datavorous/entrance-exam-dataset huggingface dataset:
The energy gap in case of which of the following is maximum?

Options:
A Germanium
B Iron
C Copper
D Aluminium

The correct answer is:
Germanium

Solution:
The energy gap of germanium is maximum 0.7 eV and energy gap of metal is zero

Topic: Physics, Semiconductors, NEET
The energy (in eV) required to excite an electron from n=2 to n=4 state in hydrogen atom is

Options:
A -0.85
B +4.25
C -3.4
D +2.55

The correct answer is:
+2.55

Solution:
We know that,
En=13.6n2 eV
According to question, the required energy is
E=E2-E4
13.6 ( 2 ) 2 â 13.6 4 2 = 13.6 4 â 13.6 16 eV
=13.6Ã4-13.616 eV
=40.816=+2.55 eV

Topic: Physics, Atomic Physics, NEET
The energy levels in an atom A,B and C are in increasing order EA<EB<EC . The x-rays emitted in transitions C→B,B→A and C→A , have wavelengths λ1,λ2 and λ3 then which of the following relation is true?

Options:
A λ3=λ1+λ2
B λ3=λ1λ2λ1+λ2
C λ1+λ2+λ3=0
D none

The correct answer is:
λ3=λ1λ2λ1+λ2

Solution:
As EA<EB<EC
Therefore C is higher energy level & A is at lowest
EC→A=EC→B+EB→A
hcλ3=hcλ1+hcλ2
therefore
λ3=λ1λ2λ1+λ2

Topic: Physics, Atomic Physics, NEET
The energy levels of a hypothetical one-electron atom system are given by En=-16n2 eV, where n=1, 2, 3, â¦â¦. The wavelength of emitted photon corresponding to the transition from first excited level to ground level is about

Options:
A 1035 A∘
B 1220 A∘
C 3650 Aâ
D 690 Aâ

The correct answer is:
1035 A∘

Solution:
Given,
En=-16n2 eV
According to the question,
E1=-16 eV
E2=-16(2)2 eV
=-164 eV
=-4 eV
We know that,
Î»=12420ÎE
Î»=12420|E2-E1|
Î»=12420|-4--16|
Î»=1242012
Î»=1035 Ã

Topic: Physics, Atomic Physics, NEET
The Energy needed For Li g â Li 3+ g +3 e â is 19600 kJ mole-1. Li g =â520â kJmole â1 . Calculate IE2 For Li g .
ionisation energy of H = 13.6 eV

Options:
A 75.3 eV/species
B 25.30 eV/species
C 30.45 eV/species
D 62.40 eV/species

The correct answer is:
75.3 eV/species

Solution:
Li (g)→Li+3+3e- ; IPI+IPII+IPIII=203·1 ev/species
Li→Li+(g)+1e-; 5·4eV/species IPI
Li+(g)→Li2++1e-; IPII=??
Li+2→Li+3 (g); IPIII=IELi2+=13·6×32 = 122.4 eV/species

Topic: Chemistry, Classification of Elements and Periodicity in Properties, NEET
The energy of a hydrogen atom in the ground state is $-13.6 \mathrm{eV}$. The energy of a $\mathrm{He}^{+}$ion in the first excited state will be

Options:
A $-13.6 \mathrm{eV}$
B $-27.2 \mathrm{eV}$
C $-54.4 \mathrm{eV}$
D $-6.8 \mathrm{eV}$

The correct answer is:
$-13.6 \mathrm{eV}$

Solution:
Energy E of an atom with principal quantum number $\mathrm{n}$ is given by $E=\frac{-13.6}{\mathrm{n}^2} Z^2$ for first excited state $\mathrm{n}=2$ and for $\mathrm{He}^{+} \mathrm{Z}=2$
$$
\begin{aligned}
\Rightarrow \quad \mathrm{E} & =\frac{-136 \times(2)^2}{(2)^2} \\
& =-13.6 \mathrm{eV}
\end{aligned}
$$

Topic: Physics, Atomic Physics, NEET, NEET 2010 (Screening)
The energy of activation for an uncatalysed reaction is 100 kJ mol-1. Presence of a catalyst lowers the energy of activation by 75%. The log 1 0 K 2 K 1 of the ratio of rate constant of catalysed and uncatalysed reactions at 27°C is?

Assume the frequency factor is same for both reactions. (Given 2.303 × 8.314 = 19.147)

Options:
A 13.05
B 26.10
C 6.52
D None of these

The correct answer is:
13.05

Solution:
In absence of catalyst, K 1 = Ae - 1 0 0 / RT
In presence of catalyst ; K 2 = Ae - 2 5 / RT
∴ K 2 K 1 = e 7 5 / RT ⇒ log e K 2 K 1 = 7 5 RT
⇒ log 1 0 K 2 K 1 = 7 5 2.303 × R × T
⇒ log 1 0 K 2 K 1 = 7 5 2.303 × 8.314 × 1 0 - 3 × 3 0 0 = 13.057

Topic: Chemistry, Chemical Kinetics, NEET
The energy of an electron in excited hydrogen atom is -3.4 eV. Then, according to Bohr's theory the angular momentum of this electron in JS is

Options:
A 2.11×10-34
B 3×10-34
C 2×10-34
D 0.5×10-34

The correct answer is:
2.11×10-34

Solution:
En=E1n2
n2=E1En=13.63.4=4
n = 2
Angular momentum (mvr) =nh2π=2h2π=hπ
=6.62×10-343.14
=2.11×10-34 Js.

Topic: Chemistry, Structure of Atom, NEET
The energy of an electron in first Bohr's orbit of H atom is −13.6 eV. The energy value of electron in the first excited state of Li 2+ is :

Options:
A 27.2 eV
B â30.6âeV
C 30.6 eV
D −27.2 eV

The correct answer is:
−30.6 eV

Solution:
For H atom & H like species
The electronic energy in the nth orbit is
E n = - 13.6 z 2 n 2 eV
E n Li 2 + = - 13.6 Ã 9 n 2 eV = - 122.4 eV n 2
If n = 2
E 2 Li 2 + = - 122.4 2 2 = - 30.6 eV

Topic: Chemistry, Structure of Atom, NEET
The energy of an electron in the ground state $(n=1)$ for $\mathrm{He}^{+}$ion is $-x J$, then that for an electron in $n=2$ state for $\mathrm{Be}^{3+}$ ion in J is

Options:
A $-\frac{x}{9}$
B $-4 x$
C $-\frac{4}{9} \mathrm{x}$
D $-x$

The correct answer is:
$-x$

Solution:
$E_n=-R_H\left(\frac{z^2}{n^2}\right) J$
For $\mathrm{He}^{+}(n=1)$,
$\begin{aligned}
& E_n=-x=-R_H\left(\frac{2^2}{1^2}\right)=-4 R_H \\
& \therefore \quad R_H=\frac{x}{4} \\
&
\end{aligned}$
For $\mathrm{Be}^{3+}(n=2)$,
$\begin{aligned}
E_n & =-R_H\left(\frac{z^2}{n^2}\right) J \\
& =-\frac{x}{4} \times\left(\frac{4 \times 4}{2 \times 2}\right)=-x J
\end{aligned}$

Topic: Chemistry, Structure of Atom, NEET, NEET 2024
The energy of He+ ion in the first excited state will be

Options:
A -13.6 eV
B -14.4 eV
C -6.8 eV
D -27.2 eV

The correct answer is:
-13.6 eV

Solution:
E=-13.6n2Z2 eV
E=-13.62222=-13.6 eV

Topic: Physics, Atomic Physics, NEET
The energy of hydrogen atom in excited state is -3.4 eV. the angular momentum of electron is

Options:
A 3h2π
B hπ
C 2h5π
D h5π

The correct answer is:
hπ

Solution:
En=-13.6n2Z2eV
n 2 = 13.6 E n Z 2 =− 13.6 3.4 ⋅ 1 2 =4
n = 2
Angular momentum =nh2π=2h2π
=hπ .

Topic: Chemistry, Structure of Atom, NEET
The energy of second Bohr orbit of the hydrogen atom is -328 kJ mol-1 ; hence the energy of fourth Bohr orbit would be

Options:
A -41Â kJÂ mol-1
B -1312Â kJÂ mol-1
C -164Â kJÂ mol-1
D -82 kJ mol-1

The correct answer is:
-82 kJ mol-1

Solution:
The energy of second Bohr orbit of hydrogen atom E2 is -328 kJ mol-1 because
E 2 =− 1312 2 2 kJ mol −1 =328  kJ mol −1
â´ En=-1312n2kJ mol-1
If n=4
â´ E4=-131242kJ mol-1
=-82 kJ mol-1

Topic: Chemistry, Structure of Atom, NEET
The energy of second Bohr orbit of the hydrogen atom is $-328 \mathrm{~kJ} \mathrm{~mol}^{-1}$; hence the energy of fourth Bohr orbit would be

Options:
A $-41 \mathrm{~kJ} \mathrm{~mol}^{-1}$
B $-82 \mathrm{~kJ} \mathrm{~mol}^{-1}$
C $-164 \mathrm{~kJ} \mathrm{~mol}^{-1}$
D $-1312 \mathrm{~kJ} \mathrm{~mol}^{-1}$

The correct answer is:
$-82 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Solution:
$\mathrm{Z}=1$ for hydrogen ; $n=2$
$$
\begin{aligned}
& \mathrm{E}_n=-k\left(\frac{\mathrm{Z}}{n}\right)^2 \\
& \mathrm{E}_2=\frac{-k \times 1}{4} \\
& \mathrm{E}_2=-328 \mathrm{KJ} \mathrm{mol}^{-1} ; k=4 \times 328 \\
& \mathrm{E}_4=\frac{-k \times 1}{16} \\
& \mathrm{E}_4=-4 \times 328 \times \frac{1}{16} \\
& \mathrm{E}_4=-82 \mathrm{KJ} \mathrm{mol}^{-1}
\end{aligned}
$$
Related Theory
The total mechanical energy of an electron in a Bohr orbit is the sum of its kinetic and potential energies. The lowest energy is called the ground state.
Caution
It is given that the energy of the electron in the first Bohr $\mathrm{H}$-atom is $-13.6 \mathrm{eV}$

Topic: Chemistry, Structure of Atom, NEET, NEET 2005
The energy of the em waves is of the order of 15 keV. To which part of the spectrum does it belong?

Options:
A Infra-red rays
B Ultraviolet rays
C γ -rays
D X-rays

The correct answer is:
X-rays

Solution:
Wavelength of the ray
λ=hcE=1240015000=0.826 Å
X-ray wavelengths are in the range 0.5 – 2.5 Å, so it is an X-ray.

Topic: Physics, Electromagnetic Waves, NEET, NEET 2015 (Phase 2)
The energy released by the fission of one 92U235 atom is 200 MeV. Calculate the energy released in kWh, when 1 g of U235 undergoes fission.

Options:
A 2.270×104 kWh
B 2.278×104 kWh
C 2.268×102 kWh
D 2.288×104 kWh

The correct answer is:
2.278×104 kWh

Solution:
Now, number of atoms in 1g of U235
=Avogadro numberatomic weight=6.023 ×1023235
Energy released per fission = 200 MeV
Therefore, the energy released on the fission of 1 g of U235
=6.023 ×1023×200235=5.126 ×1023 MeV
=5.126 ×1023×1.6 ×10-13 J=8.2 ×1010 W s ∵ 1 J=1 W s
=8.2 ×1073600=2.278 ×104 kWh

Topic: Physics, Nuclear Physics, NEET
The energy released during the fission of 1 kg of U235 is E1 and that produced during the fusion of 1 kg of hydrogen is E2. If energy released per fission of Uranium -235 is 200 MeV and that per fusion of hydrogen is 24.7 MeV, then the ratio E2E1 is

Options:
A 2
B 7
C 10
D 20

The correct answer is:
7

Solution:
Number of fissions that can occur in 1 kg of U235 is n1=NA235 kg :NA is the Avogadro number.
Number of fusions that can occur for 1 kg of hydrogen is n2=NA/1 kg4, because 4 hydrogen nuclei fuse to form one helium nuclei.
E2E1=n2Ã24.7 MeVn1Ã200MeV=235Ã24.74Ã200=7.25 (approx)
=7

Topic: Physics, Nuclear Physics, NEET
The energy required for transition in a Hydrogen atom from 1st energy level to the 2nd energy level is E. Then, which of the following transitions is possible for the same energy in Helium atom?

Options:
A 2→3
B 2→4
C 1→2
D 1→3

The correct answer is:
2→4

Solution:
For Hydrogen: E=13.6111-122eV â¦.(i)
For Helium: E=13.6221n12-1n22eV â¦(ii)
From eq (i) & (ii)
13.6 34=13.6 4 1n12-1n22
1n12-1n22=316
Thus, n1=2 & n2=4 are satisfying the equation.

Topic: Physics, Atomic Physics, NEET
The energy required to break one bond in DNA is 10–20 J. This value in eV is nearly

Options:
A 0.6
B 0.06
C 0.006
D 6

The correct answer is:
0.06

Solution:
E=10-201.6×10-19eV

=0.625×10-1

=0.0625 eV

Topic: Physics, Work Power Energy, NEET, NEET 2020 (Phase 1)
The energy required to charge a parallel plate condenser of plate separation d and plate area of cross-section $A$ such that the uniform electric field between the plates is $E$, is

Options:
A $\frac{1}{2} \varepsilon_0 E^2 / A d$
B $\varepsilon_0 E^2 / A d$
C $\varepsilon_0 E^2 A d$
D $\frac{1}{2} \varepsilon_0 E^2 A d$

The correct answer is:
$\varepsilon_0 E^2 A d$

Solution:
Energy given by the cell
$E=C V^2$
Here, $C=$ capacitance of condenser $=\frac{A \varepsilon_0}{d}$ $V=$ potential difference across the plates $=E d$ Therefore,
$\begin{aligned}
E & =\left(\frac{A \varepsilon_0}{d}\right)(E d)^2 \\
& =A \varepsilon_0 E^2 d
\end{aligned}$

Topic: Physics, Capacitance, NEET, NEET 2008 (Screening)
The energy required to remove an electron from the surface of Sodium metal is 2.3 eV. What is the longest wavelength of radiation with which it can show photoelectric effect?

Options:
A 5.4Ã10-17 m
B 5.4Ã10-8 m
C 5.4Ã10-7 m
D 5.4Ã10-9 m

The correct answer is:
5.4Ã10-7 m

Solution:
Energy required to show photo emission =2.3 eV
=2.3 Ã1.6Ã10-19 J
Thus E = hcÎ» or Î»=hcE=6.626Ã10-34Ã3Ã1082.3Ã1.6Ã10-19
= 5.4 Ã10-7 m.
Thus, longest wavelength to show photoelectric effect =5.4 Ã10-7 m.

Topic: Chemistry, Structure of Atom, NEET
The energy spectrum of Î²-particles [number N E as a function of Î²-energy E] emitted from a radioactive source is

Options:
A
B
C
D

The correct answer is:

Solution:
In β - decay, energy emission is continuous and energy is shared between β - particles and daughter nuclei.
Thus certain β -particles have zero energy and certain have maximum energy.
Option (1) is most appropriate

Topic: Physics, Nuclear Physics, NEET
The energy that is released by the hydrolysis of ATP by actin is used for:

Options:
A actin filament assembly
B actin filament disassembly
C actin-myosin assembly
D actin-myosin disassembly

The correct answer is:
actin filament assembly

Solution:
Actin monomers or the G actin polymerizes to form actin filaments or F actin. The first step in the polymerization is the formation of dimers and trimers. This further grows to form the long chains. The actin monomers requires ATP molecule to get polymerized and to form the actin filament.

Topic: Biology, Locomotion and Movement, NEET
The energy that should be added to an electron to reduce its de-Broglie wavelength from 1 nm to 0.5 nm is

Options:
A four times the initial energy
B equal to the initial energy
C twice the initial energy
D thrice the initial energy

The correct answer is:
thrice the initial energy

Solution:
De-Broglie Wavelength
λ=h2mE
∴ λ1λ2=E2E1
⇒ 1×10-90.5×10-9=E2E1
⇒ 2=E2E1
⇒ E2E1=4
∴ E2=4E1
∴ Energy to be added =E2-E1
=4E1-E1=3E1

Topic: Physics, Dual Nature of Matter, NEET
The energy that will be ideally radiated by a 100 kW transmitter in 1 hour is

Options:
A 36×104 J
B 36×105 J
C 1×105 J
D 36×107 J

The correct answer is:
36×107 J

Solution:
Energy =Pt

=100×103×1 hr

=105×3600 =36×107 J

Topic: Physics, Work Power Energy, NEET, NEET 2022 (Phase 1)
The energy-releasing metabolic process in which substrate is oxidised without an external electron acceptor is called

Options:
A glycolysis
B fermentation
C aerobic respiration
D photorespiration

The correct answer is:
fermentation

Solution:
In fermentation, the incomplete oxidation of glucose is achieved under anaerobic conditions by set of reactions, where pyruvic acid is converted to $\mathrm{CO}_2$ and ethanol. The enzyme pyruvic acid, decarboxylase and alcohol dehydrogenase catalyse these reactions. In animal cells also, like muscles during exercise, where oxygen is inadequate for cellular respiration, pyruvic acid is reduced to lactic acid by lactate dehydrogenase. The reducing agent is $\mathrm{NADH}+\mathrm{H}^{+}$, which is reoxidised to $\mathrm{NAD}^{+}$, in both the processes.

Topic: Biology, Respiration in Plants, NEET, NEET 2010 (Screening)
The energy-releasing process in which the substrate is oxidized without an external electron acceptor is called

Options:
A fermentation
B photorespiration
C aerobic respiration
D glycolysis

The correct answer is:
fermentation

Solution:
Fermentation is an anaerobic pathway which takes place in certain bacteria and fungi which do not involve electron transport chain. In this process, substrate level phosphorylation takes place in glycolysis producing 2 ATP molecules.

Topic: Biology, Respiration in Plants, NEET, NEET 2008 (Screening)
The engine of a car produces an acceleration of 6 ms-2 in the car. If this car pulls another car of the same mass, then the acceleration would be

Options:
A 6 ms-2
B 12 ms-2
C 3 ms-2
D 1.5 ms-2

The correct answer is:
3 ms-2

Solution:
Let, mass of car=m
ââââââForce applied by engine =6m
When two cars are pulled,
m+ma=6m
or 2ma=6m or a=3 ms â2

Topic: Physics, Laws of Motion, NEET
The enolic form of ethyl acetoacetate as below has:

Options:
A 18 sigma bonds and 2 pi-bonds
B 16 sigma bonds and 1 pi-bonds
C 9 sigma bonds and 2 pi-bonds
D 9 sigma bonds and 1 pi-bonds

The correct answer is:
18 sigma bonds and 2 pi-bonds

Solution:
18 Ï and 2Ï bonds in both keto and enol form of ethyl acetoacetate.

Topic: Chemistry, Chemical Bonding and Molecular Structure, NEET, NEET 2015 (Phase 1)
The enolic form of ethyl acetoacetate as below has

Options:
A 16 sigma bonds and 1 pi-bond
B 9 sigma bonds and 2 pi-bonds
C 9 sigma bonds and 1 pi-bond
D 18 sigma bonds and 2 pi-bonds

The correct answer is:
18 sigma bonds and 2 pi-bonds

Solution:
As 1 H atom =1σ ,
1 single bond =1σ
1 double bond =1σ,1π

It has 18σ and 2π .

Topic: Chemistry, Carboxylic Acid Derivatives, NEET
The enthalpies of combustion of carbon and carbon monoxide are -393.5 and -283 kJmol-1 respectively. The enthalpy of formation of carbon monoxide per mole is

Options:
A –676.5 kJ
B –110.5 kJ
C 110.5 kJ
D 676.5 kJ

The correct answer is:
–110.5 kJ

Solution:
(i) C(s)+O2(g)âCO2(g)ÎH=-393.5 kJ
(ii) CO(g)+12O2(g)âCO2(g)ÎH=-283.0 kJ
eq. (i) and (ii) gives
(iii) C(s)+12O2(g)âCO(g)ÎH=-110.5 kJ
This equation (iii) also represents formation of one mole of CO and thus enthalpy change is the heat of formation of CO (g).

Topic: Chemistry, Thermodynamics (C), NEET
The enthalpies of formation of N2O and NO are 28 and 90 kJ mol-1 respectively. The enthalpy of the reaction 2N2O(g)+O2(g)→4NO(g) is equal to

Options:
A 8 kJ
B 88 kJ
C -16 kJ
D 304 kJ

The correct answer is:
304 kJ

Solution:
N 2  +  1 2 O 2 → N 2 O;  ΔH = 28 kJâ¦(i)
1 2 N 2 â+â 1 2 O 2 âNO;ââÎHâ=â90âkJ â¦(ii)
By equation [4 Ã (ii)] â [2 Ã (i)],
2N2O+O2â4NO
ÎH=360-56
ÎH=304 kJ.

Topic: Chemistry, Thermodynamics (C), NEET
The enthalpy and entropy change for the reaction

B r 2 (l)+C l 2 (g)â2BrCl(g)

are 30âkJâmo l â1 ââandââ105âJ K â1 âmo l â1 respectively. The temperature at which the reaction will be in equilibrium is

Options:
A 285.7 K
B 273 K
C 450 K
D 300 K

The correct answer is:
285.7 K

Solution:
We know that

ÎG=ÎHâTÎS

When the reaction is in equilibrium, ÎG=0

0=ÎHâTÎSâT= ÎH ÎS T= 30Ã1000 105 =285.7âK

Topic: Chemistry, Thermodynamics (C), NEET
The enthalpy and entropy change for the reaction:
$$
\mathrm{Br}_{2(\mathrm{l})}+\mathrm{Cl}_{2(g)} \rightarrow 2 \mathrm{BrCl}_{(g)}
$$
are $30 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $105 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}$ respectively. The temperature which the reaction will be in equilibrium is:

Options:
A $300 \mathrm{~K}$
B $285.5 \mathrm{~K}$
C $273 \mathrm{~K}$
D $450 \mathrm{~K}$

The correct answer is:
$285.5 \mathrm{~K}$

Solution:
For the reaction
$$
\begin{gathered}
\mathrm{Br}_2(l)+\mathrm{Cl}_2(g) \rightarrow 2 \mathrm{BrCl}(g) \\
\Delta \mathrm{H}=30 \mathrm{~kJ} / \mathrm{mol} \\
\Delta \mathrm{S}=105 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}
\end{gathered}
$$
For at equilibrium $\Delta \mathrm{G}=0$
$$
\begin{aligned}
& \therefore \quad \Delta \mathrm{G}=\Delta \mathrm{H}-\mathrm{T} \Delta \mathrm{S} \\
& \Delta \mathrm{H}=\mathrm{T} \Delta \mathrm{S} \\
& \mathrm{T}=\frac{\Delta \mathrm{H}}{\Delta \mathrm{S}}=\frac{30 \times 1000 \mathrm{~J} \mathrm{~mol}^{-1}}{105 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}} \\
& =285.7 \mathrm{~K} \\
&
\end{aligned}
$$

Topic: Chemistry, Thermodynamics (C), NEET, NEET 2006
The enthalpy and entropy change for the reaction Br 2 ( l )+ Cl 2 ( g )â2 BrCl( g ) are 30 kJ mol-1 and 105 JK-1 mol-1 respectively. The temperature at which the reaction will be in equilibrium is

Options:
A 273 K
B 450 K
C 300 K
D 285.7 K

The correct answer is:
285.7 K

Solution:
We know that
âG=âH-TâS
When the reaction is in equilibrium, âG=0
0=âH-Tâ S â T=âHâS
T= 30Ã1000 105=285.7 K

Topic: Chemistry, Thermodynamics (C), NEET
The enthalpy and entropy change for the reaction :

Br 2 l + Cl 2 g →2BrCl g

are 30 kJ mol-1 and 105 J K-1 mol-1 respectively. The temperature at which the reaction will be in equilibrium is

Options:
A 300 K
B 285.7 K
C 273 K
D 450 K

The correct answer is:
285.7 K

Solution:
Br 2 l + Cl 2 g →2BrCl g
Î H = 3 0 kJ mol - 1 , Î S = 1 0 5 J K - 1 mol - 1
Î S = Î H T i.e., 1 0 5 = 3 0 T Ã 1 0 0 0
â´ T = 3 0 Ã 1 0 0 0 1 0 5 = 285.7 K

Topic: Chemistry, Thermodynamics (C), NEET
The enthalpy and entropy of a reaction are -5.0 kJ/mol and -20 JK-1 mol-1 respectively and are independent of temperature. What is the highest temperature upto which the reaction is feasible?

Options:
A 250 K
B 240 K
C 275 K
D 300 K

The correct answer is:
250 K

Solution:
Both ÎH and ÎS are negative and, therefore, the given reaction will be spontaneous (ÎG<0) if
ÎH-TÎS<0
or -5.0 kJ-T(-20 JK-1)<0
or -5.0Ã103 J+T(20 JK-1)<0
i.e., T<5Ã103 J20 JK-1<250 K .

Topic: Chemistry, Thermodynamics (C), NEET
The enthalpy change for a reaction does not depend upon

Options:
A use of different reactants for the same products
B the nature of intermediate reactions steps
C the differences in the initial and final temperature of involved substances
D the physical state of reactants and products

The correct answer is:
the nature of intermediate reactions steps

Solution:
The enthalpy change for a reaction does not depend upon the nature of intermediate reactions steps.
Heat of reaction depends upon (i) physical state of reactants and product, (ii) conditions of constant P or V and (iii) Temperature.

Topic: Chemistry, Thermodynamics (C), NEET
The enthalpy change involved in the oxidation of glucose is –2880kJmol-1 . Twenty five percent of this energy is available for muscular work. If 100 kJ of muscular work is needed to walk one kilometer, what is the maximum distance that a person will be able to walk after eating 120 g of glucose?

Options:
A 7.9 km
B 9.7 km
C 4.8 km
D 8.4 km

The correct answer is:
4.8 km

Solution:
Total energy involved in the oxidation of glucose =–2880 kJmol-1 .
As twenty five percent of this energy is available for muscular work.
So energy available for muscular work =2880×25100=720 kJmol-1
Energy available for muscular work from 120 g glucose =720×Wglucosemolecularweight=720×120180=480kJ
As 100 kJ energy is enough for = 1 km
so 480 kJ energy is enough for =480×1100=4.8km

Topic: Chemistry, Thermodynamics (C), NEET
The enthalpy changes for given reaction at 298 K is - x cal/mol. If the reaction occurs spontaneously at 298 K, the entropy change at that temperature

Options:
A can be negative, but numerically larger than x/298 cal  mol −1   K −1
B can be negative, but numerically smaller than x/298 cal  mol −1   K −1
C cannot be negative
D cannot be positive

The correct answer is:
can be negative, but numerically smaller than x/298 cal  mol −1   K −1

Solution:
ΔG=ΔH-TΔS
For a spontaneous process, ΔH-TΔS<0 .
So, ΔS >  ΔH T  =  − x cal mol 298 K
Hence, ΔS is numerically smaller than x 298  cal  mol −1   K −1 .

Topic: Chemistry, Thermodynamics (C), NEET
The enthalpy of atomisation for the reaction CH4(g)→ C(g) + 4H (g) is 1665 kJ mol–1. What is the bond energy of C–H bond?

Options:
A 416.2 kJ mol–1
B 466.4 kJ mol–1
C 368.3 kJ mol–1
D 536.8 kJ mol–1

The correct answer is:
416.2 kJ mol–1

Solution:
Correct Option is : (A)
416.2 kJ mol–1

Topic: Chemistry, Thermodynamics (C), NEET
The enthalpy of fusion of water is $1.435 \mathrm{kcal} / \mathrm{mol}$. The molar entropy change for the melting of ice at $0^{\circ} \mathrm{C}$ is

Options:
A $10.52 \mathrm{cal} /(\mathrm{mol} \mathrm{K})$
B $21.04 \mathrm{cal} /(\mathrm{mol} \mathrm{K})$
C $5.260 \mathrm{cal} /(\mathrm{mol} \mathrm{K})$
D $0.526 \mathrm{cal} /(\mathrm{mol} \mathrm{K})$

The correct answer is:
$5.260 \mathrm{cal} /(\mathrm{mol} \mathrm{K})$

Solution:
Molar entropy change for the melting of ice,
$\begin{aligned}
\Delta S_{\text {melt }} & =\frac{\Delta H_{\text {fusion }}}{T} \\
& =\frac{1.435 \mathrm{kcal} / \mathrm{mol}}{(0+273) \mathrm{K}} \\
& =5.26 \times 10^{-3} \mathrm{kcal} / \mathrm{mol} \mathrm{K} \\
& =5.26 \mathrm{cal} / \mathrm{mol} \mathrm{K}
\end{aligned}$

Topic: Chemistry, Thermodynamics (C), NEET, NEET 2012 (Screening)
The enthalpy of hydrogenation of cyclohexene is $-119.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$. If resoance energy of benzene is -150 $\mathrm{kJ} \mathrm{mol}^{-1}$, its enthalpy of hydrogenation would be:

Options:
A $-358.5 \mathrm{~kJ} \mathrm{~mol}^{-1}$
B $-508.9 \mathrm{~kJ} \mathrm{~mol}^{-1}$
C $-208.1 \mathrm{~kJ} \mathrm{~mol}^{-1}$
D $-269.9 \mathrm{~kJ} \mathrm{~mol}^{-1}$

The correct answer is:
$-208.1 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Solution:
$\therefore$ Heat of hydrogenation of cyclohaxene $=-119.5 \mathrm{~kJ} / \mathrm{mol}$
$\therefore$ Heat of hydrogenation of benzene $=3 \times-119.5=-358.5 \mathrm{~kJ} / \mathrm{mol}$
Resonance energy
$=$ Observed $\Delta \mathrm{H}-$ Calculated $\Delta \mathrm{H}$
$=-150.4=-358.5-x$
$x=-208.1 \mathrm{~kJ}$
Related Theory
The energy calculated for a resonance hybrid is lower than the energies of any of the alternative structures; the molecule is then said to be stabilized by resonance.

Topic: Chemistry, Thermodynamics (C), NEET, NEET 2006
The enthalpy of hydrogenation of cyclohexene is - 119.5 kJ mol-1. If resonance energy of benzene is -150.4 kJ mol-1, its enthalpy of hydrogenation would be

Options:
A - 358.5 kJ mol-1
B - 508.9 kJ mol-1
C - 208.1 kJ mol-1
D - 269.9 kJ mol-1

The correct answer is:
- 208.1 kJ mol-1

Solution:
Enthalpy of hydrogenation of benzene
= 3 Δ H Hydrogenation - R.E
= 3 × - 1 1 9 · 5 - - 1 5 0 · 4 = - 3 5 8 · 5 + 1 5 0 · 4
= - 2 0 8 · 1 kJ mol - 1

Topic: Chemistry, Thermodynamics (C), NEET