Physical Chemistry 2 Quiz: Arrhenius Equation
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Arrhenius EquationQuestion 1 of 13

An Arrhenius plot of ln⁡k\ln k vs 1/T1/T yields a straight line with slope −8450 K-8450 \, \text{K} and y-intercept 28.428.4. At what temperature does the rate constant equal the pre-exponential factor?

At infinite temperature where entropic effects dominate completely
At zero temperature where quantum tunneling becomes significant
This condition cannot be achieved at any finite positive temperature
At approximately 8450 K where thermal energy overcomes activation barrier
At room temperature where classical transition state theory applies
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Physical Chemistry 2 Quiz

Physical Chemistry 2 Quiz: Arrhenius Equation

Practice Arrhenius Equation in Physical Chemistry 2 with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Arrhenius Equation, giving you a quick way to practice the rules, question types, and explanations that matter most for Physical Chemistry 2.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

An Arrhenius plot of ln⁡k\ln k vs 1/T1/T yields a straight line with slope −8450 K-8450 \, \text{K} and y-intercept 28.428.4. At what temperature does the rate constant equal the pre-exponential factor?

  1. At infinite temperature where entropic effects dominate completely
  2. At zero temperature where quantum tunneling becomes significant
  3. This condition cannot be achieved at any finite positive temperature (correct answer)
  4. At approximately 8450 K where thermal energy overcomes activation barrier
  5. At room temperature where classical transition state theory applies
Explanation: When you encounter an Arrhenius plot question, you're dealing with the fundamental relationship ln⁡k=ln⁡A−EaRT\ln k = \ln A - \frac{E_a}{RT}, where kk is the rate constant, AA is the pre-exponential factor, EaE_a is activation energy, RR is the gas constant, and TT is temperature. The question asks when k=Ak = A. Setting these equal: A=Ae−Ea/RTA = A e^{-E_a/RT}. Dividing both sides by AA gives 1=e−Ea/RT1 = e^{-E_a/RT}. Taking the natural logarithm: 0=−EaRT0 = -\frac{E_a}{RT}. For this equation to be satisfied, either Ea=0E_a = 0 (no activation barrier) or T=∞T = \infty (infinite temperature). From your slope of −8450 K-8450 \, \text{K}, the activation energy is Ea=8450R≈70.3 kJ/molE_a = 8450R \approx 70.3 \, \text{kJ/mol}, which is definitely not zero. Since infinite temperature is physically impossible, the condition k=Ak = A cannot be achieved at any finite positive temperature, making answer C correct. Answer A is wrong because infinite temperature is unattainable, not just a matter of entropy dominating. Answer B misunderstands the physics—at zero temperature, reaction rates approach zero, and quantum tunneling doesn't make k=Ak = A. Answer D incorrectly suggests that at T=8450 KT = 8450 \, \text{K}, this condition is met, but this temperature has no special significance for the k=Ak = A condition. Remember: when analyzing Arrhenius behavior, kk only approaches AA as T→∞T \to \infty, which is why AA represents the theoretical maximum rate constant.

Question 2

A complex reaction mechanism involves a temperature-dependent equilibrium constant Keq=e−ΔH/(RT)+ΔS/RK_{eq} = e^{-\Delta H/(RT) + \Delta S/R} where ΔH=−25 kJ/mol\Delta H = -25 \, \text{kJ/mol} and ΔS=−80 J/(mol\cdotpK)\Delta S = -80 \, \text{J/(mol·K)}, followed by a rate-determining step with Ea=65 kJ/molE_a = 65 \, \text{kJ/mol}. What is the overall apparent activation energy at 350 K?

  1. 90 kJ/mol, indicating strong unfavorable entropic contributions to kinetics
  2. 40 kJ/mol, indicating favorable pre-equilibrium compensating activation barrier (correct answer)
  3. 65 kJ/mol, indicating negligible pre-equilibrium effects on kinetics
  4. 115 kJ/mol, indicating severe unfavorable thermodynamic and kinetic barriers
  5. 15 kJ/mol, indicating highly favorable pre-equilibrium dominating kinetics
Explanation: When you encounter a complex reaction mechanism with a pre-equilibrium followed by a rate-determining step, you need to consider how the temperature-dependent equilibrium constant affects the overall kinetics. The apparent activation energy combines contributions from both the equilibrium and the rate-determining step. For a mechanism with pre-equilibrium, the apparent activation energy is: Eapp=Ea−ΔHE_{app} = E_a - \Delta H, where EaE_a is the activation energy of the rate-determining step and ΔH\Delta H is the enthalpy change of the pre-equilibrium. Here, Ea=65E_a = 65 kJ/mol and ΔH=−25\Delta H = -25 kJ/mol, so: Eapp=65−(−25)=65+25=40E_{app} = 65 - (-25) = 65 + 25 = 40 kJ/mol The negative ΔH\Delta H means the pre-equilibrium is exothermic, which actually helps overcome the activation barrier of the subsequent step. Choice A (90 kJ/mol) incorrectly adds the magnitudes and misinterprets the entropic effect on activation energy. Choice C (65 kJ/mol) ignores the pre-equilibrium contribution entirely, assuming only the rate-determining step matters. Choice D (115 kJ/mol) incorrectly adds all energy terms as barriers, failing to recognize that the exothermic pre-equilibrium reduces the apparent activation energy. Choice B correctly calculates 40 kJ/mol and properly identifies that the favorable (exothermic) pre-equilibrium compensates for part of the activation barrier. Remember: in pre-equilibrium mechanisms, an exothermic first step (negative ΔH\Delta H) effectively reduces the apparent activation energy, while an endothermic first step increases it.

Question 3

A Marcus-type reaction shows temperature-dependent reorganization energy λ(T)=λ0+βT\lambda(T) = \lambda_0 + \beta T where λ0=0.8 eV\lambda_0 = 0.8 \, \text{eV}, β=2×10−4 eV/K\beta = 2 \times 10^{-4} \, \text{eV/K}, and the activation energy is Ea=λ(T)/4E_a = \lambda(T)/4. What is the apparent pre-exponential factor if the true collision frequency is 1011 s−110^{11} \, \text{s}^{-1} and there's a temperature-independent transmission coefficient of 0.6?

  1. 6.0×1010 s−16.0 \times 10^{10} \, \text{s}^{-1} with moderate electronic coupling efficiency (correct answer)
  2. 1.2×1011 s−11.2 \times 10^{11} \, \text{s}^{-1} with good electronic coupling efficiency
  3. 3.6×1010 s−13.6 \times 10^{10} \, \text{s}^{-1} with reduced electronic coupling efficiency
  4. 8.4×1010 s−18.4 \times 10^{10} \, \text{s}^{-1} with high electronic coupling efficiency
  5. 4.8×1010 s−14.8 \times 10^{10} \, \text{s}^{-1} with moderate electronic coupling efficiency
Explanation: When you encounter Marcus-type electron transfer reactions with temperature-dependent reorganization energy, you need to understand how the apparent pre-exponential factor differs from the true collision frequency due to electronic coupling effects. The apparent pre-exponential factor is determined by multiplying the true collision frequency by the transmission coefficient: Aapp=κ×ZA_{app} = \kappa \times Z, where κ\kappa is the transmission coefficient and ZZ is the collision frequency. Given that the transmission coefficient is 0.6 (temperature-independent) and the true collision frequency is 1011 s−110^{11} \, \text{s}^{-1}: Aapp=0.6×1011 s−1=6.0×1010 s−1A_{app} = 0.6 \times 10^{11} \, \text{s}^{-1} = 6.0 \times 10^{10} \, \text{s}^{-1} The transmission coefficient of 0.6 indicates moderate electronic coupling efficiency - not too high (which would approach 1.0) but reasonable for electron transfer. Answer B (1.2×10111.2 \times 10^{11}) incorrectly doubles the collision frequency, suggesting a misunderstanding of how transmission coefficients work. Answer C (3.6×10103.6 \times 10^{10}) uses an incorrect transmission coefficient of 0.36, while Answer D (8.4×10108.4 \times 10^{10}) uses 0.84, both missing the given value of 0.6. The temperature-dependent reorganization energy and activation energy information are red herrings here - they don't affect the pre-exponential factor calculation, only the exponential term in the rate expression. Study tip: In Marcus theory problems, distinguish between parameters that affect the pre-exponential factor (collision frequency, transmission coefficient) versus those affecting the activation barrier (reorganization energy, driving force). The pre-exponential factor depends only on collision dynamics and electronic coupling strength.

Question 4

A catalyzed reaction pathway reduces the activation energy from 120 kJ/mol to 75 kJ/mol at 350 K. If the uncatalyzed reaction has a half-life of 2.5 hours, what will be the approximate half-life of the catalyzed reaction?

  1. 18 minutes with the catalyst providing moderate rate enhancement
  2. 42 minutes with the catalyst providing significant rate enhancement
  3. 6.8 minutes with the catalyst providing substantial rate enhancement (correct answer)
  4. 1.2 minutes with the catalyst providing dramatic rate enhancement
  5. 35 seconds with the catalyst providing extreme rate enhancement
Explanation: When you encounter activation energy problems involving catalysts, you're dealing with the Arrhenius equation and how dramatically small energy changes affect reaction rates. The key relationship is that reaction rate constants follow k=Ae−Ea/RTk = Ae^{-E_a/RT}, where lowering activation energy exponentially increases the rate. To solve this, you need to find the ratio of rate constants. Since kcat/kuncat=e−(Ea,cat−Ea,uncat)/RTk_{cat}/k_{uncat} = e^{-(E_{a,cat} - E_{a,uncat})/RT}, you can calculate: kcat/kuncat=e−(75,000−120,000)/(8.314×350)=e45,000/2,910=e15.46≈5.3×106k_{cat}/k_{uncat} = e^{-(75,000 - 120,000)/(8.314 \times 350)} = e^{45,000/2,910} = e^{15.46} ≈ 5.3 \times 10^6 Since half-life is inversely proportional to the rate constant for first-order reactions (t1/2=ln⁡(2)/kt_{1/2} = \ln(2)/k), the catalyzed half-life becomes: t1/2,cat=2.5 hours÷5.3×106≈0.11 hours=6.6 minutest_{1/2,cat} = 2.5 \text{ hours} ÷ 5.3 \times 10^6 ≈ 0.11 \text{ hours} = 6.6 \text{ minutes} Answer C (6.8 minutes) is correct and represents substantial rate enhancement, which matches the massive rate constant increase. Answer A (18 minutes) underestimates the exponential effect of the 45 kJ/mol energy reduction. Answer B (42 minutes) suggests only modest enhancement, ignoring the exponential nature of the Arrhenius equation. Answer D (1.2 minutes) overcalculates the effect, possibly from computational errors in the exponential. Remember: activation energy changes have exponential effects on rates. A 45 kJ/mol reduction at moderate temperatures produces million-fold rate increases, not just small improvements. Always double-check your exponential calculations in Arrhenius problems.

Question 5

A reaction has an activation energy of 85.2 kJ/mol and a rate constant of 2.4×10−3 s−12.4 \times 10^{-3} \, \text{s}^{-1} at 298 K. When the temperature is increased to 318 K, the rate constant becomes 1.8×10−2 s−11.8 \times 10^{-2} \, \text{s}^{-1}. What is the pre-exponential factor AA for this reaction?

  1. 4.2×1011 s−14.2 \times 10^{11} \, \text{s}^{-1}
  2. 7.8×1010 s−17.8 \times 10^{10} \, \text{s}^{-1}
  3. 2.1×1012 s−12.1 \times 10^{12} \, \text{s}^{-1} (correct answer)
  4. 1.5×1011 s−11.5 \times 10^{11} \, \text{s}^{-1}
  5. 9.3×1011 s−19.3 \times 10^{11} \, \text{s}^{-1}
Explanation: When you encounter temperature-dependent rate constant problems, you're dealing with the Arrhenius equation: k=Ae−Ea/RTk = Ae^{-E_a/RT}. The pre-exponential factor AA represents the maximum possible rate constant when all collisions have sufficient energy. To find AA, rearrange the Arrhenius equation: A=keEa/RTA = ke^{E_a/RT}. You can use either temperature condition since AA is constant. Using the data at 298 K: A=(2.4×10−3)×e85,200/(8.314×298)A = (2.4 \times 10^{-3}) \times e^{85,200/(8.314 \times 298)} First, calculate the exponent: 85,2008.314×298=34.37\frac{85,200}{8.314 \times 298} = 34.37 Then: A=2.4×10−3×e34.37=2.4×10−3×8.7×1014=2.1×1012 s−1A = 2.4 \times 10^{-3} \times e^{34.37} = 2.4 \times 10^{-3} \times 8.7 \times 10^{14} = 2.1 \times 10^{12} \, \text{s}^{-1} This confirms answer C is correct. Answer A (4.2×10114.2 \times 10^{11}) is too small by about a factor of 5, likely from calculation errors in the exponential term. Answer B (7.8×10107.8 \times 10^{10}) is off by roughly an order of magnitude, suggesting confusion with unit conversions or mathematical mistakes. Answer D (1.5×10111.5 \times 10^{11}) is also too small, possibly from errors in handling the large exponential value. Study tip: Always double-check your exponential calculations in Arrhenius problems—small errors in the exponent create huge differences in the final answer. Also remember that AA should be much larger than your rate constants since it represents the theoretical maximum rate.

Question 6

A reaction exhibits non-Arrhenius behavior where the activation energy appears to decrease linearly with temperature according to Ea(T)=E0−αTE_a(T) = E_0 - \alpha T where E0=150 kJ/molE_0 = 150 \, \text{kJ/mol} and α=0.12 kJ/(mol\cdotpK)\alpha = 0.12 \, \text{kJ/(mol·K)}. What is the apparent activation energy at 400 K?

  1. 102 kJ/mol, indicating significant temperature-dependent barrier modification (correct answer)
  2. 198 kJ/mol, indicating moderate temperature-dependent barrier enhancement
  3. 150 kJ/mol, indicating no temperature dependence in this regime
  4. 78 kJ/mol, indicating substantial temperature-dependent barrier reduction
  5. 126 kJ/mol, indicating moderate temperature-dependent barrier modification
Explanation: When you encounter non-Arrhenius behavior with temperature-dependent activation energies, you're dealing with complex reaction mechanisms where the energy barrier itself changes with temperature. This often occurs in solution reactions, enzymatic processes, or reactions involving multiple pathways. To find the apparent activation energy at 400 K, substitute directly into the given equation: Ea(400)=E0−αT=150−(0.12)(400)=150−48=102 kJ/molE_a(400) = E_0 - \alpha T = 150 - (0.12)(400) = 150 - 48 = 102 \text{ kJ/mol} Answer A (102 kJ/mol) is correct. The significant decrease from the base value of 150 kJ/mol to 102 kJ/mol represents substantial temperature-dependent barrier modification, where higher temperatures fundamentally alter the reaction pathway or mechanism. Answer B (198 kJ/mol) incorrectly adds the temperature correction instead of subtracting it, misunderstanding the negative sign in the linear relationship. Answer C (150 kJ/mol) represents the base activation energy E0E_0 but ignores the temperature dependence entirely—this would only be correct at 0 K. Answer D (78 kJ/mol) appears to result from calculation errors, possibly doubling the temperature correction. The key insight is that the 48 kJ/mol reduction represents "significant" rather than "substantial" modification because it's about a 32% decrease from the baseline—meaningful but not extreme. Study tip: For non-Arrhenius kinetics, always substitute the given temperature into the provided equation first, then evaluate whether the change represents significant, moderate, or substantial modification based on the percentage change from the baseline value.

Question 7

An enzyme-catalyzed reaction follows Arrhenius behavior up to 45°C, above which the rate constant decreases with increasing temperature. The activation energy below 45°C is 25 kJ/mol. If the rate constant at 45°C is 8.5×103 s−18.5 \times 10^{3} \text{ s}^{-1} and at 55°C it is 6.2×103 s−16.2 \times 10^{3} \text{ s}^{-1}, what is the apparent activation energy for the enzyme denaturation process?

  1. −15.2 kJ/mol-15.2 \text{ kJ/mol}
  2. −28.7 kJ/mol-28.7 \text{ kJ/mol} (correct answer)
  3. −42.1 kJ/mol-42.1 \text{ kJ/mol}
  4. −55.6 kJ/mol-55.6 \text{ kJ/mol}
Explanation: Above 45°C, enzyme denaturation dominates, giving negative apparent activation energy. Using ln⁡(k2k1)=EaR(1T1−1T2)\ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right) with k1=8.5×103k_1 = 8.5 \times 10^3 at T1=318T_1 = 318 K and k2=6.2×103k_2 = 6.2 \times 10^3 at T2=328T_2 = 328 K: ln⁡(6.2×1038.5×103)=Ea8.314(1318−1328)\ln\left(\frac{6.2 \times 10^3}{8.5 \times 10^3}\right) = \frac{E_a}{8.314}\left(\frac{1}{318} - \frac{1}{328}\right). This gives −0.314=Ea8.314×9.58×10−5-0.314 = \frac{E_a}{8.314} \times 9.58 \times 10^{-5}, so Ea=−28,700 J/mol=−28.7 kJ/molE_a = -28,700 \text{ J/mol} = -28.7 \text{ kJ/mol}. Choice A uses incorrect temperature conversion. Choice C doubles the correct value. Choice D uses the original activation energy incorrectly.

Question 8

A reaction exhibits non-Arrhenius behavior where the effective activation energy varies with temperature according to Eeff(T)=E0−αTE_{eff}(T) = E_0 - \alpha T, where E0=80E_0 = 80 kJ/mol and α=0.05\alpha = 0.05 kJ/(mol·K). If the pre-exponential factor is 2.0×1012 s−12.0 \times 10^{12} \text{ s}^{-1}, what is the rate constant at 400 K?

  1. 1.8×102 s−11.8 \times 10^{2} \text{ s}^{-1}
  2. 9.1×101 s−19.1 \times 10^{1} \text{ s}^{-1}
  3. 7.2×104 s−17.2 \times 10^{4} \text{ s}^{-1}
  4. 3.6×103 s−13.6 \times 10^{3} \text{ s}^{-1} (correct answer)
Explanation: When you encounter non-Arrhenius behavior, you're dealing with reactions where the activation energy itself changes with temperature, rather than following the standard Arrhenius equation. This often occurs in complex reactions or those involving multiple pathways. To find the rate constant, you'll use the modified Arrhenius equation: k=Aexp⁡(−Eeff(T)RT)k = A \exp\left(-\frac{E_{eff}(T)}{RT}\right). First, calculate the effective activation energy at 400 K: Eeff(400)=80−0.05×400=80−20=60 kJ/molE_{eff}(400) = 80 - 0.05 \times 400 = 80 - 20 = 60 \text{ kJ/mol}. Convert to J/mol: 60,000 J/mol. Now substitute into the equation: k=2.0×1012exp⁡(−60,0008.314×400)=2.0×1012exp⁡(−18.05)=2.0×1012×1.8×10−8=3.6×103 s−1k = 2.0 \times 10^{12} \exp\left(-\frac{60,000}{8.314 \times 400}\right) = 2.0 \times 10^{12} \exp(-18.05) = 2.0 \times 10^{12} \times 1.8 \times 10^{-8} = 3.6 \times 10^{3} \text{ s}^{-1} Answer A (1.8×1021.8 \times 10^{2}) likely comes from using the original E0=80E_0 = 80 kJ/mol without accounting for the temperature dependence. Answer B (9.1×1019.1 \times 10^{1}) may result from calculation errors in the exponential term. Answer C (7.2×1047.2 \times 10^{4}) could arise from incorrect unit conversions or sign errors in the temperature correction term. The key strategy here is recognizing that non-Arrhenius behavior requires you to first calculate the temperature-dependent activation energy before applying the Arrhenius equation. Always check your units carefully—activation energies are typically given in kJ/mol but R uses J/(mol·K).

Question 9

Two parallel reactions have the same pre-exponential factor but different activation energies: Reaction 1 has Ea1=45E_{a1} = 45 kJ/mol and Reaction 2 has Ea2=65E_{a2} = 65 kJ/mol. At what temperature will the rate constant of Reaction 1 be exactly 100 times larger than that of Reaction 2?

  1. 289 K289 \text{ K}
  2. 334 K334 \text{ K}
  3. 578 K578 \text{ K} (correct answer)
  4. 412 K412 \text{ K}
Explanation: Since both reactions have the same A, we have k1k2=e−Ea1/RTe−Ea2/RT=e(Ea2−Ea1)/RT=100\frac{k_1}{k_2} = \frac{e^{-E_{a1}/RT}}{e^{-E_{a2}/RT}} = e^{(E_{a2}-E_{a1})/RT} = 100. Taking ln: (65000−45000)8.314T=ln⁡(100)=4.605\frac{(65000-45000)}{8.314T} = \ln(100) = 4.605. Solving: T=200008.314×4.605=578 KT = \frac{20000}{8.314 \times 4.605} = 578 \text{ K}. Choice A uses ln⁡(10)\ln(10) instead of ln⁡(100)\ln(100). Choice B incorrectly subtracts activation energies in the wrong order. Choice D results from using activation energies in kJ instead of J in the calculation.

Question 10

A reaction has an activation energy of 85 kJ/mol at 298 K. If the rate constant increases by a factor of 15 when the temperature is raised to 318 K, what is the pre-exponential factor A if the rate constant at 298 K is 2.5×10−4 s−12.5 \times 10^{-4} \text{ s}^{-1}?

  1. 1.2×1011 s−11.2 \times 10^{11} \text{ s}^{-1} (correct answer)
  2. 3.8×1010 s−13.8 \times 10^{10} \text{ s}^{-1}
  3. 6.7×109 s−16.7 \times 10^{9} \text{ s}^{-1}
  4. 2.1×1012 s−12.1 \times 10^{12} \text{ s}^{-1}
Explanation: Using the Arrhenius equation k=Ae−Ea/RTk = A e^{-E_a/RT}, we can solve for A using A=keEa/RTA = k e^{E_a/RT}. At 298 K: A=(2.5×10−4)×e(85000)/(8.314×298)=(2.5×10−4)×e34.3=(2.5×10−4)×(4.8×1014)=1.2×1011 s−1A = (2.5 \times 10^{-4}) \times e^{(85000)/(8.314 \times 298)} = (2.5 \times 10^{-4}) \times e^{34.3} = (2.5 \times 10^{-4}) \times (4.8 \times 10^{14}) = 1.2 \times 10^{11} \text{ s}^{-1}. Choice B uses incorrect exponential calculation. Choice C incorrectly uses the rate constant at 318 K instead of 298 K. Choice D results from using activation energy in kJ instead of J.

Question 11

An Arrhenius plot (ln k vs. 1/T) for a reaction gives a straight line with slope = -8420 K and y-intercept = 28.5. If this reaction is carried out at 350 K in the presence of a catalyst that reduces the activation energy by 25%, what will be the new rate constant?

  1. 2.8×109 s−12.8 \times 10^{9} \text{ s}^{-1} (correct answer)
  2. 1.4×108 s−11.4 \times 10^{8} \text{ s}^{-1}
  3. 6.2×107 s−16.2 \times 10^{7} \text{ s}^{-1}
  4. 3.5×106 s−13.5 \times 10^{6} \text{ s}^{-1}
Explanation: From the Arrhenius plot: slope = −Ea/R=−8420-E_a/R = -8420, so Ea=8420×8.314=70.0E_a = 8420 \times 8.314 = 70.0 kJ/mol. The y-intercept gives ln⁡A=28.5\ln A = 28.5, so A=e28.5=3.5×1012A = e^{28.5} = 3.5 \times 10^{12}. With catalyst: Ea,cat=0.75×70.0=52.5E_{a,cat} = 0.75 \times 70.0 = 52.5 kJ/mol. At 350 K: kcat=(3.5×1012)×e−52500/(8.314×350)=(3.5×1012)×e−18.0=2.8×109 s−1k_{cat} = (3.5 \times 10^{12}) \times e^{-52500/(8.314 \times 350)} = (3.5 \times 10^{12}) \times e^{-18.0} = 2.8 \times 10^{9} \text{ s}^{-1}. Choice B uses the original activation energy. Choice C uses 50% reduction instead of 25%. Choice D incorrectly calculates the pre-exponential factor.

Question 12

A student measures rate constants for a reaction at different temperatures and creates an Arrhenius plot. The data points deviate from linearity at high temperatures, showing a smaller slope than expected. Which explanation is most consistent with this observation?

  1. The reaction mechanism changes to one with higher activation energy at elevated temperatures due to competing pathways.
  2. Thermal decomposition of reactants becomes significant, reducing the effective concentration and apparent rate constant.
  3. The reaction approaches the diffusion-controlled limit where molecular collisions become the rate-determining step. (correct answer)
  4. Experimental error due to increased evaporation of solvent at higher temperatures leads to concentration uncertainties.
Explanation: At very high temperatures, reactions can approach the diffusion-controlled limit where the rate is limited by how fast molecules can encounter each other, not by activation energy. This causes deviation from Arrhenius behavior with a smaller apparent activation energy (smaller slope). Choice A would increase the slope, not decrease it. Choice B would affect concentration but not fundamentally change the temperature dependence pattern. Choice D describes experimental error rather than a fundamental physical phenomenon that causes systematic deviation from Arrhenius behavior.

Question 13

For a reaction following Arrhenius behavior, the rate constant doubles every 15 K increase in temperature around 400 K. What is the apparent activation energy for this process?

  1. 38.2 kJ/mol38.2 \text{ kJ/mol}
  2. 52.6 kJ/mol52.6 \text{ kJ/mol} (correct answer)
  3. 67.1 kJ/mol67.1 \text{ kJ/mol}
  4. 74.8 kJ/mol74.8 \text{ kJ/mol}
Explanation: Using the two-temperature form of Arrhenius equation: ln⁡(k2k1)=EaR(1T1−1T2)\ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R}\left(\frac{1}{T_1} - \frac{1}{T_2}\right). Given k2/k1=2k_2/k_1 = 2, T1=400T_1 = 400 K, T2=415T_2 = 415 K: ln⁡(2)=Ea8.314(1400−1415)=Ea8.314×9.04×10−5\ln(2) = \frac{E_a}{8.314}\left(\frac{1}{400} - \frac{1}{415}\right) = \frac{E_a}{8.314} \times 9.04 \times 10^{-5}. Solving: Ea=0.693×8.3149.04×10−5=52,600 J/mol=52.6 kJ/molE_a = \frac{0.693 \times 8.314}{9.04 \times 10^{-5}} = 52,600 \text{ J/mol} = 52.6 \text{ kJ/mol}. Choice A uses ln⁡(1.5)\ln(1.5) instead of ln⁡(2)\ln(2). Choice C uses incorrect temperature difference calculation. Choice D uses 10 K instead of 15 K temperature increase.