datavorous/entrance-exam-dataset huggingface dataset:
The rate constants $\mathrm{k}_1$ and $\mathrm{k}_2$ for two different reactions are $10^{16} \cdot \mathrm{e}^{-2000 / T}$ and $10^{15} \cdot \mathrm{e}^{-1000 / T}$ respectively. The temperature at which $\mathrm{k}_1=\mathrm{k}_2$ is

Options:
A $\frac{1000}{2.303} \mathrm{~K}$
B $1000 \mathrm{~K}$
C $\frac{2000}{2.303} \mathrm{~K}$
D $2000 \mathrm{~K}$

The correct answer is:
$\frac{1000}{2.303} \mathrm{~K}$

Solution:
$$
K_1=10^{16} \mathrm{e}^{-\frac{2000}{T}}
$$
or $\log \mathrm{K}_1=16-\frac{2000}{2.303 \mathrm{~T}}$
$$
K_2=10^{15} e^{\frac{1000}{T}}
$$
or $\log K_2=15-\frac{1000}{2.303 \mathrm{~T}}$
If $\mathrm{T}=\frac{1000}{2.303} \mathrm{~K}$, then $\mathrm{K}_1=\mathrm{K}_2$

Topic: Chemistry, Chemical Kinetics, NEET, NEET 2008 (Mains)
The rate equation for the reaction $2 \mathrm{~A}+\mathrm{B} \longrightarrow \mathrm{C}$ is found to be: rate $\mathrm{k}[\mathrm{A}][\mathrm{B}]$. The correct statement in relation to this reaction is that the

Options:
A
unit of $\mathrm{K}$ must $\mathrm{be} \mathrm{s}^{-1}$

B
values of $\mathrm{k}$ is independent of the initial concentration of $A$ and $B$

C
rate of formation of $C$ is twice the rate of disappearance of $A$

D
$t_{1 / 2}$ is a constant

The correct answer is:
values of $\mathrm{k}$ is independent of the initial concentration of $A$ and $B$

Solution:
No Solution Available

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2004
The rate equation for the reaction $2 A+B \rightarrow C$ is found to be: rate $=k[A][B]$. The correct statement in relation to this reaction is that the

Options:
A Rate of formation of $C$ is twice the rate of disappearance of $A$
B $t_{1 / 2}$ is a constant
C Unit of $k$ must be $s^{-1}$
D Value of $k$ is independent of the initial concentrations of $A$ and $B$

The correct answer is:
Value of $k$ is independent of the initial concentrations of $A$ and $B$

Solution:
For $2 A+B \rightarrow C$
Rate $=K[A][B]$
Value of rate constant $K=A e^{-E_a / R T}$ here $K$ is independent of the initial concentration of $A$ and $B$.

Topic: Chemistry, Chemical Kinetics, JEE Main
The rate equation of a gaseous reaction is given by $r=k[A][B]$. If the volume of the reaction vessel is suddenly reduced to $1 / 2$ of the initial volume, the reaction rate relating to the original rate will be

Options:
A $\frac{1}{4}$
B 4
C $\frac{1}{2}$
D 2

The correct answer is:
4

Solution:
The concentration of gaseous $\propto \frac{1}{\text { volume }}$
$\therefore$ If volume of vessel is reduced of initial volume to half then concentration increases to two-times.
$$
\begin{array}{llrl}
\therefore & & {\left[A^{\prime}\right]=2[A]} \\
& & {\left[B^{\prime}\right]=2[B]} \\
\text { Initially, } & r=k[A][B]
\end{array}
$$
After reduction of volume, $r_{1}=k_{2}\left[A^{\prime}\right] 2\left[B^{\prime}\right]$
$$
\begin{aligned}
&r_{1}=4 k\left[A^{\prime}\right]\left[B^{\prime}\right] \quad\left(k_{2}=k\right) \\
&r_{1}=4 r
\end{aligned}
$$
$\therefore$ Rate increases to 4 times of original rate.

Topic: Chemistry, Chemical Kinetics, JEE Main
The rate law for a reaction between reactants $A, B$ and $C$ is $r=K[A][B][C]^2$, if concentration of $A$ is halved, then rate of reaction

Options:
A increase $\frac{1}{2}$ times
B remains the same
C decreases $\frac{1}{2}$ times
D increase 2 times

The correct answer is:
decreases $\frac{1}{2}$ times

Solution:
$\mathrm{r}=\mathrm{K}[\mathrm{A}][\mathrm{B}][\mathrm{C}]^2$ of canc. of $\mathrm{A}$ is halved
$\mathrm{r}=\mathrm{K} \frac{[\mathrm{A}]}{2}[\mathrm{~B}][\mathrm{C}]^2$

Topic: Chemistry, Chemical Kinetics, MHT CET, MHT CET 2022 (08 Aug Shift 1)
The rate law for a reaction between the substances $\mathrm{A}$ and $\mathrm{B}$ is given by Rate $=\mathrm{k}[\mathrm{A}]^{\mathrm{n}}[\mathrm{B}]^{\mathrm{m}}$ On doubling the concentration of A and halving the concentration of B, the ratio of the new rate to the earlier rate of the reaction will be as

Options:
A
$(m+n)$

B
$(n-m)$

C
$2^{(\mathrm{n}-\mathrm{m})}$

D
$\frac{1}{2^{(\mathrm{m}+\mathrm{n})}}$

The correct answer is:
$2^{(\mathrm{n}-\mathrm{m})}$

Solution:
Rate $_1=\mathrm{k}[\mathrm{A}]^{\mathrm{n}}[\mathrm{B}]^{\mathrm{m}} ; \quad$ Rate $_2=\mathrm{k}[2 \mathrm{~A}]^{\mathrm{n}}[1 / 2 \mathrm{~B}]^{\mathrm{m}}$
$\therefore \frac{\text { Rate }_2}{\text { Rate }_1}=\frac{\mathrm{k}[2 \mathrm{~A}]^n[1 / 2 \mathrm{~B}]^m}{\mathrm{k}[\mathrm{A}]^n[\mathrm{~B}]^m}=[2]^n[1 / 2]^m=2^{\mathrm{n}} \cdot 2^{-m}=2^{\mathrm{n}-\mathrm{m}}$

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2003
The rate law for reaction $A+2 B=C+2 D$ will be

Options:
A Rate $=K[A][B]$
B Rate $=K[A][2 B]$
C Rate $=K[A][B]^2$
D Rate $=K \frac{[C][D]^2}{[A][B]^2}$

The correct answer is:
Rate $=K[A][B]^2$

Solution:
The rate law for an reaction can be given by; rate $=K(A)(B)^2$, i.e. the powers are raised which are given as coefficient of reactant.

Topic: Chemistry, Chemical Kinetics, JEE Main
The rate law for the reaction below is given by the expression kA[B]
A+B→Product
If the concentration of B is increased from 0.1 to 0.3 mol, keeping the value of A at 0.1 mol, the rate constant will be:

Options:
A 3k
B 9k
C k/3
D k

The correct answer is:
k

Solution:
Rate constant is independent of concentration. It depends only on temperature. But the rate of the reaction increases by three times as the order of the reaction is one.

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2016 (10 Apr Online)
The rate law for the reaction
Sucrose + Water $\xrightarrow{\left[\mathrm{H}^{+}\right]}$Glucose + Fructose is given by

Options:
A Rate $=K$ [sucrose] [water]
B Rate $=K$ [sucrose] [water $^0]$
C Rate $=K$ [sucrose $]^0[$ water $]$
D Rate $=K[\text { sucrose }]^{1 / 2}[\text { water }]^{1 / 2}$

The correct answer is:
Rate $=K$ [sucrose] [water $^0]$

Solution:
It is a pseudo-unimolecular reaction.

Topic: Chemistry, Chemical Kinetics, JEE Main
The rate of a certain biochemical reaction at physiological temperature T occurs 106 times faster with enzyme than without. The change in the activation energy upon adding enzyme is:

Options:
A -62.303RT
B -6RT
C +62.303RT
D +6RT

The correct answer is:
-62.303RT

Solution:
K=Ae-E/RT......(1)
106 K=Ae−EcRT…….(2)

On dividing eq, 2 by eq, 1
106=eE−Ec/RT

on taking logarithm both sides
6ln10=E−EcRT

2.303×6=E−EcRT
E−Ec=2.303×(6RT)
ΔEa=Ec−E=−2.303×6RT

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2020 (08 Jan Shift 1)
The rate of a chemical reaction doubles for every $10^{\circ} \mathrm{C}$ rise of temperature. If the temperature is raised by $50^{\circ} \mathrm{C}$, the rate of the reaction increases by about :

Options:
A
24 times

B
32 times

C
64 times

D
10 times

The correct answer is:
32 times

Solution:
Temperature coefficient $\mu=2$;
$$
\begin{aligned}
& \mu^{\frac{\Delta \mathrm{T}}{10}}=\frac{\mathrm{k}_2}{\mathrm{k}_1} \\
& 2^{\frac{50}{10}}=2^5=32=\frac{\mathrm{k}_2}{\mathrm{k}_1}
\end{aligned}
$$
Therefore $32 \mathrm{k}_1=\mathrm{k}_2$

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2011
The rate of a first order reaction is $1.5 \times$ $10^{-2} \mathrm{~mol} \mathrm{~L}^{-1} \mathrm{~min}^{-1}$ at $0.5 \mathrm{M}$ concentration of the reactant. The half life of the reaction is:

Options:
A $0.383 \mathrm{~min}$
B $23.1 \mathrm{~min}$
C $8.73 \mathrm{~min}$
D $7.53 \mathrm{~min}$

The correct answer is:
$23.1 \mathrm{~min}$

Solution:
Let the reaction be
$$
\begin{aligned}
& \mathrm{A} \stackrel{k}{\longrightarrow} \text { Product } \\
& \text { Rate }=k[\mathrm{~A}] \\
& 1.5 \times 10^{-2} \mathrm{~mol} \mathrm{\textrm {L } ^ { - 1 } \mathrm { min } ^ { - 1 } = k [ 0 . 5 ]} \\
& k=\frac{1.5 \times 10^{-2}}{0.5} \\
&=3 \times 10^{-2} \mathrm{~min}^{-1} \\
& t_{1 / 2}=\frac{0.693}{k} \\
& \Rightarrow \quad t_{1 / 2}=\frac{0.693}{3 \times 10^{-2}} \\
&=0.231 \times 10^2 \mathrm{~min} \\
& \Rightarrow \quad t_{1 / 2}=23.1 \mathrm{~min}
\end{aligned}
$$
Related Theory
Decay profiles for first-order reactions with large and small rate constants

Topic: Chemistry, Chemical Kinetics, JEE Main
The rate of a first order reaction is $1.5 \times 10^{-2} \mathrm{molL}^{-1} \mathrm{~min}^{-1}$ at $0.5 \mathrm{M}$ concentration of the reactant. The half life of the reaction is

Options:
A 8.73 min
B 7.53 min
C 0.383 min
D 23.1 min

The correct answer is:
23.1 min

Solution:
Rate $\left(\frac{d x}{d t}\right)=K \cdot c \quad ; 1.5 \times 10^{-2}=K \times 0.5$
For first order $K=\frac{1.5 \times 10^{-2}}{0.5}=3 \times 10^{-2} \min^{-1}$
$t_{1 / 2}=\frac{.693}{K}=\frac{.693}{3 \times 10^{-2}}=23.1 \mathrm{~min}$

Topic: Chemistry, Chemical Kinetics, JEE Main
The rate of a gaseous reaction is given by K[ A ][ B ]. If the volume of the reaction vessel is reduced to 1 4 of the initial volume, the reaction rate relative to the original rate is -

Options:
A 1 16 times
B 1 8 times
C 8times
D 16 times

The correct answer is:
16 times

Solution:
Rate, r=K[ A ][ B ]
According to the given data, the volume of the reaction vessel is reduced to 1 4 th. Hence, it can be written as
r 1 =K[4A][4B]
=16K[ A ][ B ]
=16r
â´ r 1 =16r

Topic: Chemistry, Thermodynamics (C), JEE Main
The rate of a reaction A doubles on increasing the temperature from 300 to 310 K. By how much, the temperature of reaction B should be increased from 300 K so that rate doubles if activation energy of the reaction B is twice to that of reaction A.

Options:
A 4.92 K
B 9.84 K
C 19.67 K
D 2.45 K

The correct answer is:
4.92 K

Solution:
log 2=EaR1300-1310 .......(i)
log 2=2EaR 1300-1T ......(ii)
2EaR 1300-1T=EaR 1300-1310
1300+1310=2T
⇒ T=300×310610×2
=304.92

Hence, the temperature of reaction B should be increased from 300 K by 304.92−300=4.92 K.

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2017 (08 Apr Online)
The rate of a reaction doubles when its temperature changes from 300K to 310K. Activation energy of such a reaction will be:
â
R=8.314 JK-1mol-1 and log 2=0.301

Options:
A 58.5 kJ mol -1
B 60.5 kJ mol -1
C 53.6 kJ mol -1
D 48.6 kJ mol -1

The correct answer is:
53.6 kJ mol -1

Solution:
logk2k1=−Ea2.303R(1T2−1T1)
logk2k1=Ea2.303RT2-T1T1T2
log 2=Ea2.303×8.314×10300×310
Ea=0.301×2.303×8.314×300×31010=53603.9 J/mol
Ea=53.6 kJ mol-1.

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2013 (07 Apr)
The rate of a reaction is doubled for every $10^{\circ}$ rise in temperature. The increase in reaction rate as a result of temperature rise from $10^{\circ}$ to $100^{\circ}$ is

Options:
A 112
B 512
C 400
D 614

The correct answer is:
512

Solution:
$\frac{k_t+10}{K_t}=\frac{r_t+10}{r_t}=2$
For an increase of temperature by $90^{\circ} \mathrm{C}$ i.e. 9 times, the rate increases by $2^9$ times i.e. 512.

Topic: Chemistry, Chemical Kinetics, JEE Main
The rate of a reaction quadruples when the temperature changes from 300  to 310 K . The activation energy of this reaction is:
(Assume Activation energy and pre-exponential factor are independent of temperature; ln(2)=0.693; R=8.314 J mol-1K-1)

Options:
A 53.6 kJmol-1
B 214.4 kJmol-1
C 107.2 kJmol-1
D 53.7 kJmol-1

The correct answer is:
107.2 kJmol-1

Solution:
The activation energy, rate constant and temperature can be related using Arrhenius equation.

k=Ae-EaRT

lnk2k1=EaR 1T1-1T2
ln4=EaR 1300-1310
2×0.693=EaR 10300×310
E a = 2×0.693×300×310×8.314 10 = 107.2 kJ/mol

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2017 (09 Apr Online)
The rate of a reaction

Options:
A Increases with increase in temperature
B Decreases with increase in temperature
C Does not depend on temperature
D Does not depend on concentration

The correct answer is:
Increases with increase in temperature

Solution:
Thus both rate and rate constant $K$ increase with temperature, $r=k$ (reactant) ${ }^n$ and $k=A e^{-E_\alpha / k T}$

Topic: Chemistry, Chemical Kinetics, JEE Main
The rate of change of (x2+16) with respect to xx-1 at x=3 is

Options:
A 2
B 115
C -125
D -3

The correct answer is:
-125

Solution:
Let y=x2+16 and z=xx-1
On differentiating w.r.t., x, we get,
dzdx=x-1-xx-12=-1x-12 and
dydx=1×2x2x2+16
⇒dydz=xx2+16 -1x-12
At x=3
dydz=35 (4)(-1)=-125

Topic: Mathematics, Differentiation, JEE Main
The rate of decomposition for methyl nitrite and ethyl nitrite can be given in terms of rate constant k1 and k2 respectively. The energy of activation for the two reactions are 152.30 kJ mol-1 and 157.7 kJ mol-1 as well as frequency factors are 1013 and 1014 respectively for the decomposition of methyl and ethyl nitrite. Calculate the temperature at which rate constant will be same for the two reactions.

Options:
A 256âK
B 354 K
C 282âK
D 674âK

The correct answer is:
282 K

Solution:
k=Ae-EaRT
For methyl nitrite k1=1013 e-152300/8.314ÃT
For ethyl nitrite k2=1014 e-157700/8.314ÃT
If k1=k2 then
1013 e-152300/8.314ÃT=1014 e-157700/8.314ÃT
Or 2.303logâ¡10=157700 - 1523008.314 Ã T
T=282 K

Topic: Chemistry, Chemical Kinetics, JEE Main
The rate of decomposition of detritus is accelerated

Options:
A If the detritus contains lignin chitin, tannins and cellulose.
B If the detritus is present in acidic soil
C If the moisture and aeration are at optimum
D If the temperature is below 10°C

The correct answer is:
If the moisture and aeration are at optimum

Solution:
Detritus is the dead organic material which includes the parts or bodies of dead organisms and their faecal matter. These organic content are broken down to inorganic substances by the process called decomposition. It is brought about by decomposers. Decomposition is a mainly an oxygen dependent process and its rate is dependent on chemical composition and climatic factors. Temperature and moisture play an important role in rate of decomposition. Enough supply of oxygen and proper moisture accelerates the decomposition process.

Topic: Biology, Ecosystem, NEET
The rate of decomposition of detritus is affected by the abiotic factors like availability of oxygen, $\mathrm{pH}$ of the soil substratum, temperature etc. Discuss.

Options:

None

Solution:
Decomposition of dead remains or matter of animal and plant is exceuted by microorganisms like bacteria, fungi etc. Abiotic factors like termperature, humidity, $\mathrm{pH}$ of soil and light etc. affect the growth rate of these decomposers. The soil $\mathrm{pH}$ affects the composition of acidophilic and basophilic microorganisms. Aerobic and anaerobic processes occur in absence or presence of oxygen.
Complete degradation of substances occur in presence of oxygen, while there will be an incomplete degradation in absence of oxygen. Similarly microbes can not grow optimally at high temperature but the growth of stress tolerant microorganisms is favoured by high temperature.

Topic: Biology, Ecosystem
The rate of diffusion of two gases A and B are in the ratio 1:4 assume;PA=PB. In other experiment a mixture contains these gases A and B in the mole ratio 2:3. The ratio of mole fraction of the gases A and B in the diffused mixture is –

Options:
A 1 : 6
B 1 : 12
C 1 : 18
D 1 : 24

The correct answer is:
1 : 6

Solution:
1st experiment â r A r B = M B M A = 1 4
2nd experiment â r A r B = 2 3 M B M A âµâ r A r B = n a n B
=23Ã14=16

Topic: Chemistry, States of Matter, JEE Main
The rate of disintegration of a radioactive substance falls from 403 dps to 53 dps in 6 hours. The half-life of the radioactive substance is

Options:
A 67 hours
B 2 hours
C 3 hours
D 1 hour

The correct answer is:
2 hours

Solution:
A=A0e-λt

⇒100=800e-λ(6×60)

⇒e-350x=18

⇒-360λ=ln18=-ln8

⇒λ=ln23360=ln2120

⇒T1/2=ln2λ=ln2(ln2/120)=120 minutes

⇒T1/2=2 hours

Topic: Physics, Nuclear Physics, JEE Main
The rate of formation of new organic matter by rabbit in a grassland, is called

Options:
A net productivity
B secondary productivity
C net primary productivity
D gross primary productivity

The correct answer is:
secondary productivity

Solution:
The amount of energy accumulated in green plants through the process of photosynthesis is known as primary productivity. It may gross PP (i.e., the total organic matter synthesized) or Net PP (i.e., the weight of organic matter stored). The rate of resynthesis of organic matter by the consumers is known as secondary productivity.

Topic: Biology, Ecosystem, NEET, NEET 2012 (Mains)
The rate of gas phase chemical reactions generally increases rapidly with rise in temperature. This is mainly because

Options:
A the collision frequency increases with temperature
B the fraction of molecules having energy in excess of the activation energy increases with temperature
C the activation energy decreases with temperature
D the average kinetic energy of molecules increases with temperature

The correct answer is:
the fraction of molecules having energy in excess of the activation energy increases with temperature

Solution:
No Solution

Topic: Chemistry, Chemical Kinetics, KVPY, KVPY 2011 (SB/SX)
The rate of growth of bacteria in a culture is proportional to the number of bacteria present and the bacteria count is 1000 at initial time t=0. The number of bacteria is increased by 20% in 2 hours. If the population of bacteria is 2000 after kloge65 hours, then kloge22 is equal to:

Options:
A 8
B 4
C 16
D 2

The correct answer is:
4

Solution:
Given dBdt∝B⇒dBdt=λB, λ is proportionility constant

⇒dBB=λdt

Integrate both sides with proper limit we get,

∫B1B2dBB=∫t1t2λdt

⇒lnB2B1=λt2-t1 ...i

Given that, at t=0, B=1000 & t=20% increase=1200

∴ ln12001000=2λ⇒λ=12ln65

Now an equation i becomes,

lnB2B1=12ln65t2-t1 ...ii

Given that, at t2=kln2, B2=2000 & t=0, B1=1000

From ii, ln20001000=12ln65kln65-0=12k

⇒k=2ln2

∴ kln22=2ln2ln22=4

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2021 (26 Feb Shift 1)
The rate of increase of bacteria in a certain culture is proportional to the number present. If it double in 5 hours then in 25 hours, its number would be

Options:
A 8 times the original
B 16 times the original
C 32 times the original
D 64 times the original

The correct answer is:
32 times the original

Solution:
Let $P_0$ be the initial population and let the population after $t$ years be $P$. Then $\frac{d P}{d t}=k P \Rightarrow \frac{d P}{P}=k d t$
On integrating, we have $\log P=k t+c$
At $t=0, \quad P=P_0$
$\therefore \log P_0=0+c_{;} \therefore \log P=k t+\log P_0 \Rightarrow \log \frac{P}{P_0}=k t$
When $t=5$ hrs, $P=2 P_0$
$\therefore \log \frac{2 P_0}{P_0}=5 k \Rightarrow K=\frac{\log 2}{5} ; \therefore \log \frac{P}{P_0}-\frac{\log 2}{5} t$
When $t=25$ hours, we have
$\log \frac{P}{P_0}=\frac{\log 2}{5} \times 25-5 \log 2-\log 32 \ldots \quad \therefore=32 P_0$.

Topic: Mathematics, Differential Equations, JEE Main
The rate of reaction $2 \mathrm{~N}_2 \mathrm{O}_5 \longrightarrow 4 \mathrm{NO}_2+\mathrm{O}_2$ can be written in three ways
$\begin{aligned}
\frac{-d\left[\mathrm{~N}_2 \mathrm{O}_5\right]}{d t} & =k\left[\mathrm{~N}_2 \mathrm{O}_5\right] \\
\frac{d\left[\mathrm{NO}_2\right]}{d t} & =k^{\prime}\left[\mathrm{N}_2 \mathrm{O}_5\right] \\
\frac{d\left[\mathrm{O}_2\right]}{d t} & =k^{\prime \prime}\left[\mathrm{N}_2 \mathrm{O}_5\right]
\end{aligned}$
The relationship between $k$ and $k^{\prime}$ and between $k$ and $k^{\prime \prime}$ are

Options:
A $k^{\prime}=2 k ; k^{\prime}=k$
B $k^{\prime}=2 k ; k^{\prime \prime}=k / 2$
C $k^{\prime}=2 k ; k^{\prime \prime}=2 k$
D $k^{\prime}=k ; k^{\prime \prime}=k$

The correct answer is:
$k^{\prime}=2 k ; k^{\prime \prime}=k / 2$

Solution:
$$
\begin{aligned}
& \text { Rate }=-\frac{1}{2} \frac{d\left[\mathrm{~N}_2 \mathrm{O}_5\right]}{d t}=\frac{1}{4} \frac{d\left[\mathrm{NO}_2\right]}{d t}=\frac{d\left[\mathrm{O}_2\right]}{d t} \\
& \Rightarrow \frac{1}{2} k\left[\mathrm{~N}_2 \mathrm{O}_5\right]=\frac{1}{4} k^{\prime}\left[\mathrm{N}_2 \mathrm{O}_5\right]=k^{\prime \prime}\left[\mathrm{N}_2 \mathrm{O}_5\right]
\end{aligned}
$$
$$
\begin{array}{llrl}
\Rightarrow & \frac{k}{2} & =\frac{k^{\prime}}{4}=k^{\prime \prime} \\
& \therefore & k^{\prime} & =2 k ; k^{\prime \prime}=\frac{k}{2}
\end{array}
$$

Topic: Chemistry, Chemical Kinetics, NEET, NEET 2011 (Mains)
The rate of reaction increases with rise in temperature because of

Options:
A increase in number of activated molecules
B increase in energy of activation
C decrease in energy of activation
D increase in the number of effective collisions

The correct answer is:
increase in the number of effective collisions

Solution:
For a collision to be effective, the colloiding molecules must have energy more than a particular value known as threshold energy. At room temperature most of the reactant molecules have energy less than the threshold value. When temperature is increased, the reactant molecules absorb this energy and their energy becomes equal to or greater than threshold value. This extra amount of energy absorbed by the reacting molecules is known as activation energy.
Activation energy $=$ Threshold energy $-$ Average kinetic energy of the reactants.
Evidently, the rate of reaction increases on rise in temperature, because average kinetic energy of the reactants increases and activation energy decreases. Also, on rise in temperature, the number of molecules having kinetic energy greater than threshold value increases. Thus, the increase in the rate of reaction with increase in temperature in mainly due to increase in number of effective collisions.

Topic: Chemistry, Chemical Kinetics, KCET, KCET 2013
The rate of reaction is doubled for every 10oC rise in temperature. The increase in rate as result of an increase in temperature from 10oC to 100oC is how many times of the original rate?

Options:
A 112
B 512
C 400
D 256

The correct answer is:
512

Solution:
kt+10kt=2
The increase in rate as result of increase in temperature from 10oC to 100oC is
rn=29 r1

Topic: Chemistry, Chemical Kinetics, JEE Main
The rate of the reaction, $2 \mathrm{NO}+\mathrm{Cl}_2 \longrightarrow 2 \mathrm{NOCl}$ is given by the rate equation, rate $=\mathrm{k}[\mathrm{NO}]^2\left[\mathrm{Cl}_2\right]$
The value of the rate constant can be increased by

Options:
A increasing the temperature
B increasing the concentration of NO
C increasing the concentration of the $\mathrm{Cl}_2$
D doing all of the above

The correct answer is:
increasing the temperature

Solution:
Rate constant is only affected by temperature.

Topic: Chemistry, Chemical Kinetics, NEET, NEET 2010 (Mains)
The rate of water gushing out of a pipe of radius 5 cm is 100 L min-1. The Reynolds number for the flow is of the order of [density of water = 1000 kg m-3, coefficient of viscosity of water = 1 mPa s]

Options:
A

106

B 104
C 103
D 102

The correct answer is:
104

Solution:
Rate of volume flow =Ïr2V=100 â/min

Ïr2v=1600m3/s

Reynolds number =Re=Ïâ¨DÎ·=1000.1Ã1600Ïr2Ã2r

=3.5Ã104

Topic: Physics, Mechanical Properties of Fluids, JEE Main
The rates of a certain reaction $(d c / d t)$ at different times are as follows
\(\begin{array}{c|l}
\text {Time } & \text{Rate (mole litre }{ }^{-1} \mathrm{sec}^{-1})\\
\hline 0 & 2.8 \times 10^{-2} \\
10 & 2.78 \times 10^{-2} \\
20 & 2.81 \times 10^{-2} \\
30 & 2.79 \times 10^{-2}
\end{array}\)
The reaction is

Options:
A Zero order
B First order
C Second order
D Third order

The correct answer is:
Zero order

Solution:
The concentration of reactant does not change with time for zero order reaction (unit of K suggests zero order) since reactant is in excess.

Topic: Chemistry, Chemical Kinetics, JEE Main
The rates of certain reactions \( (\mathrm{dc} / \mathrm{dt}) \) at different times are as follows:-
\(\begin{array}{|l|l|}\hline \text{Time } (\mathrm{s}) & \text{Rate }\left(\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1}\right) \\\hline 0 & 2.8 \times 10^{-2} \\\hline 10 & 2.78 \times 10^{-2} \\\hline 20 & 2.84 \times 10^{-2} \\\hline 30 & 2.79 \times 10^{-2} \\\hline\end{array}\)
The reaction is of

Options:
A zero order
B second order
C first order
D third order

The correct answer is:
zero order

Solution:
As we can see that rate of reaction is not changing much with time it is almost constant. So, it must be a zero-order reaction. For, zero order reaction rate of reaction is independent of the concentration of reactant.

-dCdt = k

Topic: Chemistry, Chemical Kinetics, JEE Main
The rates of diffusion of gases A and B of molecular weight 36 and 64 are in the ratio

Options:
A 9 : 16
B 4 : 3
C 3 : 4
D 16 : 9

The correct answer is:
4 : 3

Solution:
r A r B = M B M A = 6 4 3 6 = 4 3
r A : r B = 4 : 3 .

Topic: Chemistry, States of Matter, JEE Main