_2\longrightarrow N_2+2H_2O$

The kinetics of this reaction is studied by measuring volume of

$N_2$ formed at different intervals.

Let,

$V_\infty$ denotes volume of

$H_2$ formed at

$\infty$ time.

(time at which reaction is complete)

$\Rightarrow$

$V_\infty \propto$ amount of

$NH_4NO_2$ present initially

$(a)$

$\Rightarrow$

$V_\infty \propto a$

$- (i)$

Let,

$Vt$ is the volume of

$N_2$ at any time

$t$ .

$\Rightarrow$

$Vt\propto$ amount of

$NH_4NO_2$ reacted

$(x)$

$\Rightarrow$

$Vt\propto x$

$- (ii)$

Now,

$(a-x)\propto (V_\infty-Vt)$

$- (iii)$

Now, since this reaction follows first order kinetics, so,

$K=\cfrac {1}{t} ln \left(\cfrac{a}{a-x}\right)$

$- (iv)$

where,

$K$ = rate constant,

$t$ = time,

$a$ = initial concentration of reactant,

$x$ = amount of reactant reacted in time

$t$

Putting (i) & (iii) in (iv):-

$K=\cfrac {1}{t} ln \left(\cfrac {V_\infty}{V_\infty-Vt}\right)$

$- (v)$

According to question,

$V_\infty=a=35.05$

Now, at

$t=10$ min,

$V_t=6.25$ ml

Putting the values in

$(v)$ :-

$K=\cfrac {1}{10} ln \left(\cfrac {35.05}{35.05-6.25}\right)$

$=\cfrac {1}{10} ln \left(\cfrac {35.05}{20.00}\right)$

$=\cfrac {1}{10} ln (1.22)=\cfrac {1}{10}\times 0.1900$

$=0.01900$

$min^{-1}$

The spontaneous decomposition of radio nuclei is a first order rate process.U-238 disintegrates with the emission of

$\alpha$ -particles and has a half-life of

$4.5 \times 10^9$ years. If at time t=0, 1 mole of U-238 is present, what will be the number of nuclei left after 1 billion years?

0%
$5.16 \times 10^{23}$
0%
$5.16 \times 10^{20}$
0%
$5.16 \times 10^{19}$
0%
None of these

Explanation

We know that,

Radioactive disintegration follows first order kinetics.

Now, half life period of a reaction is the time taken by reaction to be half completed.

So, amount of reactant left after one half life period=

$\cfrac {[A]_0}{2}$ ;

$A_0$ is the amount present at

$t=0$

Amount present after

$2$ half lines=

$\cfrac {[A]_0}{2^2}$

In general, amount of reactant left after

$n$ half lives=

$\cfrac {[A]_0}{2^n}$

Now, Half life of

$U-238=4.5\times 10^{9}$ yrs

Given time,

$t=1$ billion years

$=10^{9}$ years

So, no. of half lives=

$\cfrac {10^9}{4.5\times 10^9}=\cfrac {1}{4.5}$

Amount of reactant left=

$\cfrac {[A]_0}{2^{4.5}}$

$\cfrac {[A]_0}{22.68}$

Now, at

$t=0$ , amount of

$U-238$ present=

$1$ mole=

$6.022\times10^{23}$ molecules (nuclei)

$\therefore$ No. of nuclei left after

$1$ billion years=

$\cfrac {6.022\times 10^{23}}{22.68}$

$=0.27\times 10^{23}$

A first-order reaction is

$40$ % complete after

$8$ min. How long will it take before it is

$90$ % complete?

0%
40.20 min
0%
36.36 min
0%
24.52 min
0%
None of these

Explanation

For first order reaction we have:

$K= \cfrac {1}{t}$

$ln \cfrac {a}{a-x}$

$(i)$

where,

$K$ = rate constant of reaction

$t$ = time

$a$ = initial concentration of the reactant

$x$ = amount of reactant reacted in time

$t$

Since, the reactant has reacted

$40$ % in

$8$ min, the remaining reactant is

$60$ %

$\Rightarrow$ At

$t=8$ min ,

$(a-x)=0.6a$

Now, we have to find time when the reaction is

$90$ % complete i.e.

$(a-x)=0.1a$

So Case I:

$t=8$ min;

$(a-x)=0.6a$

Case II:

$t=?$ ;

$(a-x)=0.1a$

Since the reaction is of first order

$K$ is same in both cases so using the formula

$(i)$ :-

$\cfrac {1}{2}\times ln \left(\cfrac {1}{0.6}\right)=\cfrac {1}{t}\times ln \left(\cfrac {1}{0.1}\right)$

$\Rightarrow$

$t=8\times \cfrac {ln(1/0.1)}{ln(1/0.6)}=8\times \cfrac {2.302}{0.511}=36.36$ min

Answer: (B)

$36.36$ min

Which among the following is wrong for

$1^{st}$ order reaction?

0%
$t_{99.9\%}$ $= \dfrac {5}{2}t_{99\%}$
0%
$t_{99.9\%}$ $= 10t_{1/2}$
0%
$t_{99.9\%}$ $= 3t_{90\%}$
0%
$t_{96.875\%}$ $= 5t_{1/2}$

A reaction takes place in thee steps.The rate constants are

$K_1,K_2$ and

$k_3$ . The over all rate constant

$k=\cfrac{K_1K_3}{K_2}$ . If (energy of activation)

$E_1, E_2$ and

$E_3$ are

$100, 20$ and

$40 kJ$ .The overall energy of activation is:

0%
30
0%
40
0%
50
0%
60

In a reaction carried out at 400 k,

$0.0001\%$ of the total number of collisions are effective. The energy of activation of the reaction is:

0%
zero
0%
7.37 k cal/mol
0%
9.212 k cal/mol
0%
11.05 k cal/mol

Explanation

We know that, Arrhenius equation for calculation of energy of activation of reaction with rate constant

$K$ and temeperature

$T$ is

$K=A$

$e^{-Ea/RT}$

where,

$Ea$ = Arrhenius activation energy

$A$ = pre exponential factor (frequency factor)

Now,

$e^{-Ea/K_BT}$ = Fraction of collision having more than activation energy

where,

$K_B=$ Boltzmann constant

Given,

$T=400$

$K$ and effective collision=

$0.0001$ %

$\Rightarrow$ Effective Collision=

$e^{-Ea/K_BT}$

$\Rightarrow$

$0.0001$ %=

$e^{-Ea/1.3\times 10^{-23}\times 400}$

$\Rightarrow$

$10^{-6}$ =

$e^{-Ea/1.3\times 10^{-23}\times 400}$

$\Rightarrow$

$2.303\times \log 10^{-6}$ =

$\cfrac {-Ea}{1.3\times 10^{-23}\times 400}$

$\Rightarrow$

$2.303\times (-6)$ =

$\cfrac {-Ea}{1.3\times 10^{-23}\times 400}$

$\Rightarrow$

$Ea$ =

$1.3\times 10^{-23}\times 400\times6\times 2.303=7.19\times 10^{-20}$

$J/mol$

For the following reaction at a particular temperature, according to the equations,

$2N_2O_5 \rightarrow 4NO_2 + O_2$

$2NO_2+1/2O_2 \rightarrow N_2O_5$
the activation energies are

$E_1$ and

$E_2$ respectively, then:

0%
$E_1 > E_2$
0%
$E_1 < E_2$
0%
$E_1=2E_2$
0%
$\sqrt{E_1E_2^2}=1$

How much faster would a reaction proceed at 298 K than at 273 K, if the activation energy is

$65 kJ mol^{-1}$ ?

0%
22 times
0%
11 times
0%
33 times
0%
5.5 times

Explanation

$\because k = Ae^{-Ea/RT}$
$\implies \dfrac{k_{25}}{k_0} = \dfrac{A_{25} e^{65000/8.314 \times 298}}{A_{0} e^{65000/8.314 \times 275}}$
$\implies \dfrac{e^{-26.24}}{e^{-28.64}}$
$\implies e^{2.40} = 11$

Thus, the reaction will proceed 11 times faster.

For the equilibrium,

$A(g) \rightleftharpoons B(g), \triangle H$ is

$-40\ kJ/mol$ . If the ratio of the activation energies of the forward

$(E_{f})$ and reverse

$(E_{b})$ reactions is

$2/3$ then:

0%
$E_{f} = 60\ kJ/mol; E_{b} = 100\ kJ/mol$
0%
$E_{f} = 30\ kJ/mol; E_{b} = 70\ kJ/mol$
0%
$E_{f} = 80\ kJ/mol; E_{b} = 120\ kJ/mol$
0%
$E_{f} = 70\ kJ/mol; E_{b} = 30\ kJ/mol$

Explanation

The given reaction is,

$A(g)\rightleftharpoons B(g)$

$\triangle H= -40 KJ/mol$

$\cfrac {E_f}{E_b}=\cfrac {2}{3} \Rightarrow E_f=\cfrac {2}{3} E_b$

From graph we have,

$\triangle H=E_b-E_f$

$\Rightarrow -40=E_b-\cfrac {2}{3}E_b$

$\Rightarrow -40= \cfrac {1}{3}E_b \Rightarrow E_b=-120 KJ/mol$

$E_f=E_b-\triangle H$

$=-120- (-40)$

$-80$

$KJ/mol$ .

The reaction at hydrogen and iodine monochloride is represented by the equation is

$H_{2} (g) + 2ICl (g) \rightarrow 2HCl (g) + I_{2} (g)$
This reaction is first order in

$H_{2}(g)$ and also first - order in ICI(g). which of these proposed mechanism can be consistent with the given information about this reaction ?
Mechanism I :

$H_{2} (g) + 2ICl (g) \rightarrow 2HCl (g) + I_{2} (g)$
Mechanism II :

$H_{2} (g) + ICl (g) \overset{slow}{\rightarrow} HCl (g) + HI (g)$

$HI (g) + ICl (g) \overset{fast}{\rightarrow} HCl (g) + I_{2} (g)$ :

0%
I only
0%
II only
0%
Both I and II
0%
Neither I nor II

Explanation

According to question the given reaction is first order in

$H_2(g)$ and also first order in

$ICl(g)$ so, the slowest (rate determining) step in the reaction should involve

$H_2(g)$ &

$ICl(g)$ one mole each.

The mechanism

$II$ is seeming consistent with the conditions stated as:-

$H_2(g)+ICl(g)\longrightarrow HCl(g)+HI(g)$ (slow)

$Rate=K[H_2][ICl]$

$HI(g)+ICl(g)\longrightarrow HCl(g)+I_2(g)$ (fast)

Find the values of

$A, B$ and

$C$ in the following table for the reaction

$X + Y\rightarrow Z$ . The reaction is of first order w.r.t

$X$ and zero w.r.t.

$Y$ .

Exp.	$[X] (mol\ L^{-1})$	$[Y] (mol\ L^{-1})$	Initial rate $(mol\ L^{-1}s^{-1})$
$1.$	$0.1$	$0.1$	$2\times 10^{-2}$
$2.$	$A$	$0.2$	$4\times 10^{-2}$
$3,$	$0.4$	$0.4$	$B$
$4.$	$C$	$0.2$	$2\times 10^{-2}$

0%
$A = 0.2\ mol\ L^{-1}, B = 8\times 10^{-2} mol\ L^{-1}s^{-1}, C = 0.1\ mol\ L^{-1}$
0%
$A = 0.4\ mol\ L^{-1}, B = 4\times 10^{-2} mol\ L^{-1}s^{-1}, C = 0.2\ mol\ L^{-1}$
0%
$A = 0.2\ mol\ L^{-1}, B = 2\times 10^{-2} mol\ L^{-1}s^{-1}, C = 0.4\ mol\ L^{-1}$
0%
$A = 0.4\ mol\ L^{-1}, B = 2\times 10^{-2} mol\ L^{-1}s^{-1}, C = 0.4\ mol\ L^{-1}$

Explanation

$Rate = k[X][Y]^{0}$

Rate is independent of the conc. of

$Y$ and it depends only on the conc. of

$X$ and it is the first-order reaction.

From exp.

$(1), 2\times 10^{-2} = k(0.1) ..... (i)$

From exp.

$(2), 4\times 10^{-2} = k(A) ....(ii)$

Dividing (ii) and (i),

$\dfrac {4\times 10^{-2}}{2\times 10^{-2}} = \dfrac {k(A)}{k(0.1)} = \dfrac {A}{0.1}$

$\Rightarrow 2\times 0.1 = A$

$\Rightarrow A = 0.2\ mol\ L^{-1}$

From exp.

$(3), B = k(0.4) .... (iii)$

Dividing (iii) and (i),

$\dfrac {B}{2\times 10^{-2}} = \dfrac {k(0.4)}{k(0.1)} = 4$

$\Rightarrow B = 4\times 2\times 10^{-2} = 8\times 10^{-2} mol\ L^{-1} s^{-1}$

From exp.

$(4), 2\times 10^{-2} = k(C) .....(iv)$

Dividing (iv) and (i),

$\dfrac {2\times 10^{-2}}{2\times 10^{-2}} = \dfrac {k(C)}{k(0.1)} = \dfrac {C}{0.1}$

$\Rightarrow C = 0.1\ mol\ L^{-1}$ .

The activation energy for a reaction is

$9.0 k cal/mol$ . The increase in the rate constant when its temperature is increased from

$298 K$ to

$308 K$ is:

0%
$63$ %
0%
$50$ %
0%
$100$ %
0%
$10$ %

Explanation

$2.303 \log \dfrac{K_2}{K_1} = \dfrac{E_a}{R} \left[\dfrac{T_2 - T_1}{T_1T_2}\right]$

$\log \dfrac{K_2}{K_1} = \dfrac{9.0 \times 10^3}{2.303 \times 2} \left[\dfrac{308 - 298}{308 \times 298}\right]$

$\dfrac{K_2}{K_1} = 1.63; K_2 = 1.63 \,K_1; \dfrac{1.63 K_1 - K_1}{K_1} \times 100 = 63.0 \%$

For a reaction

$A_{2} + B_{2}\rightleftharpoons 2AB$ the figure shows the path of the reaction in absence and presence of a catalyst. What will be the energy of activation for forward

$(E_{f})$ and backward

$(E_{b})$ reaction in presence of a catalyst and

$\triangle H$ for the reaction? the dotted curve is the path of reaction in presence of a catalyst.

0%
$E_{f} = 60\ kJ/mol, E_{b} = 70\ kJ/ mol, \triangle H = 20\ kJ/ mol$
0%
$E_{f} = 20\ kJ/mol, E_{b} = 20\ kJ/ mol, \triangle H = 50\ kJ/ mol$
0%
$E_{f} = 70\ kJ/mol, E_{b} = 20\ kJ/ mol, \triangle H = 10\ kJ/ mol$
0%
$E_{f} = 10\ kJ/mol, E_{b} = 20\ kJ/ mol, \triangle H = -10\ kJ/ mol$

Explanation

$E_{f} = 70 - 60= 10\ kJ/mol$

$E_{b} = 70 - 50 = 20\ kJ/ mol$

$\triangle H = 50 - 60 = -10\ kJ/ mol$

Option D is correct

The energy diagram of a reaction

$P + Q \rightarrow R + S$ is given. What are

$A$ and

$B$ in the graph?

0%
$A \rightarrow$ activation energy, $B \rightarrow$ heat of reaction
0%
$A \rightarrow$ threshold energy, $B \rightarrow$ heat of reaction
0%
$A \rightarrow$ heat of energy, $B \rightarrow$ activation reaction
0%
$A \rightarrow$ potential energy, $B \rightarrow$ energy of reaction

Explanation

The heat of reaction is given by:-

$\triangle H=\quad { Ea }_{ f }\quad -\quad { Ea }_{ b }$ .

Let's suppose

$O$ is the highest point of the curve then

$RO$ is forward activation energy and

$PO$ is backward activation energy and the difference of these two will give the heat of reaction, [Indicated in the figure]

Hence, the correct answer is option

$\text{A}$ .

The activation energy for the reaction-

$H_{2}O_{2} \rightarrow H_{2}O + \dfrac {1}{2} O_{2}$
is

$18\ K\ cal/mol$ at

$300\ K$ . calculate the fraction of molecules of reactants having energy equal to or greater than activation energy?
Anti log

$(-13.02) = 9.36\times 10^{-14}$

0%
$9.36\times 10^{-14}$
0%
$1.2\times 10^{-12}$
0%
$4.2\times 10^{-30}$
0%
$5.2\times 10^{-15}$

Explanation

We know that Arrhenius equation for calculation of energy of activation of a reaction is given as:

$K=A$

$e^{-Ea/RT}$

$- (i)$

where,

$K$ = Rate constant

$A$ = pre exponential factor or frequency factor

$Ea$ = Activation energy

$R$ = Ideal gas constant

$T$ = Temperature in K (Kelvin)

Now, the fraction of moleucle having energy equal to or greater than activation energy is given by:-

$f= e^{-Ea/RT}$

$- (ii)$

Given,

$Ea=18 K Cal/mol$ ,

$T=300 K$

$R=2 CalK^{-1}mol^{-1}$

Using these values in equation

$(ii)$

So,

$p=e^{-18\times 1000/2\times 300}$

$[\because 1KCal=1000Cal]$

Taking

$\log$ on both sides:-

$\log f=\cfrac {18\times 1000}{2\times 300}=-30$

$\Rightarrow f=$

$Antilog$

$(-30)= 10^{-30}$

The following data were obtained during the first order thermal composition of

$SO_{2}Cl_{2}$ at a constant volume.

$SO_{2}Cl_{2(g)} \rightarrow SO_{2(g)} + Cl_{2(g)}$

Experiment	$Time/ s^{-1}$	Total pressure (atm)
$1$ $2$	$0$ $100$	$0.5$ $0.6$

What is the rate of reaction when total pressure is

$0.65\ atm$ ?

0%
$0.35\ atm\ s^{-1}$
0%
$2.235\times 10^{-3}\ atm\ s^{-1}$
0%
$7.8\times 10^{-4}\ atm\ s^{-1}$
0%
$1.55\times 10^{-4}\ atm\ s^{-1}$

Explanation

$\underset {p_o\\ {p_o-p}}{SOCl_2} \rightarrow \underset {0\\p}{SO_2} + \underset {0\ at\ t=0\\p\ at\ t=t}{Cl_2}$

Pressure at time

$t, P_{t} = p_{0} - p + p + p = p_{0} + p$

Pressure of reactants at time

$t, p_{0} - p = 2p_{0} - p_{t}\propto R$

$k = \dfrac {2.303}{t} \log \dfrac {p_{0}}{2p_{0} - P_{t}}$

$= \dfrac {2.303}{100}\log \dfrac {0.5}{2\times 0.5 - 0.6} = \dfrac {2.303}{100}\log 1.25$

$= 2.2318\times 10^{-3} s^{-1}$

Pressure of

$SO_{2}Cl_{2}$ at time

$t(p_{SO_{2}Cl_{2}})$

$= 2p_{0} - P_{t} = 2\times 0.50 - 0.65\ atm = 0.35\ atm$

Rate at that time

$= k\times p_{SO_{2}Cl_{2}}$

$= (2.2318\times 10^{-3})\times (0.35) = 7.8\times 10^{-4} atm\ s^{-1}$ .

Consider figure and mark the correct option.

0%
Activation energy of forward reaction is $E_{1} + E_{2}$ and product is less stable than reactant
0%
Activation energy of forward reaction is $E_{1} + E_{2}$ and product is more stable than reactant
0%
Activation energy of forward and backward reaction is $E_{1} + E_{2}$ and product is more stable than reactant
0%
Activation energy of backward reaction is $E_{1}$ and product is more stable than reactant

Explanation

The product is less stable since the forward reaction is not favorable. The reaction is endothermic (

$\Delta H>0$ ).

Option A is correct.

For reaction

$A \rightarrow B$ , the rate law expression is

$-\dfrac{d[A]}{dT}=k[A]^{1/2}$ . If initial concentration of

$[A]$ is

$[A]_0$ then, which of the following statement is incorrect?

0%
The integrated rate expression is $k=2/t(A^{1/2}_0-A^{1/2})$
0%
The graph of $\sqrt{A}$ vs $t$ will be
0%
The half life period $t_{1/2}=\dfrac{k}{2[A]_o^{1/2}}$
0%
The time taken for $75\%$ completion of reaction $t_{3/4}=\dfrac{\sqrt{A_o}}{K}$

Explanation

From the rate law,

$R=K[A]^{1/2}$

Thus order of Reaction is,

$n=1/2$

General formula for

$t_{1/2}$

$t_{1/2}=\cfrac {2^{n-1}-1}{K[A_o]^{n-1}}$ or

$t_{1/2} \propto \cfrac {1}{[A]^{n-1}}$

For,

$n= 1/2$

$t_{1/2} \propto \cfrac {1}{[A_o]^{1/2-1}} \Rightarrow t_{1/2} \propto \cfrac {1}{[A_o]^{-1/2}}$

or,

$t_{1/2} \propto [A_o]^{1/2}$

Thus, statement C is incorrect.

The fraction of collisions that posses the energy

$E_a$ is given by:

0%
$f = e^{\frac{-Ea}{RT}}$
0%
$f = e^{\frac{Ea}{RT}}$
0%
$f = e^{- Ea.RT}$
0%
$f = e^{Ea.RT}$

Explanation

$k=Pze^{\cfrac {-Ea}{RT}}$

$\cfrac kA$ or

$\cfrac k{Pz}=$ Fraction of collision possessing

$E_a$

$\therefore$ Fraction of collision possessing energy

$E_a= \cfrac kA$

$=\cfrac {Ae^{-E_a/RT}}{A}$

$f=e^{-E_a/RT}$

For an exothermic reaction

$A \rightarrow B$ , the activation energy is

$65\ kJ\ mol^{-1}$ and heat of reaction

$-42\ kJ\ mol^{-1}$ . The activation energy for the reaction

$B \rightarrow A$ would be :-

0%
$23\ kJ\ mol^{-1}$
0%
$107\ kJ\ mol^{-1}$
0%
$65\ kJ\ mol^{-1}$
0%
$42\ kJ\ mol^{-1}$

Explanation

$A \rightarrow B$

$(E_a)_1=65 \ kJ/mol$

$\triangle H=-42 \ kJ/mol$

Now, For

$B \rightarrow A$

$(E_a)_2=(E_a)_1- \triangle H$

$=65 -(-42)=107 \ kJ/mol$

The first order rate constant k is related to temperature as log k

$=15.0-(10^6/T)$ . Which of the following pair of value is correct?

0%
$A=10^{15}$ and $E=1.9\times 10^4$ KJ
0%
$A=10^{-15}$ and $E=40$ KJ
0%
$A=10^{15}$ and $E=40$ KJ
0%
$A=10^{-15}$ and $E=1.9\times 10^4$ KJ

Explanation

$log \ K= 15.0 - \cfrac {10^6}{T}$

We know that, $log \ K=log \ A- \cfrac {E_a}{RT}$

Comparing, we get

$log \ A=15.0$

$\Rightarrow A=10^{15}$

$\cfrac {E_a}{RT}= \cfrac {10^6}{T}$

$\Rightarrow E_a=1.9 \times 10^4 \ kJ$

For a chemical reaction to occur, all of the following must happen except:

0%
A large enough number of collisions must occur
0%
Chemical bonds in the reactants must break
0%
Reactant particles must collide which enough energy for change to occur
0%
Reactant particles must collide with correct orientation

The activation energy of a reaction is

$58.3\ kJ/mole$ . The ratio of the rate constants at

$305K$ and

$300K$ is about:

[

$R=8.3\ J{k}^{-1}{mol}^{-1}$ and Antilog

$0.1667=1.468$ ]

0%
$1.25$
0%
$1.75$
0%
$1.5$
0%
$2.0$

Explanation

$log\quad \frac { { K }_{ 2 } }{ { K }_{ 1 } } =\frac { { E }_{ o } }{ 2.303\quad X\quad R } \quad \left[ \frac { { T }_{ 2 }-{ T }_{ 1 } }{ { T }_{ 1 }{ T }_{ 2 } } \right]$

$=\quad \frac { 58.3 }{ 2.303\quad X\quad 8.3 } \left[ \frac { 305-300 }{ 305\quad X\quad 300 } \right]$

$=\quad 3.0499\quad X\quad 5.4644\quad X\quad 1{ 0 }^{ -5 }$

$=1.666\quad X\quad 1{ 0 }^{ -4 }$

$log\quad \frac { { K }_{ 2 } }{ { K }_{ 1 } } =0.1666\quad X\quad 1{ 0 }^{ -3 }$

$\frac { { K }_{ 2 } }{ { K }_{ 1 } } =antilog(0.1666)$

$=1.468\approx 1.5$

Hence, 1.5 is the answer.

A reaction takes place in three steps with the rate constant

$k_1, k_2$ and

$k_3$ . The overall rate constant

$k=\dfrac{k_1(k_2)^{1/2}}{k_3}$ . If activation energies are

$40, 30$ and

$20\ kJ$ for step

$I, II$ and

$III$ respectively, the overall activation energy of reaction will be:

0%
10
0%
15
0%
30
0%
35

Explanation

Over all activation energy

$=E_{1}+\dfrac{E_{2}}{2}-E_{3}$

$=40+\dfrac{30}{2}-20$

$=35 KJ$

If the activation energy is 65 kJ then how much time faster a reaction proceed at ${ 25 }^{ o }C$ than ${ 0 }^{ o }C$ ?

0%
12
0%
11
0%
13
0%
15

Which of the following reaction has zero activation energy?

0%
${ CH }_{ 4 }+\overset { \bullet }{ C } l\rightarrow \overset { \bullet }{ C } { H }_{ 3 }HCl$
0%
${ Cl }_{ 2 }\rightarrow 2Cl$
0%
$\overset { \bullet }{ C } { H }_{ 3 }+\overset { \bullet }{ C } { H }_{ 3 }+{ CH }_{ 3 }+{ CH }_{ 3 }$
0%
$\overset { \bullet }{ C } { H }_{ 3 }+Cl-Cl\rightarrow { CH }_{ 3 }+Cl+\overset { \bullet }{ C } l$

Explanation

Among the given reactions

${ \dot { CH } }_{ 3 }+{ \dot { CH } }_{ 3 }\rightarrow { CH }_{ 3 }-{ CH }_{ 3 }$ has zero activation energy.

All the other reactions are following a free radial which needs energy in the form

$U.V$ light which is high activation energy.

Where as in the above reaction the free radicals are bonded together by lowering their energy thus by increasing the stability.

So, the reaction

${ \dot { CH } }_{ 3 }+{ \dot { CH } }_{ 3 }\rightarrow { CH }_{ 3 }-{ CH }_{ 3 }$ has zero activation energy.

Thermal decomposition of a compound is of first order. If

$50$ percent of a sample of compound is decomposed in

$120$ mins, how long will it take for

$90$ percent of compound to decompose?

0%
$299$ mins.
0%
$399$ mins.
0%
$99$ mins.
0%
$9.9$ mins.

Explanation

According to half time period

$k=\dfrac{0.693}{{{t}_{\dfrac{1}{2}}}}$

$\Rightarrow \dfrac{0.693}{120}=5.77\times {{10}^{-3}}{{\min }^{-1}}$

Here,

$\left( {{t}_{\dfrac{1}{2}}}=120\ \min t \right)$

Now a first order reaction

$k=\dfrac{2.303}{t}\log \dfrac{1}{a-x}$

$a=100,\ x=90$

$\therefore \ \ 5.77\times {{10}^{-3}}=\dfrac{2.303}{t}\log \dfrac{100}{100-90}$

$t=\dfrac{2.303}{5.77\times {{10}^{-3}}}\log 10$

$=399\ \min .$

This is the required answer.

For a 1st order reaction:

$A\rightarrow Products$ ; the rate of reaction at

$[A]=0.2M$ is

$1\times {10}^{-3}$

$mol$

${L}^{-1}s^{-1}$ reaction will be

$75$ % completed in

0%
$138.6s$
0%
$277.2s$
0%
$442s$
0%
$822s$

The activation energy of a certain uncatalysed reaction at

$300K$ is

$76kJ{mol}^{-1}$ . The activation energy is lowered by

$19kJ{mol}^{-1}$ by the use of catalyst. By what factor, the rate of catalysed reaction is increased?

0%
${e}^{4.6}$
0%
${e}^{5.6}$
0%
${e}^{6.6}$
0%
${e}^{7.6}$

For the equilibrium,

$Ag \rightleftharpoons B(g),\Delta H$ is

$-40kJmol^{-1}$ . If the ratio of activation energies of the forward

$(E_{f})$ and

$(E_{b})$ reaction is

$(2/3)$ then:

0%
$E_{f}=60KJ \ mol^{-1}:E_{b}=100kJ\ mol^{-1}$
0%
$E_{f}=30KJ \ mol^{-1}:E_{b}=70kJ\ mol^{-1}$
0%
$E_{f}=80KJ \ mol^{-1}:E_{b}=120kJ\ mol^{-1}$
0%
$E_{f}=70KJ \ mol^{-1}:E_{b}=30kJ\ mol^{-1}$

What is the activation energy for this change?

0%
$62.4\ kcal$
0%
$2.60\ kcal$
0%
$25.9\ kcal$
0%
None of these

A reaction proceeds through the formation of an intermediate B in a unimolecular reaction

The integrated rate law for this reaction is?

0%
$[A] = [A]_0e^{-k_at}$
0%
$[A] = [A]_0(e^{-k_at} - e^{-k_bt})$
0%
$[A] = \dfrac{[A]_0}{2}(1 + \dfrac{k_a e^{-k_b t} - k_b e^{-k_a t}}{k_a - k_b})$
0%
$[A]= [A]_0(1 + e^{-k_a t} - e^{-k_b t})$

The rate equation for the reaction 2A+B

$\longrightarrow$ C is rate =K[A][B].
the correct statement about this is :

0%
K is independent of [A] and [B]
0%
${t}_{1/2}$ is constant
0%
Unit of K is ${sec}^{-1}$
0%
Rate of formation of C is twice the rate of disappearance of A

Which of the following reactants can be used to distinguish between benzaldehyde and Benzyl alcohol?

0%
${KMnO}4$
0%
${Cr}_2{O}_4$
0%
Sodium metal
0%
All

It is often stated that, near room temperature, a reaction rate doubles if the temperature increases by

${ 10 }^{ \circ }C$ Calculate the activation energy of a reaction that obeys this rule exactly.

0%
$12.4 kcal$
0%
$24.8 kcal$
0%
$6.7 kcal$
0%
$49.6 kcal$

An adiabatic container fitted with a movable adiabatic piston (operating at 1 atm ) is filled with 2 litre of 2 M

$H_2O_2$ (aq) solution at 300 K. If

$H_2O_2$ dissociates following first order decay with a half life of 10 min then change in internal energy in first 20 min will be [R = 2 cal/mol K.]:

0%
900 Cal
0%
450 Cal
0%
1800 Cal
0%
225 Cal

The slope in the activation energy curve is 5.42

$\times$

$10^{3}$ . The value of the activation energy is approximately

0%
104 J $mol^{-1}$
0%
104 MJ $mol^{-1}$
0%
104 KJ $mol^{-1}$
0%
104 J $mol^{-1}K^{-1}$

Consider the data below for a reaction

$A\rightarrow B$

Time (sec)	0	10	20	30
Rate	$1.60\times 10^{-2}$	$1.60\times 10^{-2}$	$1.60\times 10^{-2}$	$1.59\times 10^{-2}$

from the above data the order of reaction is:

0%
Zero
0%
1
0%
2
0%
3

If a reaction

$A + B \rightarrow C$ is exothermic to the extent of

$30 kJ mol^{-1}$ and the forward reaction has an activation energy,

$X\ kJ mol^{-1}$ the activation energy for the reverse reaction is :

0%
$30 \ kJ mol^{-1}$
0%
$X - 30 kJ mol^{-1}$
0%
$X \ kJ mol^{-1}$
0%
$X + 30 \ kJ mol^{-1}$

Decomposition of

$H_2O_2$ follows a first order reaction. In fifty minutes the concentration of

$H_2O_2$ decreases from

$0.5$ to

$0.125\ M$ in one such decomposition. When the concentration of

$H_2O_2$ reaches

$0.05\ M$ , the rate of formation of

$O_2$ , will be:

0%
$6.93\times 10^{-2}\ mol\ min^{-3}$
0%
$6.93\times 10^{-4}\ mol\ min^{-1}$
0%
$2.66\ L\ min^{-1}$ at $STP$
0%
$1.34\times 10^{-2}\ mol\ min^{-1}$

$SO_2CI_2\rightleftharpoons SO_2+CI_2$ is the first order as gas raection with

$K=2.2\times10^{-5} sec^{-1}$ at

$320^oC$ .The percentage of

$SO_2CI_2$ decomposed on heating for 90 minutes is

0%
$1.118$
0%
$0.1118$
0%
$18.11$
0%
$11.30$

If the half-life of a first-order reaction is 100 seconds then the time required for completion of 90 % of the reaction is ;

0%
693 sec
0%
90 sec
0%
332 sec
0%
10 sec

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

0:0:1

Practice Class 12 Engineering Chemistry Quiz Questions and Answers

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

0:0:1

Practice Class 12 Engineering Chemistry Quiz Questions and Answers

Support mcqexams.com by disabling your adblocker.