datavorous/entrance-exam-dataset huggingface dataset:
For a mixture of two volatile,completely miscible liquids A and B, with P A ∘ = 500 torr and P B ∘ = 800 torr , What is the composition of last droplet of liquid remaining in equilibrium with vapour?
Provided the initial ideal solution has a composition of x A  = 0.6 and x B  = 0.4

Options:
A x A = 0.6 ; x B = 0.4
B x A = 0.5 ; x B = 0.5
C x A = 0.7 ; x B = 0.3
D x A = 0.3 ; x B = 0.7

The correct answer is:
x A = 0.7 ; x B = 0.3

Solution:
P A o = 5 0 0 torr ; P B o = 8 0 0 torr
When most of the liquid has vaporized xA=0.6 (given) would be yA=0.6
y A = P A o x A P A o x A + P B o 1 - x A
â 0.6 = 5 0 0 . x A 5 0 0 x A + 8 0 0 1 - x A
x A = 0.70 ; x B = 0.30

Topic: Chemistry, Solutions, JEE Main
For a natural number n, let αn=19n-12n. Then, the value of 31α9-α1057α8 is ______

Options:

The correct answer is:
4

Solution:
If r1 and r2 are roots of the quadratic ax2+bx+c=0

then r1n=31r1n-2-228r1n-1 and r2n=31r2n-2-228r2n-1

So, r1n-r2n=31r1n-2-r2n-2-228r1n-1-r2n-1

Or an-31an-1+228an-2=0 where an=r1n-r2n

Since 19 & 12 are the roots of the quadratic x2-31x+228=0

we know, an-31an-1+228an-2=0

Now, for n=10, α10-31α3+228α8=0

⇒31α9-α1057α8=4

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2022 (25 Jun Shift 1)
For a parallel beam of monochromatic light of wavelength λ, diffraction is produced by a single slit whose width a is of the order of the wavelength of the light. If D is the distance of the screen from the slit, the width of the central maxima will be:

Options:
A 2Dλa
B Dλa
C Daλ
D 2Daλ

The correct answer is:
2Dλa

Solution:
Linear width of central maximum.
D 2θ=2Dθ=2Dλa

Topic: Physics, Wave Optics, JEE Main
For a particle executing SHM, the displacement x is given by x=Acosâ¡Ït. Identify the graph which represents the variation of potential energy (PE) as a function of time t and displacement x.

Options:
A I, III
B II, IV
C II, III
D I, IV

The correct answer is:
I, III

Solution:
Potential energy as a function of displacement :
PE= 1 2 k y 2
â PE vs y graph will be an upwards parabola.
â graph III
Potential energy as a function of time:
PE= 1 2 k y 2 = 1 2 k ( A cos 2 Ït ) 2
âPE= 1 2 k A 2 cos 2 Ït= 1 2 k A 2 ( 1+cos2Ït ) 2
â PE vs time is a cosine wave with angular frequency 2Ïâgraph I.

Topic: Physics, Oscillations, JEE Main
For a particle in a uniform circular motion, the acceleration a â at a point PR, Î¸ on the circle of radius R is (here Î¸ is measured from the y-axis)

Options:
A v 2 R i ^ + v 2 R j ^
B - v 2 R cos θ i ^ + v 2 R sin θ j ^
C - v 2 R sin Î¸ i ^ - v 2 R cos Î¸ j ^
D - v 2 R cos θ i ^ - v 2 R sin θ j ^

The correct answer is:
- v 2 R sin Î¸ i ^ - v 2 R cos Î¸ j ^

Solution:
​

Topic: Physics, Laws of Motion, JEE Main
For a particle in uniform circular motion the acceleration $\vec{a}$ at a point $P(R, \theta)$ on the circle of radius $\mathrm{R}$ is (here $\theta$ is measured from the $x$-axis)

Options:
A
$-\frac{v^2}{R} \cos \theta \hat{i}+\frac{v^2}{R} \sin \theta \hat{j}$

B
$-\frac{v^2}{R} \sin \theta \hat{i}+\frac{v^2}{R} \cos \theta \hat{j}$

C
$-\frac{v^2}{R} \cos \theta \hat{i}-\frac{v^2}{R} \sin \theta \hat{j}$

D
$\frac{v^2}{R} \hat{i}+\frac{v^2}{R} \hat{j}$

The correct answer is:
$-\frac{v^2}{R} \cos \theta \hat{i}-\frac{v^2}{R} \sin \theta \hat{j}$

Solution:
For a particle in uniform circular motion,
$$
\begin{aligned}
& \overrightarrow{\mathrm{a}}=\frac{\mathrm{v}^2}{\mathrm{R}} \text { towards centre of circle } \\
& \therefore \quad \overrightarrow{\mathrm{a}}=\frac{\mathrm{v}^2}{\mathrm{R}}(-\cos \theta \hat{\mathrm{i}}-\sin \theta \hat{\mathrm{j}}) \\
& \text { or } \quad \vec{a}=-\frac{v^2}{R} \cos \theta \hat{i}-\frac{v^2}{R} \sin \theta \hat{j} \\
&
\end{aligned}
$$

Topic: Physics, Rotational Motion, JEE Main, JEE Main 2010
For a particle in uniform circular motion, the acceleration a→ at any point PR,θ on the circular path of radius R is (when θ is measured from the positive x-axis and v is uniform speed):

Options:
A -v2Rsinθi^+v2Rcosθj^
B -v2Rcosθi^+v2Rsinθj^
C -v2Rcosθi^-v2Rsinθj^
D -v2Ri^+v2Rj^

The correct answer is:
-v2Rcosθi^-v2Rsinθj^

Solution:
Acceleration of a particle in uniform circular motion is directed towards the centre of the circle and its magnitude is v2R. The radius vector isr→=Rcosθi^+Rsinθj^

Therefore, a→=-v2Rr^=v2R−cosθi^−sinθj^

Topic: Physics, Work Power Energy, JEE Main, JEE Main 2022 (25 Jun Shift 2)
For a particle performing uniform cicular motion, choose the correct statement(s) from the following

Options:
A
Magnitude of particle velocity (speed) remains constant

B
Particle velocity remains directed perpendicular to radius vector

C
Direction of acceleration keeps changing as particle moves

D
Angular momentum is constant in magnitude but direction keeps changing

The correct answers are:
Magnitude of particle velocity (speed) remains constant
,
Particle velocity remains directed perpendicular to radius vector
,
Direction of acceleration keeps changing as particle moves

Solution:
According to the question, for a particle performing uniform circular motion.
In uniform circular motion :
(i) Speed will always constant.
(ii) Velocity will be tangential in the direction of motion at a particular point. If a string breaks up the particle move tangentially.
(iii) The direction of acceleration $a=\left(\frac{v^2}{r}\right)$ is always be towards centre of the circular path.
(iv) Angular momentum $L=m v r$ is constant in magnitude and direction/out of the plane perpendicularly. So direction of $(L)$ does not change. So does not verify option (d).

Topic: Physics, Motion In Two Dimensions
For a particular reversible reaction at temperature $T, \Delta H$ and $\Delta S$ were found to be both $+v e$. If $T_e$ is the temperature at equilibrium, the reaction would be spontaneous when

Options:
A
$T_e>T$

B
$\mathrm{T}>\mathrm{T}_{\mathrm{e}}$

C
$T_e$ is 5 times $T$

D
$T =T_e$

The correct answer is:
$\mathrm{T}>\mathrm{T}_{\mathrm{e}}$

Solution:
$$
\Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S}
$$
at equilibrium, $\Delta \mathrm{G}=0$
for a reaction to be spontaneous $\Delta$ G should be negative
$$
\therefore \mathrm{T}>\mathrm{T}_{\mathrm{e}}
$$

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2010
For a perfectly black body, its absorptive power is

Options:
A $1$
B $0.5$
C $0$
D Infinity

The correct answer is:
$1$

Solution:
$1$

Topic: Physics, Thermal Properties of Matter, JEE Main
For a periodic motion represented by the equation y=sinωt+cosωt the amplitude of the motion is

Options:
A 1
B 0.5
C 2
D 2

The correct answer is:
2

Solution:
Given the equation of motion is

y=sinωt+cosωt ...1

Simplify equation (1) to obtain the standard form:

y=2sinωtcosπ4+cosωtsinπ4=2sinωt+π4 ...2

Comparing equation (2) with standard equation of SHM y=Asinωt+ϕ, we get amplitude of the given oscillation as 2.

Topic: Physics, Oscillations, JEE Main, JEE Main 2023 (10 Apr Shift 2)
For a photoelectric effect setup, in the following diagram if \( V_{2}>V_{1} \) then

Options:
A \( \lambda_{1}=\sqrt{\lambda_{2}} \)
B \( \lambda_{1} < \lambda_{2} \)
C \( \lambda_{1}=\lambda_{2} \)
D \( \lambda_{1}>\lambda_{2} \)

The correct answer is:
\( \lambda_{1}>\lambda_{2} \)

Solution:
∵V2>V1 λ1>λ2

Topic: Physics, Dual Nature of Matter, JEE Main
For a photosensitive surface, the work function is $3.3 \times 10^{-19} \mathrm{~J}$. Find threshold frequency. (Take $\mathrm{h}=6.6 \times 10^{-34} \mathrm{Js}$ )

Options:
A $5 \times 10^{14} \mathrm{~Hz}$
B $0.5 \times 10^{14} \mathrm{~Hz}$
C $25 \times 10^{14} \mathrm{~Hz}$
D $2.5 \times 10^{14} \mathrm{~Hz}$

The correct answer is:
$5 \times 10^{14} \mathrm{~Hz}$

Solution:
Given $-\varphi=3.3 \times 10^{-19} \mathrm{~J}$ and $\mathrm{h}=6.6 \times 10^{-34} \mathrm{~J}$
The work function is written as -
$\begin{aligned}& v_0=\varphi / \mathrm{h} \\& \Rightarrow \nu_0=\frac{3.3 \times 10^{-19}}{6.6 \times 10^{-34}}=0.5 \times 10^{15}=5 \times 10^{14} \mathrm{~Hz}\end{aligned}$

Topic: Physics, Dual Nature of Matter, JEE Main
For a plane electromagnetic wave propagating in x-direction, which one of the following combinations gives the correct possible directions for the electric field (E) and magnetic field (B) respectively?

Options:
A j^+k^, j^+k^
B -j^+k^, -j^-k^
C j^+k^, -j^-k^
D -j^+k^, -j^+k^

The correct answer is:
-j^+k^, -j^-k^

Solution:
Direction =E→×B→

(i) (j^+k^)×(j^+k^)=0

(ii) (-j^+k^)×(-j^-k^)

=0+ı^-(-ı^)+0

=2ı^

(iii) (j^+k^)×(-j^-k^)=0

(iv) (-j^+k^)×(-j^+k^)

=0

Topic: Physics, Electromagnetic Waves, JEE Main
For a plane eletromagnetic wave, the electric field is given by $\overrightarrow{\mathrm{E}}=90 \sin \left(0.5 \times 10^{3} \mathrm{x}+1.5 \times 10^{11} \mathrm{t}\right) \mathrm{k} . \mathrm{v} / \mathrm{m} .$ The corresponding magnetic field $\vec{B}$ will be

Options:
A $\vec{B}=3 \times 10^{-7} \sin \left(0.5 \times 10^{3} x+1.5 \times 10^{11} \mathrm{t}\right) \hat{i} \mathrm{~T}$
B $\vec{B}=3 \times 10^{-7} \sin \left(0.5 \times 10^{3} x+1.5 \times 10^{11} \mathrm{t}\right) \hat{j} \mathrm{~T}$
C $\vec{B}=27 \times 10^{9} \sin \left(0.5 \times 10^{3} x+1.5 \times 10^{11} \mathrm{t}\right) \hat{j} \mathrm{~T}$
D $\overrightarrow{\mathrm{B}}=3 \times 10^{-7} \sin \left(0.5 \times 10^{3} \mathrm{x}+1.5 \times 10^{11} \mathrm{t}\right) \hat{\mathrm{k}} \mathrm{T}$

The correct answer is:
$\vec{B}=3 \times 10^{-7} \sin \left(0.5 \times 10^{3} x+1.5 \times 10^{11} \mathrm{t}\right) \hat{j} \mathrm{~T}$

Solution:
$E=90 \sin \left(0.5 \times 10^{3} x+1.5 \times 10^{11} t\right) \hat{k} v / m$
Direction of propagation is $-\hat{i}$
$\& E_{0}=90$
$\frac{\mathrm{E}_{0}}{\mathrm{~B}_{0}}=\mathrm{C} \Rightarrow \mathrm{B}_{0}=\frac{\mathrm{E}_{0}}{\mathrm{C}}=\frac{90}{3 \times 10^{8}} \mathrm{~T}=3 \times 10^{-7} \mathrm{~T}$
$\overrightarrow{\mathrm{E}}$ is along $\hat{\mathrm{k}} \& \overrightarrow{\mathrm{E}} \times \overrightarrow{\mathrm{B}}$ should be along $-\hat{\mathrm{i}} \Rightarrow \overrightarrow{\mathrm{B}}$ is should be along $\hat{\mathrm{j}}$ Hence equation of magnetic field is
$\mathrm{B}=3 \times 10^{-7} \sin \left(0.5 \times 10^{3} \mathrm{x}+1.5 \times 10^{11} \mathrm{t}\right) \hat{\mathrm{j}} \mathrm{T}$

Topic: Physics, Electromagnetic Waves, JEE Main
For a population at carrying capacity:

Options:
A Mortality rate = Birth rate
B Mortality rate > Birth rate
C Mortality rate < Birth rate
D None of the above

The correct answer is:
Mortality rate = Birth rate

Solution:
Carrying capacity is the maximum population that can be sustained is a particular area. In other words, the number organisms can survive normally in a particular area. The main limiting factor for the carrying capacity is the resources available in the area or the environmental factors. If the environmental conditions are not met then the population will keep decreasing until the resources rebound. At carrying capacity, the population stabilizes and have the rate of mortality same as the rate of natality.

Topic: Biology, Organisms and Populations, NEET
For a positive integer n, 1+1xn is expanded in increasing powers of x. If three consecutive coefficients in this expansion are in the ratio, 2: 5: 12, then n is equal to

Options:

The correct answer is:
118

Solution:
Cr-1n:Crn:Cr+1n=2:5:12

⇒CrnCr-1n=52 and Cr+1nCrn=125

⇒n-r+1r=52 and n-rr+1=125

⇒2n-7r+2=0 and 5n-17r-12=0

On solving n=118 and r=34

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2020 (02 Sep Shift 2)
For a process to occur spontaneously, the system free energy or chemical potential of the system :

Options:
A Must decrease
B Must increase
C Must remain stable
D None of the given options

The correct answer is:
Must decrease

Solution:
A spontaneous process is that for which the free energy of the system must decrease. It means ΔG=-ve.

Topic: Chemistry, Thermodynamics (C), JEE Main
For a radioactive material, the half-life is 10 minutes. If initially there are 600 number of nuclei, the time taken (in minutes) for the disintegration of 450 nuclei is

Options:
A 30
B 10
C 20
D 15

The correct answer is:
20

Solution:
Given,

N0=600; N′=450
N=N0-N′
N=150
Number of remaining nuclei,
NN0=12tt1/2 (Here, when, t=t12⇒N=N02)

⇒150600=12t10
⇒ 14=12t10 ⇒122=12t10
2=t10⇒t=20 minutes

Topic: Physics, Nuclear Physics, NEET, NEET 2018
For a radioactive sample, the initial activity of the material was 8 counts and after 3 h it becomes 1 count,.The half-life of the sample is

Options:
A 2 h
B 1 h
C 3 h
D 4 h

The correct answer is:
1 h

Solution:
Here A0 = 8 counts, A=1 count and t=3 h.
AA0=12n
â 18=12n
or 123=12nân=3
So, T1/2 = tn = 33= 1 h

Topic: Physics, Nuclear Physics, JEE Main
For a radioactive sample, the initial activity of the material was 8 cps and after 3 h it becomes 1 cps. The half-life of the sample is

Options:
A 2 h
B 1 h
C 3 h
D 4 h

The correct answer is:
1 h

Solution:
Here A0 = 8 counts, A=1 count and t=3 h.
AA0=12n
â 18=12n
or 123=12nân=3
So, T1/2 = tn = 33= 1 h

Topic: Physics, Nuclear Physics, JEE Main
For a reaction $\frac{1}{2} A \rightarrow 2 B$, rate of disappearance of ' $A$ ' is related to the rate of appearance of 'B' by the expression

Options:
A
$-\frac{\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}=\frac{1}{2} \frac{\mathrm{d}[\mathrm{B}]}{\mathrm{dt}}$

B
$-\frac{\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}=\frac{1}{4} \frac{\mathrm{d}[\mathrm{B}]}{\mathrm{dt}}$

C
$-\frac{\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}=\frac{\mathrm{d}[\mathrm{B}]}{\mathrm{dt}}$

D
$-\frac{\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}=4 \frac{\mathrm{d}[\mathrm{B}]}{\mathrm{dt}}$

The correct answer is:
$-\frac{\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}=\frac{1}{4} \frac{\mathrm{d}[\mathrm{B}]}{\mathrm{dt}}$

Solution:
$$
\begin{aligned}
& \frac{1}{2} \mathrm{~A} \longrightarrow 2 \mathrm{~B} \\
& \frac{-2 \mathrm{~d}[\mathrm{~A}]}{\mathrm{dt}}=+\frac{\mathrm{d}[\mathrm{B}]}{2 \mathrm{dt}} \\
& \frac{-\mathrm{d}[\mathrm{A}]}{\mathrm{dt}}=\frac{1}{4} \frac{\mathrm{d}[\mathrm{B}]}{\mathrm{dt}}
\end{aligned}
$$

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2008
For a reaction, $2 A+B \longrightarrow$ products, If concentration of $B$ is kept constant and concentration of $A$ is doubled then rate of reaction is

Options:
A doubled
B quadrupled
C halfed
D remain same

The correct answer is:
quadrupled

Solution:
The rate of reaction for
$2 A+B \longrightarrow \text { products is rate }=k[A]^2[B]$
If concentration of $A$ is doubled,
$\begin{aligned}
& \text { Rate }^{\prime}=[2 A]^2[B]=4 \mathrm{k}[A]^2[B] \\
& \text { Rate }^{\prime}=4 \text { Rate. }
\end{aligned}$

Topic: Chemistry, Chemical Kinetics, COMEDK, COMEDK 2023
For a reaction 4 Ms+nO2g→2M2Ons

The free energy change is plotted as a function of temperature. The temperature below which the oxide is stable could be inferred from the plot as the point at which :

Options:
A the slope change from negative to positive
B the free energy change show a change from negative to positive value
C the slope changes from positive to negative
D the slope changes from positive to zero

The correct answer is:
the free energy change show a change from negative to positive value

Solution:
For oxide to be stable its ΔG of formation value should be negative.

Gibbs free energy, also known as the Gibbs function, Gibbs energy, or free enthalpy, is a quantity that is used to measure the maximum amount of work done in a thermodynamic system when the temperature and pressure are kept constant.

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2020 (06 Sep Shift 2)
For a reaction A→2B+C the half lives are 100 s and 50 s when the concentration of reactant A is 0.5 and 1.0 mol L= respectively. The order of the reaction is

Options:

The correct answer is:
2

Solution:
The relation between half-life and initial concentration of nth order reaction can be written as follows,

t12∝1A0n-1

100∝10.5n-1

50∝11n-1

21=10.5n-1

21=2n-1

n-1=1

n=2

order =2

For the second order reactions half-life is inversely proportional to initial concentration.

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2022 (26 Jul Shift 1)
For a reaction, $A+2 B \rightarrow$ Products, when concentration of $B$ alone is increased half-life remains the same. If concentration of $A$ alone is doubled, rate remains the same. The unit of rate constant for the reaction is

Options:
A $\mathrm{s}^{-1}$
B $\mathrm{L} \mathrm{mol}^{-1} \mathrm{~s}^{-1}$
C $\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1}$
D $\mathrm{atm}^{-1}$

The correct answer is:
$\mathrm{s}^{-1}$

Solution:
For the reaction,
$A+2 B \longrightarrow \text { Products }$
Half-life $\left(t_{1 / 2}\right)$ remains same as $[B]$ increases for lst order, $t_{1 / 2} \propto \frac{l}{[B]}$
$\therefore$ w.r.t. $B$, reaction is of first order.
Also, as $[A]$ is doubled, rate remains constant.
$\therefore$ It is zero order w.r.t. $A$.
$\therefore$ Rate law expression is $k[A]^{0}[B]^{1}$
$\therefore$ Overall order $=0+1=1$
$\therefore$ Unit of $k=\mathrm{s}^{-1}$

Topic: Chemistry, Chemical Kinetics, KCET, KCET 2021
For a reaction, A+B2+⟶B+A2+, at 250C E0=0.2955V. The value of Keq is

Options:
A 10
B 1010
C -10
D 10-10

The correct answer is:
1010

Solution:
Given that,

For a reaction,

A+B2+⟶B+A2+,E°=0.2955V

T=25°C, ⇒T=298 K n=2

Using Nernst equation,

E°=2.303RTnFlogkeq

E°=0.0591nlogkeqat 25oC

⇒ 0.2955=0.05912logkeq

keq=1010

Therefore, the value of Keq is 1010.

Topic: Chemistry, Electrochemistry, JEE Main
For a reaction AâB,â E a =10 kJ/mol,âÎH=5âkJ/mol. Thus potential energy profile for this reaction is

Options:
A
B
C
D

The correct answer is:

Solution:
ÎH=(Ea)Æâ(Ea)b
5=10â(Ea)b
(E a ) b =5 kJ/mol.
(Ea)Æ>(Ea)b hence option B is correct

Topic: Chemistry, Chemical Equilibrium, JEE Main
For a reaction $A+B \longrightarrow P$, the following data are provided
The rate constant for this reaction in standard unit is

Options:
A 5
B 1.2
C $2.4 \times 10^{-4}$
D 50

The correct answer is:
50

Solution:
(d) $A+B \longrightarrow P$
According to rate law expression
$$
\text { Rate }=k[A]^\alpha[B]^\beta(K \text { is constant }) \ldots (i)
$$
Taking ratio of (1) and (2) entry,
$$
\begin{aligned}
\frac{4 \times 10^{-2}}{2 \times 10^{-2}} & =\frac{[0.04]^\beta[0.02]^\alpha}{[0.02]^\beta[0.02]^\alpha} \\
2 & =[2]^\beta \\
{[2]^1 } & =[2]^\beta \\
\beta & =1
\end{aligned}
$$
Taking ratio of (2) and (3) entry,
$$
\begin{aligned}
\frac{8 \times 10^{-2}}{4 \times 10^{-2}} & =\frac{[0.04]^\alpha[0.04]^\beta}{[0.02]^\alpha[0.04]^\beta} \\
{[2] } & =[2]^\alpha \Rightarrow \alpha=1
\end{aligned}
$$
Put value of $\alpha$ and $\beta$ in (1) entry
$$
\begin{aligned}
2 \times 10^{-2} & =k[A]^1[B]^1 \\
k & =\frac{2 \times 10^{-2}}{0.02 \times 0.02} \\
& =\frac{2 \times 10^{-2} \times 10^4}{4}=\frac{100}{2}=50
\end{aligned}
$$

Topic: Chemistry, Chemical Kinetics, JEE Main
For a reaction, $\mathrm{A}+\mathrm{B} \rightarrow$ products, the rate of the reaction at various concentrations are given below
The rate law for the above reaction is

Options:
A $r=k[A][B]^{2}$
B $r=k[A]^{3}[B]$
C $\mathrm{r}=\mathrm{k}[\mathrm{A}]^{2}[\mathrm{~B}]^{2}$
D $r=k[A]^{2}[B]$

The correct answer is:
$r=k[A]^{2}[B]$

Solution:
Let order of reaction wrt A and $\mathrm{B}$ is $\mathrm{m}$ and $\mathrm{n}$ respectively. Then
$$
\begin{aligned}
\text { rate } &=\mathrm{k}[\mathrm{A}]^{\mathrm{m}}[\mathrm{B}]^{\mathrm{n}} \\
2 &=\mathrm{k}[0.2]^{\mathrm{m}}[0.2]^{\mathrm{n}}
\end{aligned}
$$
$$
4=k[0.2]^{m}[0.4]^{n}
$$
$$
36=\mathrm{k}[0.6]^{\mathrm{m}}[0.4]^{\mathrm{n}}
$$
On comparing Eqs. (i) and (ii), we get
$$
\begin{aligned}
& \frac{4}{2} &=\frac{[0.4]^{\mathrm{n}}}{[0.2]^{\mathrm{n}}} \\
\Rightarrow & 2 &=2^{\mathrm{n}} \\
\Rightarrow & \mathrm{n} &=1
\end{aligned}
$$
Again on comparing Eqs. (ii) and (iii), we get
$$
\begin{aligned}
& & \frac{36}{4} &=\frac{[0.6]^{\mathrm{m}}}{[0.2]^{\mathrm{m}}} \\
\Rightarrow & & 9=3^{2}=3^{\mathrm{m}} \\
\Rightarrow & & \mathrm{m}=2
\end{aligned}
$$
$\therefore$ Rate law for the given reaction will be
$$
\mathrm{r}=\mathrm{k}[\mathrm{A}]^{2}[\mathrm{~B}]
$$

Topic: Chemistry, Chemical Kinetics, JEE Main
For a reaction, Ag→Al; ΔH=-3RT. The correct statement for the reaction is

Options:
A ΔH=ΔU=O
B ΔH<ΔU
C ΔH>ΔU
D ΔH=ΔU≠O

The correct answer is:
ΔH>ΔU

Solution:
∆ H=∆ U+∆ngRT
∆ ng=np-nr=0-1=-1

∆ H=∆ U+-1RT∆ H=∆ U-RT

Given, ∆H=-3RT

-3RT= ∆ U+-1RT∆ U=-3RT+RT=-2RT
∆ H>∆ U

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2017 (08 Apr Online)
For a reaction $A \rightarrow$ Products, a plot of $\log t_{1 / 2}$ versus $\log a_0$ is shown in the figure. If the initial concentration of $A$ is represented by $a_0$, the order of the reaction is

Options:
A
one

B
zero

C
two

D
three

The correct answer is:
zero

Solution:
Plot given is for zero order reaction.

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2012 (19 May Online)
For a reaction, activation energy $\left(E_a\right)=0$ and rate constant $(K)=3.2 \times 10^6 \mathrm{~s}^{-1}$ at $300 \mathrm{~K}$. What is the value of the rate constant at $310 \mathrm{~K}$

Options:
A $3.2 \times 10^{-12} \mathrm{~s}^{-1}$
B $3.2 \times 10^6 \mathrm{~s}^{-1}$
C $6.4 \times 10^{12} \mathrm{~s}^{-1}$
D $6.4 \times 10^6 s^{-1}$

The correct answer is:
$3.2 \times 10^6 \mathrm{~s}^{-1}$

Solution:
When $E_a=0$ rate constant is independent of temperature.

Topic: Chemistry, Chemical Kinetics, JEE Main
For a reaction, consider the plot of ln k versus 1/T given in the figure. If the rate constant of this reaction at 400 K is 10-5s-1 , then the rate constant at 500 K is:

Options:
A 10-4s-1
B 10-6s-1
C 2x10-4s-1
D 4x10-4s-1

The correct answer is:
10-4s-1

Solution:
ln k=ln A-EaRT=ln A-46061T1-1T2∴EAR=4606
ln k10-5=4606×500-400500×400
ln k10-5=4606×12000=2.303
ln k10-5=ln 10
k=10-4s-1

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2019 (12 Jan Shift 2)
For a reaction if $K_p \gt K_c$, the forward reaction is favoured by

Options:
A Low pressure
B High pressure
C High temperature
D Low temperature

The correct answer is:
Low pressure

Solution:
For a reaction if $\mathrm{K}_{\mathrm{p}} \gt \mathrm{K}_{\mathrm{c}}$, the forward reaction is favoured by low pressure.
$\mathrm{K}_{\mathrm{p}}=\mathrm{K}_{\mathrm{c}}(\mathrm{RT})^{\Delta \mathrm{n}}$
Since $K_p \gt K_c$
$\therefore \Delta \mathrm{n} \gt \mathrm{o}$, i.e. product side contains more particles than the reactant side.

Topic: Chemistry, Chemical Equilibrium, JEE Main
For a reaction, if $K_{P}>K_{C}$, the forward reaction is favoured by

Options:
A low pressure
B high pressure
C high temperature
D low temperature

The correct answer is:
low pressure

Solution:
$K_{p}=K_{C}(R T)^{\Delta n}$
If $\Delta n$ is positive than $K_{p}>K_{C}$ also, gaseous moles of product $>$ moles of reactant.
$\therefore$ Forward reaction is favoured by low pressure, according to Le-Chatelier's principle.

Topic: Chemistry, Chemical Equilibrium, JEE Main
For a reaction in which all reactants and products are liquids, which one of the following equations is most applicable?

Options:
A ∆H<∆E
B ∆H=∆S
C ∆H≈∆E
D Total W=0

The correct answer is:
∆H≈∆E

Solution:
As all reactant and product are liquid
âng=0
ΔH=ΔE− Δn g RT
ΔH=ΔE ∵ Δn g =0

Topic: Chemistry, Thermodynamics (C), JEE Main
For a reaction \(\mathrm{A} \xrightarrow{\mathrm{K}_1} \mathrm{~B} \xrightarrow{\mathrm{K}_2} \mathrm{C}\)
If the rate of formation of \(B\) is set to be zero then the concentration of \(B\) is given by :

Options:
A \(\left(\mathrm{K}_1+\mathrm{K}_2\right)[\mathrm{A}]\)
B \(\left(\mathrm{K}_1 / \mathrm{K}_2\right)[\mathrm{A}]\)
C \(\left(\mathrm{K}_1-\mathrm{K}_2\right)[\mathrm{A}]\)
D \(\mathrm{K}_1 \mathrm{~K}_2[\mathrm{~A}]\)

The correct answer is:
\(\left(\mathrm{K}_1 / \mathrm{K}_2\right)[\mathrm{A}]\)

Solution:
Rate of formation of $B$ is
$\begin{aligned} & \frac{\mathrm{d}[\mathrm{B}]}{\mathrm{dt}}=\mathrm{k}_1[\mathrm{~A}]-\mathrm{k}_2[\mathrm{~B}] \\ & 0=\mathrm{k}_1[\mathrm{~A}]-\mathrm{k}_2[\mathrm{~B}] \\ & \left(\frac{\mathrm{k}_1}{\mathrm{k}_2}\right)[\mathrm{A}]=[\mathrm{B}]\end{aligned}$

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2024 (08 Apr Shift 2)