CUET 2026 Chemistry May 26 Shift 2 Question Paper- Download Answer Key and Solution PDF

Concept:

According to Arrhenius equation: \[ k = A e^{-E_a/RT} \]

where:

\(k\) = Rate constant
\(A\) = Frequency factor
\(E_a\) = Activation energy
\(R\) = Universal gas constant
\(T\) = Absolute temperature


The equation clearly shows that the rate constant depends exponentially on activation energy.

Step 1: Analyze the exponential term.


The exponential factor is: \[ e^{-E_a/RT} \]

When activation energy \(E_a\) increases: \[ -\frac{E_a}{RT} \]

becomes more negative.

As a result: \[ e^{-E_a/RT} \]

decreases rapidly.

Hence: \[ k \]

also decreases exponentially.

Step 2: Physical interpretation.


Higher activation energy means reactant molecules require more energy to cross the energy barrier.

Therefore:

Fewer molecules can form activated complex
Number of effective collisions decreases
Reaction becomes slower


Hence, rate constant decreases.

Therefore, the correct answer is: \[ \boxed{(C) Decrease exponentially} \] Quick Tip: Remember: \[ k = A e^{-E_a/RT} \] Larger activation energy \(\Rightarrow\) smaller rate constant Higher temperature \(\Rightarrow\) larger rate constant

Concept:

For first order reactions: \[ t = \frac{2.303}{k}\log\frac{a}{a-x} \]

and: \[ t_{1/2} = \frac{0.693}{k} \]

Step 1: Interpret \(75%\) completion.


If \(75%\) reaction is completed, then: \[ 25% \]

reactant remains.

Hence: \[ \frac{a}{a-x}=4 \]

Step 2: Apply integrated rate equation.

\[ 40=\frac{2.303}{k}\log4 \]

Since: \[ \log4=0.6021 \]
\[ 40=\frac{2.303\times0.6021}{k} \]
\[ 40=\frac{1.386}{k} \]
\[ k=\frac{1.386}{40} \]

Step 3: Calculate half-life.

\[ t_{1/2}=\frac{0.693}{k} \]

Substituting: \[ t_{1/2} = \frac{0.693}{1.386/40} \]
\[ t_{1/2}=20 min \]

Hence: \[ \boxed{20 min} \] Quick Tip: For first order reactions: \[ 75% completion=2t_{1/2} \] because after two half-lives only \(25%\) reactant remains.

Concept:

The standard EMF of a galvanic cell is: \[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \]

The electrode with higher reduction potential acts as cathode.

Step 1: Identify cathode and anode.


Given: \[ E^\circ_{Zn^{2+}/Zn}=-0.76\,V \]
\[ E^\circ_{Cu^{2+}/Cu}=+0.34\,V \]

Since copper has higher reduction potential: \[ Cathode=Cu \]

and: \[ Anode=Zn \]

Step 2: Apply EMF formula.

\[ E^\circ_{cell} = 0.34-(-0.76) \]
\[ E^\circ_{cell} = 0.34+0.76 \]
\[ E^\circ_{cell}=1.10\,V \]

Hence: \[ \boxed{1.10\,V} \] Quick Tip: In galvanic cells: \[ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} \] Always subtract oxidation electrode potential from reduction electrode potential.

Concept:

The Nernst equation relates electrode potential with ionic concentration and reaction quotient.

General form: \[ E = E^\circ - \frac{RT}{nF}\ln Q \]

At \(298\,K\): \[ \frac{2.303RT}{F}=0.0591 \]

Thus: \[ E = E^\circ - \frac{0.0591}{n}\log Q \]

Step 1: Understand the terms in equation.


Where:

\(E^\circ\) = Standard electrode potential
\(n\) = Number of electrons transferred
\(Q\) = Reaction quotient


The equation predicts how electrode potential changes with concentration.

Step 2: Check all options.


Option (A): \[ E=E^\circ-\dfrac{0.0591}{n}\log Q \]

Correct standard form at \(298\,K\).

Other options contain incorrect signs or incorrect placement of \(n\).

Hence: \[ \boxed{(A)} \] Quick Tip: At \(298\,K\): \[ E=E^\circ-\frac{0.0591}{n}\log Q \] This formula is extremely important for electrochemistry numericals.

Concept:

Colour in transition metal compounds may arise due to:

\(d-d\) transitions
Charge transfer transitions


In permanganate ion: \[ MnO_4^- \]

manganese exists in: \[ +7 \]

oxidation state.

Step 1: Find electronic configuration of manganese.


Atomic number of manganese: \[ 25 \]

Electronic configuration: \[ Mn=[Ar]\,3d^5 4s^2 \]

For: \[ Mn^{7+} \]

all seven valence electrons are removed.

Thus: \[ Mn^{7+}=[Ar]\,3d^0 \]

There are no electrons in \(d\)-orbitals.

Step 2: Check possibility of \(d-d\) transition.


For \(d-d\) transition:

Electrons must be present in \(d\)-orbitals.


But: \[ Mn^{7+}=3d^0 \]

Therefore: \[ d-d \]

transition is impossible.

Step 3: Identify actual reason for violet colour.


In permanganate ion:

Electrons transfer from oxygen ligand orbitals to empty metal orbitals.


This process is known as: \[ Charge transfer transition \]

Charge transfer transitions are usually very intense, which explains the deep violet colour of: \[ KMnO_4 \]

Hence, the correct answer is: \[ \boxed{Charge transfer transition} \] Quick Tip: Compounds having \(d^0\) or \(d^{10}\) configuration generally do not show \(d-d\) transitions. Their colour usually arises from charge transfer transitions.

Concept:

The rate law of a reaction expresses how the rate depends upon concentration of reactants.

For a reaction: \[ Rate=k[A]^n \]

where:

\(k\) = rate constant
\(n\) = order of reaction with respect to \(A\)


If concentration changes, the rate changes according to the power of concentration.

Step 1: Write the given rate law.


Given: \[ Rate=k[A]^2 \]

This means reaction is second order with respect to \(A\).

Step 2: Determine new concentration.


Suppose initial concentration is: \[ [A] \]

New concentration after doubling: \[ [A]'=2[A] \]

Step 3: Substitute into rate equation.


New rate: \[ Rate' = k(2[A])^2 \]
\[ Rate' = k(4[A]^2) \]
\[ Rate' = 4k[A]^2 \]

But: \[ k[A]^2=original rate \]

Therefore: \[ New rate=4\times original rate \]

Hence, the rate becomes four times.

Therefore, the correct answer is: \[ \boxed{(B) Four times} \] Quick Tip: For reactions: \[ Rate\propto[A]^n \] If concentration becomes \(m\) times: \[ New rate=m^n \] times the original rate.

Concept:

In IUPAC nomenclature:

Longest carbon chain containing functional group is selected.
Alcohol group gets lowest possible number.
Alkyl substituents are named with their positions.


Step 1: Identify longest carbon chain.


Given structure: \[ CH_3-CH(CH_3)-CH_2-OH \]

The longest continuous chain contains: \[ 3 carbon atoms \]

Hence parent chain is: \[ propane \]

Step 2: Locate alcohol group.


The: \[ -OH \]

group is attached to carbon number \(1\).

Thus: \[ propan-1-ol \]

Step 3: Identify substituent.


There is one methyl group attached to carbon number \(2\).

Hence the complete name becomes: \[ 2-Methylpropan-1-ol \]

Therefore, the correct answer is: \[ \boxed{(B) 2-Methylpropan-1-ol} \] Quick Tip: While naming alcohols: Give lowest number to \(-OH\) group. Select longest chain containing the functional group.

Concept:

A single bond contains: \[ 1\sigma \]

bond.

A double bond contains: \[ 1\sigma + 1\pi \]

bond.

Ethene structure is: \[ H_2C=CH_2 \]

Step 1: Count C-H sigma bonds.


Each carbon atom forms two C-H single bonds.

Total C-H bonds: \[ 4 \]

Each single bond contributes: \[ 1\sigma \]

Therefore: \[ 4\sigma \]

bonds are present from C-H bonds.

Step 2: Count bonds in carbon-carbon double bond.


The: \[ C=C \]

double bond contains: \[ 1\sigma +1\pi \]

bond.

Thus: \[ 1\sigma \]

and: \[ 1\pi \]

bond are contributed.

Step 3: Calculate total bonds.


Total sigma bonds: \[ 4+1=5 \]

Total pi bonds: \[ 1 \]

Hence: \[ \boxed{5\sigma,\ 1\pi} \] Quick Tip: Remember: \[ Single bond=1\sigma \] \[ Double bond=1\sigma+1\pi \] \[ Triple bond=1\sigma+2\pi \]

Concept:

Hybridization is the mixing of atomic orbitals having comparable energies to form equivalent hybrid orbitals.

Carbon electronic configuration: \[ 1s^2\,2s^2\,2p^2 \]

In methane: \[ CH_4 \]

carbon forms four equivalent sigma bonds.

Step 1: Determine number of sigma bonds around carbon.


In methane: \[ CH_4 \]

carbon forms: \[ 4 \]

single covalent bonds with hydrogen atoms.

Thus total sigma bonds around carbon: \[ 4 \]

Step 2: Relate sigma bonds with hybridization.


Hybridization pattern: \[ 2 sigma bonds \rightarrow sp \]
\[ 3 sigma bonds \rightarrow sp^2 \]
\[ 4 sigma bonds \rightarrow sp^3 \]

Since carbon forms four sigma bonds: \[ Hybridization=sp^3 \]

Step 3: Determine geometry.

\(sp^3\) hybridization leads to: \[ Tetrahedral geometry \]

with bond angle: \[ 109.5^\circ \]

Hence, the correct answer is: \[ \boxed{sp^3} \] Quick Tip: Shortcut: \[ Steric number=4 \Rightarrow sp^3 \] Methane is the most common example of \(sp^3\) hybridization.

Concept:

Electrical conductivity in aqueous solutions depends upon:

Presence of ions
Number of free moving charged particles


Substances are classified as:

Electrolytes
Non-electrolytes


Electrolytes dissociate into ions in aqueous solution and conduct electricity.

Step 1: Analyze glucose solution.


Glucose is a covalent compound.

In water: \[ Glucose \rightarrow No ions \]

Hence glucose solution conducts electricity very poorly.

Step 2: Analyze urea solution.


Urea is also a covalent molecular compound.

It dissolves in water but does not ionize.

Therefore: \[ Poor conductor \]

Step 3: Analyze sodium chloride solution.


Sodium chloride is an ionic compound.

In water: \[ NaCl \rightarrow Na^+ + Cl^- \]

Large number of ions are produced.

These ions carry electric current efficiently.

Hence: \[ Strong conductor \]

Step 4: Analyze sugar solution.


Sugar solution also does not produce ions.

Hence conductivity is negligible.

Therefore, sodium chloride solution conducts electricity most effectively.

Thus, the correct answer is: \[ \boxed{(C) Sodium chloride solution} \] Quick Tip: Strong electrolytes completely ionize in water and conduct electricity efficiently. Examples: \[ NaCl,\ HCl,\ KOH \] Non-electrolytes like sugar and glucose do not produce ions.