JEE Chemistry — MCQ Practice Questions & Answers

Question 1

At critical point, the compressibility factor Z equals:

Accepted Answer

Correct answer: C. 0.27 — At the critical point, the compressibility factor is defined as Z_c = \frac{P_c V_c}{R T_c}. For a van der Waals gas, substituting the critical constants P_c = \dfrac{a}{27b^2} , V_c = 3b , and T_c = \dfrac{8a}{27Rb} gives \begin{aligned} Z_c &= \frac{P_c V_c}{R T_c} \ &= \frac{\left(\dfrac{a}{27b^2}ight)(3b)}{R \left(\dfrac{8a}{27Rb}ight)} \ &= \frac{\dfrac{3a}{27b}}{\dfrac{8a}{27b}} \ &= \frac{3}{8} \ &= 0.375. \end{aligned} For most real gases, the experimentally observed value is approximately \boxed{Z_c = \frac{P_c V_c}{R T_c} \approx 0.27.}

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A. 1

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B. 0.5

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D. 2

Question 2

The rate constant of a reaction at 25°C is k₁ and at 75°C is k₂. If Ea = 60 kJ/mol, what is k₂/k₁? (R = 8.314 J/mol·K)

Accepted Answer

Correct answer: C. 103 — **Step 1: Identify the Arrhenius two-temperature relation** The two-temperature form of the Arrhenius equation is \ln\!\left(\frac{k_2}{k_1}ight) = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}ight) where T_1 = 25 + 273 = 298\ 	ext{K} and T_2 = 75 + 273 = 348\ 	ext{K} . **Step 2: Substitute the known values** \begin{aligned} \ln\!\left(\frac{k_2}{k_1}ight) &= \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}ight) \ 6pt] &= \frac{60000\ 	ext{J mol}^{-1}}{8.314\ 	ext{J mol}^{-1}	ext{K}^{-1}}\left(\frac{1}{298\ 	ext{K}} - \frac{1}{348\ 	ext{K}}ight) \ 6pt] &= 7215.3 	imes \left(3.356	imes10^{-3} - 2.874	imes10^{-3}ight)\ 	ext{K}^{-1} \ 6pt] &= 7215.3 	imes 4.820	imes10^{-4} \ 6pt] &\approx 4.63 \end{aligned} **Step 3: Solve for k_2/k_1 ** \frac{k_2}{k_1} = e^{4.63} \approx 103 **Result:** \boxed{\dfrac{k_2}{k_1} \approx 103}

Answer

A. 10.3

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B. 25.8

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D. 258

Question 3

For an ideal gas undergoing isothermal expansion, the work done by gas is:

Accepted Answer

Correct answer: B. nRT ln(Vf/Vi) — **Step 1: Identify the process** For an isothermal process, the temperature T remains constant throughout the expansion. **Step 2: Apply the work formula** The work done by the gas during isothermal expansion is given by: W = nRT \ln\!\left(\frac{V_f}{V_i}\right) where n is the number of moles, R is the universal gas constant, T is the absolute temperature, V_f is the final volume, and V_i is the initial volume. **Step 3: Determine the sign** Since the gas expands, we have V_f > V_i , which means: \frac{V_f}{V_i} > 1 \implies \ln\!\left(\frac{V_f}{V_i}\right) > 0 Therefore, the work done by the gas is positive: \boxed{W = nRT \ln\!\left(\frac{V_f}{V_i}\right) > 0}

Answer

A. Zero

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C. nRT ln(Vi/Vf)

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D. ΔU

Question 4

The Henry's law constant for CO₂ in water at 25°C is approximately:

Accepted Answer

Correct answer: A. 1.67 × 10³ atm/(mol/L) — Henry's law states that the partial pressure of a gas above a solution is proportional to its mole fraction in the solution: P = K_h \cdot x where P is the partial pressure, x is the mole fraction of the dissolved gas, and K_h is the Henry's law constant. For 	ext{CO}_2 dissolved in water at 25^\circ	ext{C} , the Henry's law constant is approximately: \boxed{K_h \approx 1.67 	imes 10^3 \ 	ext{atm·L/mol}}

Answer

B. 2.5 × 10⁻⁴ atm/(mol/L)

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C. 1.67 × 10⁻³ atm/(mol/L)

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D. 3.5 × 10² atm/(mol/L)

Question 5

In a electrochemical cell, the standard cell potential E°cell = 0.34 V. The value of ΔG° at 298 K is approximately: (F = 96500 C/mol)

Accepted Answer

Correct answer: B. -65.6 kJ/mol — **Step 1: Write the relationship between \Delta G^\circ and E^\circ_{	ext{cell}} ** The standard Gibbs free energy change is related to the standard cell potential by: \Delta G^\circ = -nFE^\circ_{	ext{cell}} where n is the number of moles of electrons transferred, and F = 96500\ 	ext{C mol}^{-1} is the Faraday constant. **Step 2: Substitute the known values** For n = 2 (typical two-electron transfer), F = 96500\ 	ext{C mol}^{-1} , and E^\circ_{	ext{cell}} = 0.34\ 	ext{V} : \begin{aligned} \Delta G^\circ &= -nFE^\circ_{	ext{cell}} \ 6pt] &= -(2) 	imes (96500\ 	ext{C mol}^{-1}) 	imes (0.34\ 	ext{V}) \ 6pt] &= -65{,}620\ 	ext{J mol}^{-1} \ 6pt] &\approx -65.6\ 	ext{kJ mol}^{-1} \end{aligned} **Result:** \boxed{\Delta G^\circ \approx -65.6\ 	ext{kJ mol}^{-1}}

Answer

A. -32.8 kJ/mol

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C. -98.4 kJ/mol

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D. -131.2 kJ/mol

Question 6

The relationship between Kp and Kc for the reaction: 2NO₂ ⇌ N₂O₄ is:

Accepted Answer

Correct answer: B. Kp = Kc/(RT) — **Step 1: Determine \Delta n ** For the reaction 2	ext{NO}_2 ightleftharpoons 	ext{N}_2	ext{O}_4 , the change in moles of gas is: \Delta n = n_{	ext{products}} - n_{	ext{reactants}} = 1 - 2 = -1 **Step 2: Apply the K_p – K_c relationship** The general relation is: K_p = K_c(RT)^{\Delta n} Substituting \Delta n = -1 : \begin{aligned} K_p &= K_c(RT)^{-1} \ 6pt] &= \boxed{\dfrac{K_c}{RT}} \end{aligned}

Answer

A. Kp = Kc(RT)²

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C. Kp = Kc(RT)⁻¹

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D. Kp = Kc

Question 7

Which of the following statements about Le Chatelier's principle is NOT correct?

Accepted Answer

Correct answer: A. Increasing pressure favours forward reaction in the given equilibrium — Le Chatelier's principle states that when a system at equilibrium is subjected to a change in pressure, the equilibrium shifts to partially oppose that change. **Step 1: Effect of increasing pressure** Increasing pressure favours the side of the equilibrium with **fewer moles of gas**. That is, for a general reaction a\,A_{(g)} + b\,B_{(g)} ightleftharpoons c\,C_{(g)} + d\,D_{(g)}, if a + b > c + d , increasing pressure shifts the equilibrium to the **right** (forward direction). **Step 2: Applying this to the specific reaction** For the reaction 2\,	ext{NO}_2\,(g) ightleftharpoons 	ext{N}_2	ext{O}_4\,(g), the moles of gas on each side are: \Delta n_g = n_{	ext{products}} - n_{	ext{reactants}} = 1 - 2 = -1 < 0. Since \Delta n_g < 0 , increasing pressure favours the **forward reaction**, producing more 	ext{N}_2	ext{O}_4 . **Step 3: Evaluating the statement** The claim that *"increasing pressure always favours the forward reaction"* is **not universally correct**. It is only correct w

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B. Increasing temperature shifts equilibrium towards endothermic direction

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C. Removing products shifts equilibrium to the right

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D. The equilibrium constant changes with temperature

Question 8

The standard reduction potential E° for Zn²⁺/Zn is -0.76 V and for Cu²⁺/Cu is +0.34 V. The cell potential for Zn-Cu cell is:

Accepted Answer

Correct answer: A. 1.10 V — **Step 1: Identify the cathode and anode** In the Zn–Cu cell, copper undergoes reduction and zinc undergoes oxidation, so: - **Cathode (reduction):** 	ext{Cu}^{2+} + 2e^- 	o 	ext{Cu} , with E^\circ = +0.34\ 	ext{V} - **Anode (oxidation):** 	ext{Zn} 	o 	ext{Zn}^{2+} + 2e^- , with E^\circ = -0.76\ 	ext{V} **Step 2: Apply the cell potential formula** \begin{aligned} E^\circ_{	ext{cell}} &= E^\circ_{	ext{cathode}} - E^\circ_{	ext{anode}} \ &= 0.34\ 	ext{V} - (-0.76\ 	ext{V}) \ &= 0.34\ 	ext{V} + 0.76\ 	ext{V} \end{aligned} **Result:** \boxed{E^\circ_{	ext{cell}} = 1.10\ 	ext{V}}

Answer

B. 0.42 V

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C. 1.50 V

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D. -1.10 V

JEE Chemistry - MCQ Practice Questions