Arrhenius Theory

Answer: B. Eₐ is the activation energy — the minimum energy that colliding reactant molecules must possess for the reaction to proceed (NCERT Class 12 Chemistry Chapter 3, page 80).

Why A is wrong: A is wrong because the energy released during product formation relates to ΔH (enthalpy of reaction), not Eₐ. Activation energy is the barrier height above reactant energy, not the net energy change.

Why C is wrong: C is wrong because ΔH is the difference between product and reactant enthalpies. Eₐ is the height of the energy barrier — a reaction can have a small ΔH but a large Eₐ, or vice versa.

Why D is wrong: D is wrong because the average kinetic energy of molecules is (3/2)RT per mole (from kinetic theory). Eₐ is a reaction-specific constant, not a temperature-dependent average.

Answer: C. Since k = Ae^(−Eₐ/RT) and the exponential is dimensionless, A must carry the same units as k (NCERT Class 12 Chemistry Chapter 3, page 80). For a first-order reaction, A has units s⁻¹; for a second-order reaction, L mol⁻¹ s⁻¹.

Why A is wrong: A is wrong because if A were dimensionless, the equation k = A × (dimensionless) would make k dimensionless. Rate constants have definite units depending on reaction order, so A cannot be unitless.

Why B is wrong: B is wrong because J/mol are units of energy. A represents the frequency of effective collisions, not an energy quantity. The energy term Eₐ/(RT) in the exponent is dimensionless by cancellation.

Why D is wrong: D is wrong because kelvin is a temperature unit. A has the same dimensions as k (e.g., s⁻¹ for first-order), which has nothing to do with temperature units.

Answer: B. The doubling rule follows from the exponential temperature dependence in k = Ae^(−Eₐ/RT). For typical Eₐ values (50–100 kJ/mol), a 10 K rise near room temperature approximately doubles the Boltzmann factor (NCERT Class 12 Chemistry Chapter 3, page 80).

Why A is wrong: A is wrong because Le Chatelier's principle describes the shift of equilibrium position under stress. It does not predict the quantitative change in rate constant with temperature.

Why C is wrong: C is wrong because Hess's law states enthalpy change is independent of the reaction path. It is a thermodynamic principle about ΔH, not a kinetic statement about rate-constant variation.

Why D is wrong: D is wrong because the law of mass action gives the form of the equilibrium expression. It does not contain the exponential temperature dependence that explains rate doubling.

Answer: A. A is correct. From ln(k₂/k₁) = (Eₐ/R)(1/T₁ − 1/T₂): ln 2 = 0.693, and 1/300 − 1/310 = 10/93000 = 1.075 × 10⁻⁴ K⁻¹. So Eₐ = 8.314 × 0.693 / 1.075 × 10⁻⁴ = 5.36 × 10⁴ J/mol = 53.6 kJ/mol. This is the size of activation energy behind the familiar 'rate doubles for 10 K' rule.

Why B is wrong: B is wrong because 23.3 kJ/mol comes from using log₁₀ 2 = 0.301 in the natural-log equation. The log₁₀ form needs a factor of 2.303: log(k₂/k₁) = (Eₐ/2.303R)(1/T₁ − 1/T₂).

Why C is wrong: C is wrong because 6.4 kJ/mol comes from leaving R out: 0.693 / 1.075 × 10⁻⁴ = 6.4 × 10³. That quotient is Eₐ/R, in kelvin; R has to multiply back in.

Why D is wrong: D is wrong because 123 kJ/mol comes from putting a 2.303 into the natural-log form, which multiplies the right answer by 2.303. The 2.303 belongs only with log₁₀.

Answer: A. When T₂ > T₁, the correct bracket is (1/T₁ − 1/T₂), which is positive. Writing it as (1/T₂ − 1/T₁) gives a negative bracket, making the right side negative. Since Eₐ is always positive, this forces ln(k₂/k₁) to appear negative — a sign error from swapping T₁ and T₂ (related to mistake mistake: arrhenius t subtraction).

Why B is wrong: B is wrong because forgetting °C to K conversion gives incorrect temperature values, producing a wrong numerical answer. However, both temperatures would still yield a positive (1/T₁ − 1/T₂) as long as T₂ > T₁ in whatever scale is used (°C values are both positive above 0°C), so the sign would not flip.

Why C is wrong: C is wrong because using R in J mol⁻¹ K⁻¹ versus kJ mol⁻¹ K⁻¹ changes the magnitude of Eₐ by a factor of 1000, but does not flip the sign. A unit mismatch gives an Eₐ that is 1000× too large or too small, not negative.

Why D is wrong: D is wrong because using log₁₀ instead of ln scales the left side by a factor of 2.303 but does not change its sign. ln(k₂/k₁) and log₁₀(k₂/k₁) are both positive when k₂ > k₁.

Answer: B. B is correct. The ratio of the rate constants is (2.0 × 10⁻²)/(1.0 × 10⁻³) = 20, so ln(k₂/k₁) = ln 20 = 3.0. The temperature bracket is 1/500 − 1/700 = 200/350000 = 5.714 × 10⁻⁴ K⁻¹. Then Eₐ = R ln(k₂/k₁)/(1/T₁ − 1/T₂) = 8.314 × 3.0 / 5.714 × 10⁻⁴ = 4.36 × 10⁴ J/mol = 43.6 kJ/mol.

Why A is wrong: A is wrong because 10.1 kJ/mol comes from taking the ratio as 2 rather than 20 — losing a power of ten between 10⁻² and 10⁻³ — and so using ln 2 = 0.693 instead of ln 20.

Why C is wrong: C is wrong because 18.9 kJ/mol comes from using log₁₀ 20 = 1.30 in the natural-log equation without its factor of 2.303.

Why D is wrong: D is wrong because 100 kJ/mol comes from multiplying the natural-log result by 2.303. That factor converts log₁₀ to ln; applying it to an equation already in ln counts the conversion twice.

Answer: A. Eₐ/R = 83140/8.314 = 10000 K. (1/300 − 1/310) = 10/93000 = 1.075 × 10⁻⁴ K⁻¹. ln(k₂/k₁) = 10000 × 1.075 × 10⁻⁴ = 1.075. k₂/k₁ = e^1.075 ≈ 2.93. k₂ = 1.0 × 10⁻³ × 2.93 ≈ 2.9 × 10⁻³ s⁻¹.

Why B is wrong: B is wrong because 2.0 × 10⁻³ s⁻¹ gives k₂/k₁ = 2, implying ln(k₂/k₁) = 0.693. The actual ln(k₂/k₁) = 1.075, which gives a ratio of ~2.93, not 2. This error comes from assuming the rate exactly doubles per 10 K regardless of Eₐ.

Why C is wrong: C is wrong because 1.1 × 10⁻³ s⁻¹ barely changes from the original k. This would correspond to Eₐ ≈ 8 kJ/mol — far too low. A high Eₐ of 83.14 kJ/mol causes a nearly 3-fold increase over 10 K.

Why D is wrong: D is wrong because 5.9 × 10⁻³ s⁻¹ gives k₂/k₁ = 5.9, implying ln(k₂/k₁) ≈ 1.775. This would require doubling the exponent — likely from using (1/T₁ − 1/T₂) = 2.15 × 10⁻⁴ (perhaps subtracting reciprocals incorrectly or using wrong temperature values).

Answer: D. ln(10k₁/k₁) = ln 10 = 2.303. Eₐ/R = 60000/8.314 = 7215 K. 1/T₁ − 1/T₂ = 2.303/7215 = 3.192 × 10⁻⁴ K⁻¹. 1/T₂ = 1/300 − 3.192 × 10⁻⁴ = 3.333 × 10⁻³ − 0.3192 × 10⁻³ = 3.014 × 10⁻³. T₂ = 332 K.

Why A is wrong: A is wrong because 310 K gives (1/300 − 1/310) = 1.075 × 10⁻⁴, and ln(k₂/k₁) = 7215 × 1.075 × 10⁻⁴ ≈ 0.776, so k₂/k₁ ≈ 2.2 — far short of the required 10-fold increase. A 10 K rise is insufficient for this Eₐ.

Why B is wrong: B is wrong because 600 K gives (1/300 − 1/600) = 1.667 × 10⁻³, and ln(k₂/k₁) = 7215 × 1.667 × 10⁻³ ≈ 12.03, so k₂/k₁ ≈ 167,000. This enormous ratio shows 600 K massively overshoots; likely the student confused T₂ = 2T₁ with 'double the rate.'

Why C is wrong: C is wrong because 360 K gives (1/300 − 1/360) = 5.556 × 10⁻⁴, and ln(k₂/k₁) = 7215 × 5.556 × 10⁻⁴ ≈ 4.01, so k₂/k₁ ≈ 55 — far exceeding the required 10-fold. This overshoots the target temperature.