JEE Chemistry - MCQ Practice Questions

Step 1: Write the relationship between conductivity and molar conductivity

The conductivity $\kappa$ is related to the molar conductivity ${\Lambda }_m$ and the concentration $C$ by:

Step 2: Substitute the given values

With ${\Lambda }_m=426{\text{S⋅cm}}^2\text{/mol}$ and $C=0.01\text{mol/L}=0.01\times 10^{-3}{\text{mol/cm}}^3=10^{-5}{\text{mol/cm}}^3$:

Step 3: State the final result

Since the degree of dissociation is $100\%$, the molar conductivity remains ${\Lambda }_m=426{\text{S⋅cm}}^2\text{/mol}$ unchanged, while the conductivity of the solution is:

Step 1: Integrated Rate Law Method

Each reaction order corresponds to a specific linear plot. For a first-order reaction, a plot of $\ln [\text{A}]$ versus $t$ gives a straight line with slope $-k$:

For a second-order reaction, a plot of $\frac{1}{[\text{A}]}$ versus $t$ gives a straight line with slope $k$:

The order is identified by whichever plot yields a straight line.

Step 2: Differential Rate Law — Initial Rate Method

The rate is measured at the start of the reaction for different initial concentrations. The general rate law is:

Taking the ratio of two experiments:

Solving for $n$ gives the order with respect to $[\text{A}]$.

Step 3: Half-Life Method

For special cases, the half-life $t_{1/2}$ is used. The relationship between half-life and initial concentration depends on the order:

By observing how $t_{1/2}$ changes with $[\text{A}{]}_0$, the order can be identified.

Summary

The three experimental methods are:

Step 1: Recall the freezing point depression formula

The freezing point depression is given by

where $K_f$ is the cryoscopic constant, $m$ is the molality, and $i$ is the van't Hoff factor.

Step 2: Calculate moles of NaCl

Step 3: Calculate molality

The solvent mass is $100\text{ g}=0.1\text{ kg}$, so

Step 4: Determine the van't Hoff factor

NaCl dissociates completely into ${\text{Na}}^{+}$ and ${\text{Cl}}^{-}$, giving

Step 5: Calculate the freezing point depression

The closest answer is

Step 1: Write the van't Hoff equation

Step 2: Substitute the known values

Here $K_1=0.5$, $T_1=400\text{K}$, $T_2=500\text{K}$, $\Delta H^{\circ }=-92000{\text{J mol}}^{-1}$, and $R=8.314{\text{J mol}}^{-1}{\text{K}}^{-1}$.

Step 3: Evaluate the temperature term

Step 4: Compute the right-hand side

**Step 5: Solve for $K_2$

Rounding to two significant figures:

**Step 1: Find $[{\text{H}}^{+}]$ from the given pH**

Since $\text{pH}=2.5$,

Step 2: Set up the equilibrium expression

For the weak acid dissociation $\text{HA}\rightleftharpoons {\text{H}}^{+}+{\text{A}}^{-}$, the acid dissociation constant is

Step 3: Substitute the values

Result:

Step 1: Relate wavelength to crystal field splitting energy

The crystal field splitting energy $\Delta$ expressed in wavenumbers (${\text{cm}}^{-1}$) is the reciprocal of the wavelength in centimeters:

Step 2: Substitute the given value

Step 3: Convert to nanometers

Result

The wavelength of light absorbed by $[{\text{Fe(CN)}}_6{]}^{4-}$ is:

Step 1: Find the moles of excess HCl

The moles of $\text{HCl}$ and $\text{NaOH}$ present are:

The excess $\text{HCl}$ remaining after neutralisation:

**Step 2: Calculate the concentration of ${\text{H}}^{+}$

The total volume of the solution after mixing is:

Therefore, the concentration of ${\text{H}}^{+}$ is:

Step 3: Calculate the pH

However, with a more careful calculation taking into account significant figures and the precise value of the concentration:

Step 1: Recall the Gibbs free energy equation

The spontaneity of a reaction is determined by:

Step 2: Identify the given conditions

The reaction is exothermic, so:

Step 3: Analyze the condition for non-spontaneity

For the reaction to be non-spontaneous, we require $\Delta G>0$:

**Step 4: Determine the sign of $\Delta S$

Since $\Delta H<0$, the term $-T\Delta S$ must be positive and large enough to dominate at high $T$. For $-T\Delta S>0$ with $T>0$, we need:

Step 5: Verify the behavior at high temperature

As $T\to \infty$, the positive term $T|\Delta S|$ dominates, so $\Delta G>0$, confirming non-spontaneity at high temperatures.

Conclusion

For the exothermic reaction to become non-spontaneous at high temperatures, it must be true that:

Setting up the ICE table

Let $x$ be the partial pressure of ${\text{N}}_2{\text{O}}_4$ that dissociates (in atm).

For the equilibrium

the ICE table gives:

**Step 1: Write the $K_p$ expression**

Step 2: Expand and rearrange

Step 3: Solve the quadratic

Applying the quadratic formula to $4x^2+0.15x-0.15=0$:

Taking the positive root:

**Step 4: Calculate $P({\text{NO}}_2)$

Step 1: Recall the relationship between activation energies and enthalpy

For any reaction, the forward and reverse activation energies are related by:

Step 2: Identify the given values

Step 3: Substitute and solve

Interpretation: Because the forward reaction is exothermic ($\Delta H<0$), the reverse reaction is endothermic. The reverse reaction must climb a larger energy barrier, since it must overcome both the original forward activation energy and the energy released by the forward reaction.

Final Answer:

Step 1: Spontaneity Condition

A reaction is spontaneous when the Gibbs free energy change satisfies:

Step 2: Solve for the Threshold Temperature

Rearranging the inequality for $T$:

Step 3: Substitute the Given Values

Converting units so that $\Delta H$ and $\Delta S$ are consistent — $\Delta H=+150\text{kJ/mol}=+150,000\text{J/mol}$ and $\Delta S=+200\text{J/mol⋅K}$:

Step 4: Conclusion

The reaction becomes spontaneous above:

For endothermic reactions with a positive $\Delta S$, the term $T\Delta S$ grows with increasing temperature, eventually overcoming the unfavorable positive $\Delta H$, so spontaneity increases at high temperatures.

Using Arrhenius equation: ln(k₂/k₁) = (Ea/R)(1/T₁ - 1/T₂); ln(8) = (Ea/8.314)($\frac{1}{300}$ - $\frac{1}{320}$); Ea ≈ 52.8 kJ/mol

For reversible adiabatic process: q = 0 (adiabatic) and dS_surr = 0, so ΔS_total = 0, therefore ΔS_system = 0

Chemical potential μᵢ = (∂G/∂nᵢ)T,P, which is the partial molar Gibbs free energy

ΔG° = -20 kJ/mol (negative, spontaneous). Since Q (0.01) < Kp (1000), reaction shifts forward. ΔG = ΔG° + RTlnQ = -20 + 8.314×298×ln(0.01) will be more negative, confirming forward reaction.

F-C alkylation proceeds via carbocation formation. Tertiary carbocations are most stable, followed by secondary, then primary. Reactivity: Tertiary > Secondary > Primary.

Iodoform test requires CH3CO- group. Pentan-2-ol (CH3CH2CH(OH)CH3) can be oxidized to CH3COCH2CH3 which contains the required methyl ketone group.

β-Ketoacids (RCOCH2COOH) undergo easy decarboxylation through a six-membered transition state, producing ketones. This is more facile than decarboxylation of regular carboxylic acids.

The singlet at δ 3.3 ppm suggests the OH is on a quaternary carbon (no neighboring H, hence no coupling). 2,3-dimethyl-2-butanol fits: the OH is on a quaternary carbon, explaining the singlet and broad O-H in IR.

Hydroboration occurs through a concerted mechanism involving a four-membered ring transition state, resulting in syn-addition of BH across the double bond.