datavorous/entrance-exam-dataset huggingface dataset:
The radius vector and linear momentum are respectively given by vector 2i^+2j^+k^ and 2i^-2j^+k^. Their angular momentum is

Options:
A 2i^-4j^
B 4i^-8k^
C 2i^-4j^+2k^
D 4i^-8j^

The correct answer is:
4i^-8k^

Solution:
Angular momentum L → = r → × p →
=2i^+2j^+k^×2i^-2j^+k^=4i^-8k^

Topic: Physics, Mathematics in Physics, NEET
The radula is found in ___________ and helps in _____________ .

Options:
A Mollusca, feeding.
B Mollusca, respiration
C Echinodermata, excretion
D Echinodermata, circulation

The correct answer is:
Mollusca, feeding.

Solution:
The radula is a file-like rasping organ in the mouth that helps in feeding. It is found in all molluscs except bivalves.

Topic: Biology, Animal Kingdom, NEET
The random error in the arithmetic mean of 100 observations is x ; then the random error in the arithmetic mean of 400 observations would be

Options:
A 4x
B x4
C 2x
D x2

The correct answer is:
x4

Solution:
Random error is inversely proportional to the number of observations
∴ Error⁡∝1observation
E1E2=O2O1
E2=x400100
E2=x4

Topic: Physics, Units and Dimensions, NEET
The range of a projectile when launched at an angle 15o with the horizontal is 1.5 km. What will be the range of that projectile when launched with the same velocity at an angle 45o to the horizontal?

Options:
A 0.75 km
B 1.5 km
C 3.0 km
D 6.0 km

The correct answer is:
3.0 km

Solution:
At Î¸=15o
R=u2sinâ¡30og=1.5Â km
âu2g=3Â km
At Î¸=45o
R′=u2sinâ¡90og=u2g=3Â km

Topic: Physics, Motion In Two Dimensions, NEET
The range of a projectile when launched at an angle 15o with the horizontal is 1.5 km . What will be the range of that projectile when launched with the same speed at an angle 45o to the horizontal ?

Options:
A 0.75 km
B 1.5 km
C 3.0 km
D 6.0 km

The correct answer is:
3.0 km

Solution:
At θ=15o ,
R=u2sin2×15°g=1.5 km
⇒ u2g=3 km
At θ=45° ,
R′=u2sin2×45°g=u2g=3 km

Topic: Physics, Motion In Two Dimensions, NEET
The rapid change of $\mathrm{pH}$ near the stoichiometric point of an acid base titration is the basis of indicator detection, $\mathrm{pH}$ of the solution is related to ratio of the concentration of the conjugate acid (Hln) and base $\left(\mathrm{In}^{-}\right)$forms of the indicator by the expression:

Options:
A $\log \frac{\left[\mathrm{In}^{-}\right]}{[\mathrm{HIn}]}=\mathrm{pK}_{\ln }-\mathrm{pH}$
B $\log \frac{[\mathrm{HIn}]}{\left[\mathrm{In}^{-}\right]}=\mathrm{pK}_{\mathrm{ln}}-\mathrm{pH}$
C $\log \frac{[\mathrm{HIn}]}{\left[\mathrm{In}^{-}\right]}=\mathrm{pH}-\mathrm{pK}_{\mathrm{In}}$
D $\log \frac{\left[\mathrm{In}^{-}\right]}{[\mathrm{HIn}]}=\mathrm{pH}-\mathrm{pK}_{\mathrm{In}}$

The correct answer is:
$\log \frac{\left[\mathrm{In}^{-}\right]}{[\mathrm{HIn}]}=\mathrm{pH}-\mathrm{pK}_{\mathrm{In}}$

Solution:
Let us consider the formation of a salt of a weak acid and a strong base.
$$
\begin{aligned}
\mathrm{In}^{-}+\mathrm{H}_2 \mathrm{O} & \rightleftharpoons \mathrm{HIn}+\mathrm{OH}^{-} \\
\mathrm{K}_h & =\frac{[\mathrm{HIn}]\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{In}^{-}\right]}
\end{aligned}
$$
$$
\begin{aligned}
& \mathrm{HIn} \rightleftharpoons \mathrm{H}^{+}+\mathrm{In}^{-} \\
& \mathrm{H}_2 \mathrm{O} \rightleftharpoons \mathrm{H}^{+}+\mathrm{OH}^{-} \\
& \mathrm{K}_{\mathrm{ln}}=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{In}^{-}\right]}{[\mathrm{HIn}]} \\
& \mathrm{K}_{\mathrm{w}}=\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right]
\end{aligned}
$$
From (ii) and (iii),
$$
\begin{aligned}
& \frac{\mathrm{K}_{\mathrm{w}}}{\mathrm{K}_{\mathrm{In}}}=\frac{[\mathrm{HIn}]\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{In}^{-}\right]} \\
& {\left[\mathrm{OH}^{-}\right] }=\frac{\mathrm{K}_{\mathrm{w}}}{\mathrm{K}_{\mathrm{In}}} \frac{\left[\mathrm{In}^{-}\right]}{[\mathrm{HIn}]} \\
& \log \left[\mathrm{OH}^{-}\right]=\operatorname{logK}_{\mathrm{w}}-\log _{\mathrm{ln}}+\log \\
& \frac{\left[\mathrm{In}^{-}\right]}{[\mathrm{HIn}]}
\end{aligned}
$$
$$
\begin{aligned}
& -\mathrm{pOH}=-\mathrm{pK}_w+\mathrm{pK}_{\mathrm{ln}}+\log \\
& \frac{\left[\mathrm{In}^{-}\right]}{[\mathrm{HIn}]} \\
& \mathrm{pK}_w-\mathrm{pOH}=\mathrm{pK}_{\ln }+\log \frac{\left[\mathrm{In}^{-}\right]}{[\mathrm{HIn}]} \\
& \text { or, } \quad \mathrm{pH}=\mathrm{pK}_{\mathrm{ln}}+\log \frac{\left[\mathrm{In}^{-}\right]}{[\mathrm{HIn}]} \\
& \log \frac{\left[\mathrm{In}^{-}\right]}{[\mathrm{HIn}]}=\mathrm{pH}-\mathrm{pK}_{\ln } \\
&
\end{aligned}
$$

Topic: Chemistry, Ionic Equilibrium, NEET, NEET 2004
The rapid change of pH near the stoichiometric point of an acid-base titration is the basis of indicator detection. pH of the solution is related to ratio of the concentration of the conjugate acid (HIn) and base (In-) forms of the indicator by the expression

Options:
A log[In-][HIn]=pKln-pH
B log[HIn][In-]=pKln-pH
C log[HIn][In-]=pH-pKln
D log[In-][HIn]=pH-pKln

The correct answer is:
log[In-][HIn]=pH-pKln

Solution:
Let us consider the formation of a salt of a weak acid and a strong base.
In-+H2O⇌HIn+OH-
Kh[HIn][OH-][In-] …(i)
Other equations present in the solution are
HIn⇌H++In-
H2O⇌H++OH-
KIn=[H+][In-][HIn] …(ii)
Kw=[H+][OH-] …(iii)
From (ii) and (iii),
KwKIn[HIn][OH-][In-]=Kh
[OH-]=KwKIn[In-][HIn]
log[OH-]=logKw-logKIn+log[In-][HIn]
-pOH=-pKw+pKIn+log[In-][HIn]
pKw-pOH=pKIn+log[In-][HIn]
or, pH=pKIn+log[In-][HIn]
i.e. log[In-][HIn]=pH-pKIn

Topic: Chemistry, Redox Reactions, NEET
The rat in North American deserts is capable of meeting all its water requirements through its internal fat oxidation. It is an example of

Options:
A Morphological adaptation
B Behavioral adaptation
C Anatomical adaptation
D Physiological adaptation

The correct answer is:
Physiological adaptation

Solution:
ADAPTATIONS: It is any attribute of the organism (morphological, physiological, behavioural) that enables the organism to survive and reproduce in its habitat. Many adaptations have evolved over a long evolutionary time and are genetically fixed.
Physiological adaptation: Even in the absence of water, Kangaroo rat in North American deserts is capable of meeting all its water requirements from its internal fat oxidation in which water is produced as a byproduct. Also it can concentrate its urine. There are various alternatives available to organisms for coping with extremes in their environment like morphological, physiological and behavioural adaptation.

Topic: Biology, Organisms and Populations, NEET
The rate at which new born individuals are joining the population by reproduction is known as

Options:
A natality.
B mortality.
C fertility.
D vitality.

The correct answer is:
natality.

Solution:
Number of offsprings produced per unit of population by reproduction is called Natality or birth rate.

Topic: Biology, Organisms and Populations, NEET
The rate constant for a first order reaction is 60 s–1. How much time will it take to reduce the initial concentration of the reactant to its 1/16th value?

Options:
A 6.8 × 10–2s
B 4.6 × 10–2s
C 8.5 × 10–2s
D 9.4 × 10–2s

The correct answer is:
4.6 × 10–2s

Solution:
Correct Option is : (B)
4.6 × 10–2s

Topic: Chemistry, Chemical Kinetics, NEET
The rate constant for a first order reaction is 4.606×10–3 s–1. The time required to reduce 2.0 g of the reactant to 0.2 g is :

Options:
A 200 s
B 500 s
C 1000 s
D 100 s

The correct answer is:
500 s

Solution:
k=4.606×10−3s−1

k=2.303tlogC∘Ct

where C∘= initial concentration

Ct= concentration after time t

k=2.303tlog1020.2

4.606×10−3×t=2.303×log10

t=10002=500 s

Topic: Chemistry, Chemical Kinetics, NEET, NEET 2020 (Phase 1)
The rate constant for the decomposition of hydrocarbons is 2.418 × 10–5s–1at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of the pre-exponential factor?

Options:
A 4.8 × 1012s–1
B 1.2 × 1012s–1
C 6.8 × 1012s–1
D 3.9 × 1012s–1

The correct answer is:
3.9 × 1012s–1

Solution:
Correct Option is : (D)
3.9 × 1012s–1

Topic: Chemistry, Chemical Kinetics, NEET
The rate constant for the decomposition of N2O5at various temperatures is given below:
Calculate the values of A and Ea.

Options:
A A = 2.91 × 106and Ea= 102.27 kJ mol-1
B A = 5.26 × 106and Ea= 116.85 kJ mol-1
C A = 2.91 × 106and Ea= 116.85 kJ mol-1
D A = 5.26 × 106and Ea= 102.27 kJ mol-1

The correct answer is:
A = 2.91 × 106and Ea= 102.27 kJ mol-1

Solution:
Correct Option is : (A)
A = 2.91 × 106and Ea= 102.27 kJ mol-1

Topic: Chemistry, Chemical Kinetics, NEET
The rate constant for the decomposition of N2O5at various temperatures is given below:
Predict the rate constant at 30° and 50°C.

Options:
A Rate constant at 300C = 2.26 × 10-2s-1and Rate constant at 500C = 0.607 s-1
B Rate constant at 300C = 6.08 × 10-2s-1and Rate constant at 500C = 0.889 s-1
C Rate constant at 300C = 6.08 × 10-2s-1and Rate constant at 500C = 0.607 s-1
D Rate constant at 300C = 2.26 × 10-2s-1and Rate constant at 500C = 0.889 s-1

The correct answer is:
Rate constant at 300C = 6.08 × 10-2s-1and Rate constant at 500C = 0.607 s-1

Solution:
Correct Option is : (C)
Rate constant at 300C = 6.08 × 10-2s-1and Rate constant at 500C = 0.607 s-1

Topic: Chemistry, Chemical Kinetics, NEET
The rate constant for the first order decomposition of H2O2is given by the following equation: log k = 14.34 – 1.25 × 104K/T Calculate (i) Eafor this reaction and (ii) at what temperature will its half-period be 256 minutes?

Options:
A (i) 239.339 kJ mol–1and (ii) 669K
B (i) 239.339 kJ mol–1and (ii) 1028K
C (i) 364.223 kJ mol–1and (ii) 669K
D (i) 442.528 kJ mol–1and (ii) 669K

The correct answer is:
(i) 239.339 kJ mol–1and (ii) 669K

Solution:
Correct Option is : (A)
(i) 239.339 kJ mol–1and (ii) 669K

Topic: Chemistry, Chemical Kinetics, NEET
The rate constant for the first-order reaction is 60 s-1. How much time will it take to reduce the concentration of the reactant to 1/16 of initial value?

Options:
A 4.6Ã10-2 s
B 4.6Ã104 s
C 4.6Ã10-3 s
D 4.6Ã10-4 s

The correct answer is:
4.6Ã10-2 s

Solution:
t=2.303klogA0[A]
=2.30360logaa16=2.30360logâ¡16
=2.30360Ã1.204
=0.0462s
=4.6Ã10-2 s

Topic: Chemistry, Chemical Kinetics, NEET
The rate constant is given by the equation k=p.Ze-E/RT . Which factor should register a decrease for the reaction to proceed more rapidly?

Options:
A T
B Z
C E
D p

The correct answer is:
E

Solution:
k=p.Ze-E/RT
Here k= rate constant, p = orientation factor, Z = collision frequency, E = Activation energy and R = gas constant.
From this equation it is clear that on decreasing activation energy and increasing temperature, p and z values k will increase.

Topic: Chemistry, Chemical Kinetics, NEET
The rate constant K1 of a reaction is found to be double that of rate constant K2 of another reaction. The relationship between corresponding activation energies of the two reactions at same temperature ( E1 and E2 ) can be represented as

Options:
A E1>E2
B E1<E2
C E1=E2
D None of these

The correct answer is:
None of these

Solution:
K1=A1e-E1/RT
and K2=A2e-E2/RT
K1K2=A1A2e(-E1+E2)/RT
As A1 and A2 are not given here so none of the relation is correct.

Topic: Chemistry, Chemical Kinetics, NEET
The rate constant k â² of one reaction is double the rate constant k â³ of another reaction. Then the relationship between the corresponding activation energies of the two reactions E a â² and E a â²â² will be -
(Assume the pre-exponential factor & temperature to be same)

Options:
A E a â² > E a â²â²
B E a â² = E a â²â²
C E a â² < E a â²â²
D E a â² < 4 E a â²â²

The correct answer is:
E a â² < E a â²â²

Solution:
When activation energy decreases, rate constant increases. As k â² > k â³ , E a â² < E a â²â²

Topic: Chemistry, Chemical Kinetics, NEET
The rate constant of a reaction has the units as that of rate of reaction. The reaction order is

Options:
A 0
B 1
C 2
D 3

The correct answer is:
0

Solution:
Rate =k[A]n
So k=Rate[A]n=molL-1s-1(molL-1)n
=(molL-1)1-ns-1
We have (molL-1)1-ns-1=molL-1s-1
So 1-n=1
or n=0

Topic: Chemistry, Chemical Kinetics, NEET
The rate constant of a reaction is 1.5Ã10-4 s-1 at 27Â°C and 3Ã10-4 s-1 at 127Â°C . The Ea is

Options:
A 1.663Ã103Â Cal
B 3.326Ã103 cal
C 8.314Ã103 cal
D 2.255Ã103 cal

The correct answer is:
1.663×103 Cal

Solution:
log2×1041.5×104=ℇA2×2.3031300-1400
log2=ℇA2.303×2100400×300
ℇA=0.3010×2.303×2×1200
=1663.68 Cal
=1.663×103 Cal

Topic: Chemistry, Chemical Kinetics, NEET
The rate constant of the reaction

2H2O2 aq â¶ 2H2O l+ O2 gis 3 Ã10â3min-1

At what concentration of H2O2 , the rate of the reaction will be 2 Ã 10â4 Msâ1 ?

Options:
A 6.67 Ã 10â3 M
B 2 M
C 4 M
D 0.08 M

The correct answer is:
4 M

Solution:
-dH2O2dt=kH2O21
â´ 2Ã10-4=3Ã10-360ÃH2O2
or H2O2=4 M

Topic: Chemistry, Chemical Kinetics, NEET
The rate constant of the reaction A⟶B is 0.6×10-3 mole per liter per second. If the initial concentration of A is 5 M, then the concentration of B after 20 minutes is:

Options:
A 1.08 M
B 3.60 M
C 0.36 M
D 0.72 M

The correct answer is:
0.72 M

Solution:
For, Zero Order Reaction:
\(\begin{aligned}
x & =\text { K.t } \\
& =0.6 \times 10^{-3} \times 20 \times 60
\end{aligned}\)
So, \(x=0.72 \mathrm{M}\)

Topic: Chemistry, Chemical Kinetics, NEET, NEET 2015 (Phase 2)
The rate constants $k_1$ and $k_2$ for two different reactions are $10^{16} \cdot e^{-2000 / T}$ and $10^{15} \cdot e^{-1000 / T}$, respectively. The temperature at which $k_1=k_2$ is

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

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

Solution:
Key Idea : The Arrhenius equation is represented as
$k=A e^{-E_a / R T}$
In the given equations, first take log and then compare them.
$\begin{aligned}
& k_1=10^{16} e^{-2000 / T} \\
& k_2=10^{15} e^{-1000 / T}
\end{aligned}$
On taking log, we get
$\begin{aligned}
& \log k_1=\log 10^{16}-\frac{2000}{2.303 T} \\
& \log k_2=\log 10^{15}-\frac{1000}{2.303 T} \\
& \because k_1=k_2
\end{aligned}$
Hence, from Eqs (i) and (ii)
$T=\frac{1000}{2.303} \mathrm{~K}$

Topic: Chemistry, Chemical Kinetics, NEET, NEET 2008 (Screening)
The rate constants of a reaction at 500K and 700K are 0.02 s–1and 0.07 s–1respectively. What are the values of Eaand A?

Options:
A Ea= 11521.5 J and A = 2.56
B Ea= 18230.8 J and A = 2.56
C Ea= 11521.5 J and A = 1.61
D Ea= 18230.8 J and A = 1.61

The correct answer is:
Ea= 18230.8 J and A = 1.61

Solution:
Correct Option is : (D)
Ea= 18230.8 J and A = 1.61

Topic: Chemistry, Chemical Kinetics, NEET
The rate determining step in electrophilic aromatic substitution is the

Options:
A formation of intermediate arenium cation because it occurs faster
B formation of intermediate arenium cation because it is usually slow
C elimination of H+ from arenium cation which occurs faster to restore aromaticity
D formation of the electrophile

The correct answer is:
formation of intermediate arenium cation because it is usually slow

Solution:
The rate determining step in electrophilic aromatic substitution is the formation of arenium cation is the rate determinity step, determining, while elimination of H+ from arenium cation occurs to restore aromaticity and this step is very fast.

Topic: Chemistry, Hydrocarbons, NEET
The rate equation for the reaction 2A+B→C is found to be
Rate = k [A] [B].
The correct statement in relation to this reaction is that the

Options:
A value of k is independent of the initial concentrations of A and B
B rate of formation of C is twice the rate of disappearance of A
C unit of k must be s-1
D t 1/2 is a constant

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

Solution:
2A+B→C
Rate = k[A][B]
It represents second–order reaction. So B, C and D options are incorrect.
As value of k is independent of the initial concentrations of A and B so option A is correct.

Topic: Chemistry, Chemical Kinetics, NEET
The rate law for reaction between the substances A and B is given by

Rate = k A n B m

On doubling the concentration of A and halving the concentration of B, the ratio of the new rate to the earlier rate of reaction will be

Options:
A m + n
B n - m
C 2 n - m
D 2 m+n

The correct answer is:
2 n - m

Solution:
Earlier Rate=k [ A ] n   [ B ] m New rate=k [ 2a ] n   [ 1 2 b ] m New Rate Earlier Rate = [ 2 ] n   [ 1 2 ] m = [ 2 ] n−m

Topic: Chemistry, Chemical Kinetics, NEET
The rate law for the chemical reaction

2NO2 Cl →2NO2+Cl2 =k N O 2 Cl is rate The rate determining step is

Options:
A 2NO2Clâ2NO2+2Cl
B NO2+Cl2âNO2Cl+Cl
C NO2Cl+ClâNO2+Cl2
D NO2ClâNO2+Cl

The correct answer is:
NO2Cl→NO2+Cl

Solution:
Rate =k[NO2Cl]
Hence ,rate determtextning step is
NO2Cl→NO2+Cl

Topic: Chemistry, Chemical Kinetics, NEET
The rate law for the reaction RCl+NaOH(aq)→ROH+NaCl is given by
Rate =k[RCl] . The rate of the reaction will be

Options:
A Decreased on increasing the temperature of the reaction
B Halved on reducing the concentration of alkyl halide to one half
C Doubled on doubling the concentration of sodium hydroxide
D Unaffected by increasing the temperature of the reaction

The correct answer is:
Halved on reducing the concentration of alkyl halide to one half

Solution:
The given rate law suggest that rate of raction depends on concentration of [RCl] and independent of [OH-]
Rate =k[RCl] , if [RCl]=1/2 ,
then rate =intial rate/2 .

Topic: Chemistry, Chemical Kinetics, NEET
The rate law for the reaction
RCl+NaOH(aq)→ROH+NaCl
is given by
Rate =k[RCl] . The rate of the reaction will be

Options:
A Doubled on doubling the concentration of sodium hydroxide
B Halved on reducing the concentration of alkyl halide to one half
C Decreased on increasing the temperature of reaction
D Unaffected by increasing the temperature of the reaction.

The correct answer is:
Halved on reducing the concentration of alkyl halide to one half

Solution:
It is a pseudo-first-order reaction (molecularity is two but of first order) as the rate of reaction does not depend upon the concentration of aqueous NaOH as it is in excess so no effect of change of concentration of it.
Since r â [RCl], so rate is halved when the concentration of alkyl halide is halved.

Topic: Chemistry, Chemical Kinetics, NEET
The rate law of the reaction 2N2O5→4NO2+O2 is

Options:
A Rate=K[NO2]4[O2]
B Rate=K[N2O5]2
C Rate=K[N2O5]0
D Rate=K[N2O5]

The correct answer is:
Rate=K[N2O5]

Solution:
Rate law for the reaction
2N2O5→4NO2+O2
is r=k[N2O5] first order reaction.

Topic: Chemistry, Chemical Kinetics, NEET
The rate of a chemical reaction increases by catalyst due to

Options:
A Reacting with reactants
B Increasing the activation energy
C Decreasing the activation energy
D Reacting with products

The correct answer is:
Decreasing the activation energy

Solution:
Catalyst increases the rate by decreasing the activation energy of reaction.

Topic: Chemistry, Chemical Kinetics, NEET
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, NEET, NEET 2004
The rate of a first-order reaction is 0.04 mol l-1s-1 at 10 second and 0.03 mol l-1s-1 at 20 seconds after initiation of the reaction. The half-life period of the reaction is:

Options:
A 24.1 s
B 34.1 s
C 44.1 s
D 54.1 s

The correct answer is:
24.1 s

Solution:
RateTime0.04100.0320
For first order reaction r ∝ conc.
r1r2=C1C2=43
∴k=2.303t2-t1log⁡C1C2
⇒0.693t1/2=2.30320-10log⁡43
On solving t1/2=24.1 s

Topic: Chemistry, Chemical Kinetics, NEET, NEET 2016 (Phase 1)
The rate of a non-gaseous reaction does not dependent on

Options:
A Catalyst
B Pressure
C Concentration
D Temperature

The correct answer is:
Pressure

Solution:
The rate of reaction depends upon conc. of reactant, surface area of reactant, temperature, presence of light and catalyst for those reaction that occures in solution.

Topic: Chemistry, Chemical Kinetics, NEET
The rate of a reaction is doubled for every 10oC rise in temperature. The increase in reaction rate as a result of temperature rise from 10oC to 100oC is

Options:
A 112
B 512
C 400
D 614

The correct answer is:
512

Solution:
kt+10Kt=rt+10rt=2
For an increase of temperature to 90oC i.e. 9 times, the rate increases by 29 times i.e. 512.

Topic: Chemistry, Chemical Kinetics, NEET
The rate of a reaction quadruples when temperature changes from $27^{\circ} \mathrm{C}$ to $57^{\circ} \mathrm{C}$. Calculate the energy o activation.
Given $\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}, \log 4=0.6021$

Options:
A $380.4 \mathrm{~kJ} / \mathrm{mol}$
B $3.80 \mathrm{~kJ} / \mathrm{mol}$
C $3804 \mathrm{~kJ} / \mathrm{mol}$
D $38.04 \mathrm{~kJ} / \mathrm{mol}$

The correct answer is:
$38.04 \mathrm{~kJ} / \mathrm{mol}$

Solution:
$\begin{aligned} & \log \left(\frac{\mathrm{k}_2}{\mathrm{k}_1}\right)=\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{R}}\left(\frac{1}{\mathrm{~T}_1}-\frac{1}{\mathrm{~T}_2}\right) \\ & \log \left(\frac{4}{1}\right)=\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{R}}\left(\frac{1}{300}-\frac{1}{330}\right) \\ & \mathrm{E}_{\mathrm{a}}=\frac{(\log (4)) \times 2.303 \times 8.314 \times 300 \times 330}{30} \\ & =3.804 \times 10^4 \mathrm{~J} / \mathrm{mol} \\ & =38.04 \mathrm{~kJ} / \mathrm{mol}\end{aligned}$

Topic: Chemistry, Chemical Kinetics, NEET, NEET 2024
The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction assuming that it does not change with temperature.

Options:
A 26.4 kJ mol–1
B 112.2 kJ mol–1
C 85.4 kJ mol–1
D 52.8 kJ mol–1

The correct answer is:
52.8 kJ mol–1

Solution:
Correct Option is : (D)
52.8 kJ mol–1

Topic: Chemistry, Chemical Kinetics, NEET
The rate of a reaction quadruples when the temperature changes from 300â to â310â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 53.6 kJmol-1

The correct answer is:
107.2 kJmol-1

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

Topic: Chemistry, Chemical Kinetics, NEET
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, NEET
The rate of decomposition is controlled by

Options:
A chemical composition of detritus
B climatic factor
C both a) and b)
D nature of detrivores

The correct answer is:
both a) and b)

Solution:
Decomposition is largely oxygen requiring process, rate of decomposition is controlled by chemical composition of detritus and climatic factors. Warm humid condition promotes growth of microbes thus increases rate of decomposition. If detritus is rich in nitrogen and water soluble substance than rate of decomposition is fast, if detritus is rich in lgnin, cellulose than rate of decomposition is slow.

Topic: Biology, Ecosystem, NEET