JEE Main 2026 April 08, Shift 2 Question Paper with Solutions

All 72 questions from the JEE Main 2026 (April 08, Shift 2) shift — Physics (24), Chemistry (24) and Mathematics (24) — with the correct answer and a step-by-step solution for every question.

Physics PYQs 2026 Chemistry PYQs 2026 Mathematics PYQs 2026

Chemistry24 questions

Q51Single correctSome Basic Concepts in Chemistry

Match List - I with List - II.

List - I Mass of substance	List - II Number of atoms
A. 1.8 mg water	I. $2\times10^{-4}\times \mathrm{N_A}$
B. 9.8 mg sulphuric acid	II. $1.5\times10^{-4}\times \mathrm{N_A}$
C. 1.8 mg carbon	III. $3\times10^{-4}\times \mathrm{N_A}$
D. 5.85 mg salt (NaCl)	IV. $7\times10^{-4}\times \mathrm{N_A}$

(A)

A-IV, B-III, C-I, D-II

(B)

A-III, B-II, C-IV, D-I

(C)

A-III, B-IV, C-II, D-I

(D)

A-III, B-IV, C-I, D-II

SolutionAnswer: Option 3A-III, B-IV, C-II, D-I

Approach:

For each mass, the moles follow from the molar mass, multiplied by the number of atoms in one formula unit, giving the total number of atoms as a multiple of Avogadro's number

\mathrm{N_A}

, which fixes the matching List-II value.

Step 1:Each List-I mass converts to moles and then to total atoms. Water

\mathrm{H_2O}

has

M = 18

and 3 atoms per molecule;

\mathrm{H_2SO_4}

has

M = 98

and 7 atoms; carbon has

M = 12

and 1 atom;

\mathrm{NaCl}

has

M = 58.5

and 2 atoms per formula unit.

n = \dfrac{m}{M},\quad N_{\text{atoms}} = n\,(\text{atoms/unit})\,\mathrm{N_A}

Step 2:For 1.8 mg water,

n = 1.8\times10^{-3}/18 = 10^{-4}

mol, atoms

= 10^{-4}\times 3\,\mathrm{N_A} = 3\times10^{-4}\,\mathrm{N_A}

. For 9.8 mg

\mathrm{H_2SO_4}

n = 9.8\times10^{-3}/98 = 10^{-4}

mol, atoms

= 10^{-4}\times 7\,\mathrm{N_A} = 7\times10^{-4}\,\mathrm{N_A}

A:\ 3\times10^{-4}\,\mathrm{N_A};\quad B:\ 7\times10^{-4}\,\mathrm{N_A}

Step 3:For 1.8 mg carbon,

n = 1.8\times10^{-3}/12 = 1.5\times10^{-4}

mol, atoms

= 1.5\times10^{-4}\times 1\,\mathrm{N_A} = 1.5\times10^{-4}\,\mathrm{N_A}

. For 5.85 mg

\mathrm{NaCl}

n = 5.85\times10^{-3}/58.5 = 10^{-4}

mol, atoms

= 10^{-4}\times 2\,\mathrm{N_A} = 2\times10^{-4}\,\mathrm{N_A}

C:\ 1.5\times10^{-4}\,\mathrm{N_A};\quad D:\ 2\times10^{-4}\,\mathrm{N_A}

Step 4:Each mass gives an integral or half-integral multiple of

10^{-4}

mol, and every List-II value is used exactly once, giving a one-to-one assignment.

A\to \text{III},\ B\to IV,\ C\to II,\ D\to I

Final answer: A-III, B-IV, C-II, D-I

Q52Single correctSolutions

Given below are two statements:
Given: Molar mass of C, H, O, Cl are 12, 1, 16 and 35.5

\mathrm{g\,mol^{-1}}

, respectively.
Statement I: In 30%(w/w) solution of methanol in

\mathrm{CCl_4}

(at T K), the mole fraction of

\mathrm{CCl_4}

is equal to 0.33.
Statement II: Mixture of methanol and

\mathrm{CCl_4}

shows positive deviation from Raoult's law.
In the light of the above statements, choose the correct answer from the options given below:

(A)

Both Statement I and Statement II are true

(B)

Both Statement I and Statement II are false

(C)

Statement I is true but Statement II is false

(D)

Statement I is false but Statement II is true

Q53Single correctChemical Bonding and Molecular Structure

Bromine trifluoride autoionizes to form

\mathrm{BrF_2^{\oplus}}

and

\mathrm{BrF_4^{\ominus}}

. The shapes of the cation and anion are respectively ______, and ______.

(A)

bent, square planar

(B)

linear, square planar

(C)

bent, see-saw

(D)

linear, tetrahedral

Q54Single correctSolutions

Which of the following statements are not correct?
A. For water, magnitude of

\mathrm{K_b}

is more than the magnitude of

\mathrm{K_f}

.
B. The elevation in boiling point of water when a non-volatile solute is added to it is larger in magnitude than its depression in freezing point.
C. Osmotic pressure measurement is preferred over any other colligative property to determine molar mass of proteins and polymers.
D. The dimerised form of benzoic acid in benzene is

Dimerised form of benzoic acid in benzene shown as two benzoic acid molecules linked by hydrogen bonds: C6H5-C(=O)-OH ... O=C-(OH)-C6H5

(A)

A and B only

(B)

A and D only

(C)

A, B and D only

(D)

A, C and D only

Q55Single correctEquilibrium

Consider the following reactions in which all the reactants and products are present in gaseous state:

2xy \rightleftharpoons x_2 + y_2

;

K_1 = 2.5\times10^{5}

xy + \dfrac{1}{2}z_2 \rightleftharpoons \text{xyz}

;

K_2 = 5\times10^{-3}

The value of

K_3

for the equilibrium

\dfrac{1}{2}x_2 + \dfrac{1}{2}y_2 + \dfrac{1}{2}z_2 \rightleftharpoons \text{xyz}

is:

(A)

\displaystyle 2.5\times10^{-3}

(B)

\displaystyle 2.5\times10^{3}

(C)

\displaystyle 1.0\times10^{-5}

(D)

\displaystyle 5\times10^{-3}

Q56Single correctRedox Reactions and Electrochemistry

Given at 298 K:

E^{\ominus}_{\mathrm{Fe^{2+}/Fe}} = X

Volt

E^{\ominus}_{\mathrm{Fe^{3+}/Fe}} = Y

Volt
The

E^{\ominus}_{\mathrm{Fe^{3+}/Fe^{2+}}}

in Volt at 298 K is given by:

(A)

\displaystyle 2X - 3Y

(B)

\displaystyle 3Y - 2X

(C)

\displaystyle 3Y + 2X

(D)

\displaystyle Y + X

Q57Single correctChemical Kinetics

Given below are two statements:

R = 8.314\,\mathrm{J\,K^{-1}\,mol^{-1}}

and

1\,\mathrm{cal} = 4.2\,\mathrm{J}

Statement I: When

E_a = 12.6\,\mathrm{kcal/mol}

, the room temperature rate constant is doubled by a

10^\circ\text{C}

increase in temperature (298 K to 308 K).
Statement II: For a first order reaction

A \to B

, the plot of

t_{1/2}

/s versus

[A]_0

/mol

L^{-1}

is a straight line passing through the origin with positive slope. Here

[A]_0

is the initial concentration of A and

t_{1/2}

is the half life of reaction.
In the light of the above statements, choose the correct answer from the options given below:

Graph with y-axis t(1/2)/s and x-axis [A]0/mol L^-1, showing a straight line of positive slope passing through the origin (as drawn in Statement II).

(A)

Both Statement I and Statement II are true

(B)

Both Statement I and Statement II are false

(C)

Statement I is true but Statement II is false

(D)

Statement I is false but Statement II is true

Q58Single correctClassification of Elements and Periodicity in Properties

Match List - I with List - II.

List - I Electronic configuration of neutral atom (where n = 2)	List - II 1st Ionization Energy ( $\mathrm{kJ\,mol^{-1}}$ )
A. $ns^2$	I. 2080
B. $ns^2np^1$	II. 899
C. $ns^2np^3$	III. 800
D. $ns^2np^6$	IV. 1402

(A)

A-II, B-III, C-IV, D-I

(B)

A-IV, B-III, C-II, D-I

(C)

A-III, B-II, C-IV, D-I

(D)

A-III, B-II, C-I, D-IV

Q59Single correctp-Block Elements

Find the correct statements related to group 15 hydrides.
A. Reducing nature increases from

\mathrm{NH_3}

\mathrm{BiH_3}

.
B. Tendency to donate lone pair of electrons decreases from

\mathrm{NH_3}

\mathrm{BiH_3}

.
C. The stability of hydrides decreases from

\mathrm{NH_3}

\mathrm{BiH_3}

.
D. HEH bond angle decreases from

\mathrm{NH_3}

\mathrm{SbH_3}

(E = Elements of group 15).
Choose the correct answer from the options given below:

(A)

A and B only

(B)

B and C only

(C)

A, B, C and D

(D)

A, C and D Only

Q60Single correctd- and f-Block Elements

Given below are two statements:
Statement I: The number of pairs among

[\mathrm{Ti^{4+}, V^{2+}}]

[\mathrm{V^{2+}, Mn^{2+}}]

[\mathrm{Mn^{2+}, Fe^{3+}}]

and

[\mathrm{V^{2+}, Cr^{2+}}]

in which both ions are coloured is 3.
Statement II: The number of pairs among

[\mathrm{La^{3+}, Yb^{2+}}]

[\mathrm{Lu^{3+}, Ce^{4+}}]

and

[\mathrm{Ac^{3+}, Lr^{3+}}]

ions in which both are diamagnetic is 3.
In the light of the above statements, choose the correct from the options given below:

(A)

Both Statement I and Statement II are correct

(B)

Both Statement I and Statement II are incorrect

(C)

Statement I is correct but Statement II is incorrect

(D)

Statement I is incorrect but Statement II is correct

SolutionAnswer: Option 1Both Statement I and Statement II are correct

Approach:

For Statement I, an ion is coloured when it has partially filled d-orbitals allowing d–d transitions; the count is the pairs in which BOTH ions are coloured. For Statement II, an f-block ion is diamagnetic when its f-subshell is empty (

f^0

) or completely filled (

f^{14}

); the count is the pairs in which BOTH ions are diamagnetic.

Step 1:The d-configurations are

\mathrm{Ti^{4+}} = d^0

(colourless),

\mathrm{V^{2+}} = d^3

(coloured),

\mathrm{Mn^{2+}} = d^5

(coloured),

\mathrm{Fe^{3+}} = d^5

(coloured),

\mathrm{Cr^{2+}} = d^4

(coloured).

\mathrm{Ti^{4+}}=d^0,\ \mathrm{V^{2+}}=d^3,\ \mathrm{Mn^{2+}}=d^5,\ \mathrm{Fe^{3+}}=d^5,\ \mathrm{Cr^{2+}}=d^4

Step 2:For Statement I:

[\mathrm{Ti^{4+},V^{2+}}]

has

\mathrm{Ti^{4+}}

colourless (fails);

[\mathrm{V^{2+},Mn^{2+}}]

both coloured (counts);

[\mathrm{Mn^{2+},Fe^{3+}}]

both coloured (counts);

[\mathrm{V^{2+},Cr^{2+}}]

both coloured (counts). Total = 3, so Statement I is correct.

\text{both-coloured pairs} = 3

Step 3:For Statement II, the f-configurations are

\mathrm{La^{3+}} = f^0

\mathrm{Yb^{2+}} = f^{14}

\mathrm{Lu^{3+}} = f^{14}

\mathrm{Ce^{4+}} = f^0

\mathrm{Ac^{3+}} = f^0

f^0

\mathrm{Lr^{3+}} = f^{14}

. Every ion is either

f^0

f^{14}

, hence all are diamagnetic.

f^0\ \text{or}\ f^{14}\Rightarrow\ \text{diamagnetic}

Step 4:Each of the three pairs in Statement II has both ions diamagnetic, so the count is 3 and Statement II is correct. Therefore both statements are correct.

\text{both-diamagnetic pairs} = 3

Final answer: Both Statement I and Statement II are correct

Q61Single correctd- and f-Block Elements

Given below are two statements for catalytic properties of transition metals.
Statement I: First row transition metals which act as catalyst utilise their 3d electrons only for formation of bonds between reactant molecules and atoms on the surface of catalyst.
Statement II: There is increase in the concentration of reactants on the surface of catalyst which strengthens the bonds in reacting molecules.
In the light of the above statements, choose the correct answer from the options given below :

(A)

Both Statement I and Statement II are correct

(B)

Both Statement I and Statement II are incorrect

(C)

Statement I is correct but Statement II is incorrect

(D)

Statement I is incorrect but Statement II is correct

Q62Single correctPurification and Characterisation of Organic Compounds

Given below are two statements :
Statement I: Vapours of the liquid with higher boiling point condense before vapours of the liquid with lower boiling points in fractional distillation.
Statement II: The vapours rising up in the fractionating column become richer in high boiling component of the mixture.
In the light of the above statements, choose the correct answer from the options given below :

(A)

Both Statement I and Statement II are true

(B)

Both Statement I and Statement II are false

(C)

Statement I is true but Statement II is false

(D)

Statement I is false but Statement II is true

Q63Single correctSome Basic Principles of Organic Chemistry

The major product of which of the following reaction is not obtained by rearrangement reaction?

(A)

Benzene ring plus CH2CH2CH2Cl (1-chloropropane) with Anhyd. AlCl3 giving a benzene ring bearing an isopropyl group (Friedel-Crafts alkylation with carbocation rearrangement).

(B)

tert-butyl alcohol (tert-C bearing OH) with H3O+ giving 2-methylpropene (isobutylene, a terminal alkene) via E1 dehydration, no rearrangement.

(C)

n-hexane CH3(CH2)4CH3 with Anhyd. AlCl3/HCl rearranging to 2-methylpentane CH3-CH(CH3)-CH2-CH2-CH3 (isomerisation/rearrangement).

(D)

A branched primary/secondary alcohol bearing OH with H3O+ giving a more substituted alkene via carbocation rearrangement (hydride/methyl shift).

Q65Single correctHydrocarbons

n-Butane on monochlorination under photochemical condition gives an optically active compound "P". "P" on further chlorination gives dichloro compounds.
The number of dichloro compounds obtained (ignore stereoisomers) is :

(A)

(B)

(C)

(D)

Q66Single correctOrganic Compounds Containing Halogens

Given below are two statements :
Statement I: Due to increase in van der Waals forces, the order of boiling points is

\mathrm{CH_3CH_2CH_2I > CH_3CH_2I > CH_3I}

(A)

Both Statement I and Statement II are true

(B)

Both Statement I and Statement II are false

(C)

Statement I is true but Statement II is false

(D)

Statement I is false but Statement II is true

Q67Single correctOrganic Compounds Containing Nitrogen

Consider the following reaction.

A piperidine ring with a carbonyl (lactam/amide C=O on ring carbon adjacent to N), the nitrogen bearing a CH(CH3)-CHO side chain (2-methyl aldehyde). Reagents: (i) NaBH4/MeOH, (ii) NaOH(aq.), heat, (iii) H3O+ giving product P.

(A)

Intact piperidine ring with nitrogen attached to a CH(CH3)-CH2OH group (an N-substituted amino alcohol, no ring carbonyl).

(B)

Piperidine ring with an OH substituent on the carbon adjacent to nitrogen and the nitrogen bearing a CH(CH3)-CHO (aldehyde) side chain.

(C)

Open-chain structure: a carboxylic acid HOOC-(CH2)4- chain whose terminal carbon bears an NH that connects to CH(CH3)-CH2OH (amino alcohol). Ring opened, carbonyl now COOH.

(D)

Open-chain structure: HO-(CH2)5- chain whose terminal carbon bears NH connecting to CH(CH3)-CH2OH; both ends bear hydroxyl groups (a diol with secondary amine).

Q68Single correctOrganic Compounds Containing Nitrogen

Which statements are True?
A. In Hoffmann bromamide degradation, 4 moles of NaOH and 2 moles of

\mathrm{Br_2}

are consumed per mole of an amide
B. Hoffmann bromamide reaction is not given by alkyl amides
C. Primary amines can be synthesized by Hoffmann bromamide degradation.
D. Secondary amide on reaction with

\mathrm{Br_2}

and NaOH will give secondary amine.
E. The by-products of Hoffmann degradation are

\mathrm{Na_2CO_3}

and

\mathrm{H_2O}

.
Choose the correct answer from the options given below :

(A)

A, C and E only

(B)

B, C and D only

(C)

C and E only

(D)

C, D and E only

Q69Single correctBiomolecules

The incorrect statement from the following with respect to carbohydrates is :

(A)

All monosaccharides are reducing sugars.

(B)

The monosaccharide units obtained from hydrolysis of oligosaccharides are always the same.

(C)

Starch and cellulose are typical examples of polysaccharides, which are very high molecular weight compounds of more than ten monosaccharide units.

(D)

Open chain and cyclic structures co-exist at equilibrium that are responsible for certain properties as in the case of

\displaystyle \mathrm{D-(+)-}

glucose.

Q70Single correctBiomolecules

Which of the following amino acid will give violet coloured complex with neutral ferric chloride solution?

(A)

Threonine

(B)

Serine

(C)

Tyrosine

(D)

Cysteine

Q71NumericalCoordination Compounds

Number of paramagnetic complexes among the following is ______.

[\mathrm{MnBr_4}]^{2-},\ [\mathrm{NiCl_4}]^{2-},\ [\mathrm{Ni(CN)_4}]^{2-},\ [\mathrm{Ni(CO)_4}],\ [\mathrm{CoF_6}]^{3-},\ [\mathrm{Fe(CN)_6}]^{4-},\ [\mathrm{Mn(CN)_6}]^{3-},\ [\mathrm{Ti(CN)_6}]^{3-},\ [\mathrm{Cu(H_2O)_6}]^{2+},\ [\mathrm{Co(C_2O_4)_3}]^{3-}

SolutionAnswer: 6

Approach:

For each complex the central metal oxidation state, the d-electron count, the coordination geometry, and the ligand field strength (strong/weak) are determined. By crystal field theory and valence bond theory, the unpaired electrons are counted; a complex is paramagnetic when it has at least one unpaired electron.

Step 1:The task is to count complexes with one or more unpaired electrons among the ten listed. The oxidation states and d-electron counts are:

[\mathrm{MnBr_4}]^{2-}

n^{2+}

d^5

;

[\mathrm{NiCl_4}]^{2-}

i^{2+}

d^8

;

[\mathrm{Ni(CN)_4}]^{2-}

i^{2+}

d^8

;

[\mathrm{Ni(CO)_4}]

i^0

d^{10}

;

[\mathrm{CoF_6}]^{3-}

o^{3+}

d^6

;

[\mathrm{Fe(CN)_6}]^{4-}

e^{2+}

d^6

;

[\mathrm{Mn(CN)_6}]^{3-}

n^{3+}

d^4

;

[\mathrm{Ti(CN)_6}]^{3-}

i^{3+}

d^1

;

[\mathrm{Cu(H_2O)_6}]^{2+}

u^{2+}

d^9

;

[\mathrm{Co(C_2O_4)_3}]^{3-}

o^{3+}

d^6

\mathrm{Mn^{2+}}:d^5,\ \mathrm{Ni^{2+}}:d^8,\ \mathrm{Ni^0}:d^{10},\ \mathrm{Co^{3+}}:d^6,\ \mathrm{Fe^{2+}}:d^6,\ \mathrm{Mn^{3+}}:d^4,\ \mathrm{Ti^{3+}}:d^1,\ \mathrm{Cu^{2+}}:d^9

Step 2:Consider the tetrahedral halide complexes (weak-field, high spin).

[\mathrm{MnBr_4}]^{2-}

d^5

tetrahedral fills

e^2t_2^3

giving 5 unpaired electrons.

[\mathrm{NiCl_4}]^{2-}

d^8

tetrahedral fills

e^4t_2^4

giving 2 unpaired electrons. Both are paramagnetic.

[\mathrm{MnBr_4}]^{2-}:\ n=5;\quad [\mathrm{NiCl_4}]^{2-}:\ n=2

Step 3:Consider the strong-field and

d^{10}

Ni species.

[\mathrm{Ni(CN)_4}]^{2-}

d^8

with strong-field CN

^-

is square planar, all electrons paired, n=0.

[\mathrm{Ni(CO)_4}]

is N

i^0

d^{10}

, completely filled, n=0. Both are diamagnetic.

[\mathrm{Ni(CN)_4}]^{2-}:\ n=0;\quad [\mathrm{Ni(CO)_4}]:\ n=0

Step 4:Consider the octahedral complexes.

[\mathrm{CoF_6}]^{3-}

d^6

with weak-field F

^-

is high spin t

_{2g}^4e_g^2

, n=4 (paramagnetic).

[\mathrm{Fe(CN)_6}]^{4-}

d^6

with strong CN

^-

is low spin t

_{2g}^6

, n=0 (diamagnetic).

[\mathrm{Mn(CN)_6}]^{3-}

d^4

low spin t

_{2g}^4

, n=2 (paramagnetic).

[\mathrm{Ti(CN)_6}]^{3-}

d^1

, n=1 (paramagnetic).

[\mathrm{Cu(H_2O)_6}]^{2+}

d^9

, n=1 (paramagnetic).

[\mathrm{Co(C_2O_4)_3}]^{3-}

d^6

with chelating oxalate (strong field) is low spin t

_{2g}^6

, n=0 (diamagnetic).

[\mathrm{CoF_6}]^{3-}:n=4;\ [\mathrm{Fe(CN)_6}]^{4-}:n=0;\ [\mathrm{Mn(CN)_6}]^{3-}:n=2;\ [\mathrm{Ti(CN)_6}]^{3-}:n=1;\ [\mathrm{Cu(H_2O)_6}]^{2+}:n=1;\ [\mathrm{Co(C_2O_4)_3}]^{3-}:n=0

Step 5:The paramagnetic complexes (n>=1) are

[\mathrm{MnBr_4}]^{2-}

[\mathrm{NiCl_4}]^{2-}

[\mathrm{CoF_6}]^{3-}

[\mathrm{Mn(CN)_6}]^{3-}

[\mathrm{Ti(CN)_6}]^{3-}

[\mathrm{Cu(H_2O)_6}]^{2+}

. The diamagnetic complexes (n=0) are

[\mathrm{Ni(CN)_4}]^{2-}

[\mathrm{Ni(CO)_4}]

[\mathrm{Fe(CN)_6}]^{4-}

[\mathrm{Co(C_2O_4)_3}]^{3-}

. Total paramagnetic = 6.

2 + 4 = 6

Final answer: 6

Q72NumericalOrganic Compounds Containing Oxygen

'x' is the product which is obtained from benzene by reacting it with carbon monoxide and hydrogen chloride in the presence of cuprous chloride. 'y' is the major product obtained from the benzene by reacting it with ethanoyl chloride in the presence of anhydrous

\mathrm{AlCl_3}

. Product (major) obtained by heating x and y in the presence of alkali is z. Total number of

\pi\,(pi)

electrons in z is ______

Q73NumericalAtomic Structure

Consider two radiations of wavelengths
1.

\lambda_1 = 2000\ \text{\AA}

\lambda_2 = 6000\ \text{\AA}

The ratio of the energies of these two radiations

\left(\dfrac{E_1}{E_2}\right)

is ______ (Nearest integer).

Q74NumericalChemical Thermodynamics

Consider the reaction:

2\mathrm{H_2S}(g) + 3\mathrm{O_2}(g) \rightarrow 2\mathrm{H_2O}(l) + 2\mathrm{SO_2}(g)

The magnitude of enthalpy change for the reaction in

\mathrm{kJ~mol^{-1}}

is ______. (Nearest integer)
Given:

\Delta_f H^{\ominus}(\mathrm{H_2S}) = -20.1~\mathrm{kJ~mol^{-1}}

\Delta_f H^{\ominus}(\mathrm{H_2O}) = -286.0~\mathrm{kJ~mol^{-1}}

\Delta_f H^{\ominus}(\mathrm{SO_2}) = -297.0~\mathrm{kJ~mol^{-1}}

SolutionAnswer: 1126

Approach:

Hess's law in the form of standard formation enthalpies applies: the reaction enthalpy equals the sum over products minus the sum over reactants, with stoichiometric coefficients. Elemental

\text{O}_2

in its standard state has zero formation enthalpy. The magnitude is reported.

Step 1:Coefficients:

2\,\text{H}_2\text{S} + 3\,\text{O}_2 \rightarrow 2\,\text{H}_2\text{O} + 2\,\text{SO}_2

. Formation enthalpies (kJ/mol):

\text{H}_2\text{S} = -20.1

\text{H}_2\text{O} = -286.0

\text{SO}_2 = -297.0

\text{O}_2 = 0

. The target is the magnitude of

\Delta_r H

\Delta_f H:\ \mathrm{H_2S}=-20.1,\ \mathrm{H_2O}=-286.0,\ \mathrm{SO_2}=-297.0,\ \mathrm{O_2}=0

Step 2:The Hess's law expression with stoichiometric coefficients is written out.

\Delta_r H = [2\Delta_f H(\mathrm{H_2O}) + 2\Delta_f H(\mathrm{SO_2})] - [2\Delta_f H(\mathrm{H_2S}) + 3\Delta_f H(\mathrm{O_2})]

Step 3:Evaluating the product sum.

2(-286.0) + 2(-297.0) = -572.0 - 594.0 = -1166.0\ \mathrm{kJ}

Step 4:Evaluating the reactant sum (

\text{O}_2

contributes zero).

2(-20.1) + 3(0) = -40.2\ \mathrm{kJ}

Step 5:Subtracting and taking the magnitude, rounded to the nearest integer.

\Delta_r H = -1166.0 - (-40.2) = -1125.8\ \mathrm{kJ\,mol^{-1}},\quad |\Delta_r H| \approx 1126

Final answer: 1126

Q75NumericalEquilibrium

Solid carbon, CaO and

\mathrm{CaCO_3}

are mixed and allowed to attain equilibrium at T K.

\mathrm{CaCO_3(s)} \rightleftharpoons \mathrm{CaO(s)} + \mathrm{CO_2(g)} \qquad Kp_1 = 0.08\ \mathrm{atm}

\mathrm{C(s)} + \mathrm{CO_2(g)} \rightleftharpoons 2\,\mathrm{CO(g)} \qquad Kp_2 = 2\ \mathrm{atm}

\text{The partial pressure of CO is}

____

\times 10^{-1}

atm

JEE Main 2026 April 08, Shift 2 Question Paper with Solutions

Physics24 questions

Chemistry24 questions

Mathematics24 questions

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