Physical Chemistry 2 Quiz: Integrated Rate Laws And Half Life
20 questions · exam conditions
0:00
Integrated Rate Laws And Half LifeQuestion 1 of 20

Zero-order reaction, k=0.010k=0.010 M/s, [A]0=0.10[A]_0=0.10 M. What is t1/2t_{1/2}?

5.0 s
10 s
69 s
1000 s
← Back to quizzes

Physical Chemistry 2 Quiz

Physical Chemistry 2 Quiz: Integrated Rate Laws And Half Life

Practice Integrated Rate Laws And Half Life 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 Integrated Rate Laws And Half Life, giving you a quick way to practice the rules, question types, and explanations that matter most for Physical Chemistry 2.

How to use this quiz

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.

All questions

Question 1

Zero-order reaction, k=0.010k=0.010 M/s, [A]0=0.10[A]_0=0.10 M. What is t1/2t_{1/2}?

  1. 5.0 s (correct answer)
  2. 10 s
  3. 69 s
  4. 1000 s
Explanation: For a zero-order reaction, the half-life is [A]0 divided by 2k, so t1/2 = 0.10 M / (2 x 0.010 M/s) = 5.0 s. A common mistake is using the first-order formula 0.693/k, which gives about 69 s, but zero-order decay is linear and its half-life depends on the initial concentration.

Question 2

For a second-order reaction, [A]0=0.50[A]_0=0.50 M and t1/2=20t_{1/2}=20 s. [A][A] after 60 s is:

  1. 0.0625 M
  2. 0.25 M
  3. 0.125 M (correct answer)
  4. 0.50 M
Explanation: For a second-order reaction, 1/[A] = 1/[A]0 + kt. The half-life gives k = 1/(t1/2 [A]0) = 1/(20 x 0.50) = 0.1 M^-1 s^-1. At 60 s, 1/[A] = 2 + 6 = 8, so [A] = 0.125 M. The tempting 0.0625 M comes from incorrectly halving the concentration three times as if this were first order.

Question 3

A first-order plot of ln⁡[A]\ln[A] vs tt has slope −0.030 s−1-0.030\ \mathrm{s}^{-1}. What is the time for 75% consumption?

  1. 23 s
  2. 92 s
  3. 69 s
  4. 46 s (correct answer)
Explanation: The slope of a first-order ln[A] vs t plot is -k, so k = 0.030 s^-1. 75% consumption leaves 25% of the reactant, which is two half-lives. Since t1/2 = ln2 / k = 23 s, the time for two half-lives is 46 s. The tempting wrong answer is 23 s, but that is only one half-life, corresponding to 50% consumption.

Question 4

Compared to a first-order reaction with rate constant kk, one with rate constant $4k$ has a half-life that is:

  1. One-fourth as long (correct answer)
  2. One-half as long
  3. Twice as long
  4. Four times as long
Explanation: For a first-order reaction, half-life = ln 2 / k. Replacing k with 4k gives ln 2 / (4k), which is exactly one quarter of the original. The tempting wrong answer is one-half as long; that would require half-life to depend on the square root of k, but it doesn't.

Question 5

For a zero-order reaction, the second half-life is what fraction of the first?

  1. One-fourth as long
  2. One-half as long (correct answer)
  3. The same length
  4. Twice as long
Explanation: For a zero-order reaction, half-life depends on the current concentration: t1/2 = [A]/(2k). After the first half-life, only half the initial concentration remains, so the second half-life is half the first. The tempting wrong answer is that half-lives are always constant, which applies to first-order, not zero-order reactions.

Question 6

A first-order reaction has a rate constant of 1.2×10−3 s−11.2 \times 10^{-3} \text{ s}^{-1} at 298 K. If the initial concentration is 0.50 M and the reaction proceeds until only 15% of the original reactant remains, what is the total time elapsed?

  1. 1580 s (correct answer)
  2. 2420 s
  3. 1890 s
  4. 3150 s
Explanation: For a first-order reaction, ln⁡([A]t/[A]0)=−kt\ln([A]_t/[A]_0) = -kt. If 15% remains, then [A]t=0.15×0.50=0.075[A]_t = 0.15 \times 0.50 = 0.075 M. So ln⁡(0.075/0.50)=−1.2×10−3×t\ln(0.075/0.50) = -1.2 \times 10^{-3} \times t. This gives ln⁡(0.15)=−1.897=−1.2×10−3×t\ln(0.15) = -1.897 = -1.2 \times 10^{-3} \times t, so t=1.897/(1.2×10−3)=1580t = 1.897/(1.2 \times 10^{-3}) = 1580 s. Choice B uses ln⁡(0.85)\ln(0.85) instead of ln⁡(0.15)\ln(0.15). Choice C uses the wrong form of the integrated rate law. Choice D assumes second-order kinetics.

Question 7

A first-order reaction has a rate constant of 2.5×10−4 s−12.5 \times 10^{-4} \text{ s}^{-1} at 298 K. If the initial concentration is 0.80 M and you observe that the concentration has decreased to 0.20 M, what additional time is required for the concentration to reach 0.05 M?

  1. 2772 seconds
  2. 5544 seconds (correct answer)
  3. 8316 seconds
  4. 11088 seconds
  5. 13860 seconds
Explanation: When you encounter first-order reaction kinetics problems involving multiple time intervals, you need to use the integrated rate law strategically. The key insight is that you don't need to find the absolute time - you're looking for the additional time required. For first-order reactions, use the integrated rate law: ln⁡([A]t[A]0)=−kt\ln\left(\frac{[A]_t}{[A]_0}\right) = -kt First, let's find how long it took to go from 0.80 M to 0.20 M: ln⁡(0.200.80)=−(2.5×10−4)t1\ln\left(\frac{0.20}{0.80}\right) = -(2.5 \times 10^{-4})t_1 ln⁡(0.25)=−1.386=−(2.5×10−4)t1\ln(0.25) = -1.386 = -(2.5 \times 10^{-4})t_1 t1=5544 secondst_1 = 5544 \text{ seconds} Now find the total time to reach 0.05 M from the original 0.80 M: ln⁡(0.050.80)=−(2.5×10−4)ttotal\ln\left(\frac{0.05}{0.80}\right) = -(2.5 \times 10^{-4})t_{total} ln⁡(0.0625)=−2.773=−(2.5×10−4)ttotal\ln(0.0625) = -2.773 = -(2.5 \times 10^{-4})t_{total} ttotal=11088 secondst_{total} = 11088 \text{ seconds} The additional time needed is: 11088−5544=5544 seconds11088 - 5544 = 5544 \text{ seconds} Answer choice A (2772 seconds) represents half the correct time - perhaps from miscalculating one of the time intervals. Choice C (8316 seconds) might result from adding instead of subtracting incorrectly. Choice D (11088 seconds) is the total time from start to finish, not the additional time requested. Study tip: In multi-step kinetics problems, always identify exactly what time interval the question is asking for. Calculate total times first, then subtract to find additional time periods.

Question 8

A reaction exhibits the following concentration-time data that suggests zero-order kinetics initially, then transitions to first-order. If the zero-order rate constant is k0=0.5 M s−1k_0 = 0.5 \text{ M s}^{-1} and the transition occurs at [A] = 2.0 M, what is the first-order rate constant if the half-life from 2.0 M to 1.0 M is 150 seconds?

  1. 4.62×10−3 s−14.62 \times 10^{-3} \text{ s}^{-1} (correct answer)
  2. 6.93×10−3 s−16.93 \times 10^{-3} \text{ s}^{-1}
  3. 9.24×10−3 s−19.24 \times 10^{-3} \text{ s}^{-1}
  4. 1.39×10−2 s−11.39 \times 10^{-2} \text{ s}^{-1}
  5. 2.31×10−2 s−12.31 \times 10^{-2} \text{ s}^{-1}
Explanation: When you encounter reactions that exhibit changing kinetics, you're dealing with a mechanism where the rate-determining step changes as concentration decreases. This often occurs when substrate concentration drops below the saturation point of an enzyme or catalyst. For the transition from 2.0 M to 1.0 M following first-order kinetics, you can use the first-order integrated rate law. Since this represents exactly one half-life (from 2.0 M to 1.0 M in 150 seconds), you can apply the relationship t1/2=ln⁡(2)k1t_{1/2} = \frac{\ln(2)}{k_1}. Solving for the first-order rate constant: k1=ln⁡(2)t1/2=0.693150 s=4.62×10−3 s−1k_1 = \frac{\ln(2)}{t_{1/2}} = \frac{0.693}{150 \text{ s}} = 4.62 \times 10^{-3} \text{ s}^{-1} This matches answer A, which is correct. Answer B (6.93×10−3 s−16.93 \times 10^{-3} \text{ s}^{-1}) represents a common error where students use ln⁡(2)=0.693\ln(2) = 0.693 directly as the rate constant instead of dividing by the half-life time. Answer C (9.24×10−3 s−19.24 \times 10^{-3} \text{ s}^{-1}) likely results from incorrectly using the zero-order rate constant in calculations or misapplying the concentration change. Answer D (1.39×10−2 s−11.39 \times 10^{-2} \text{ s}^{-1}) appears to stem from using an incorrect time unit conversion or misunderstanding the half-life concept. Remember: when dealing with mixed-order kinetics, identify which kinetic regime applies to the specific concentration range you're analyzing. The zero-order information here is a distractor—focus only on the first-order portion for this calculation.

Question 9

A consecutive reaction A→k1B→k2CA \xrightarrow{k_1} B \xrightarrow{k_2} C has rate constants k1=0.1 s−1k_1 = 0.1 \text{ s}^{-1} and k2=0.3 s−1k_2 = 0.3 \text{ s}^{-1}. Starting with 2.0 M of A and no B or C, at what time does the concentration of intermediate B reach its maximum value?

  1. 3.66 seconds
  2. 5.49 seconds (correct answer)
  3. 7.32 seconds
  4. 10.99 seconds
  5. 14.65 seconds
Explanation: When you encounter consecutive reactions, you're dealing with an intermediate species that's both formed and consumed. The key insight is that intermediate B reaches maximum concentration when its rate of formation equals its rate of consumption. For the reaction sequence A→k1B→k2CA \xrightarrow{k_1} B \xrightarrow{k_2} C, the rate of change of B is: d[B]dt=k1[A]−k2[B]\frac{d[B]}{dt} = k_1[A] - k_2[B] At maximum [B], this rate equals zero: k1[A]−k2[B]=0k_1[A] - k_2[B] = 0 For first-order consecutive reactions, the time at which [B] reaches its maximum is given by: tmax=ln⁡(k2/k1)k2−k1t_{max} = \frac{\ln(k_2/k_1)}{k_2 - k_1} Substituting the given values: tmax=ln⁡(0.3/0.1)0.3−0.1=ln⁡(3)0.2=1.0990.2=5.49 secondst_{max} = \frac{\ln(0.3/0.1)}{0.3 - 0.1} = \frac{\ln(3)}{0.2} = \frac{1.099}{0.2} = 5.49 \text{ seconds} This confirms answer B is correct. Answer A (3.66 seconds) likely comes from incorrectly using 1k2−k1\frac{1}{k_2 - k_1} without the logarithmic term. Answer C (7.32 seconds) might result from switching the rate constants in the calculation. Answer D (10.99 seconds) appears to be the natural logarithm value (1.099) multiplied by 10 instead of divided by 0.2. Remember this formula for consecutive first-order reactions: the maximum intermediate concentration occurs at t=ln⁡(k2/k1)k2−k1t = \frac{\ln(k_2/k_1)}{k_2 - k_1}. This relationship is fundamental in understanding reaction kinetics and appears frequently in physical chemistry problems.

Question 10

A reaction mechanism involves a pre-equilibrium: A+B⇌CA + B \rightleftharpoons C (fast, KeqK_{eq}) followed by C→kDC \xrightarrow{k} D (slow). If the overall reaction appears to follow second-order kinetics in the product formation rate, and Keq=0.5 M−1K_{eq} = 0.5 \text{ M}^{-1}, what is the relationship between the observed rate constant kobsk_{obs} and the elementary rate constant kk?

  1. kobs=kk_{obs} = k
  2. kobs=0.5kk_{obs} = 0.5k
  3. kobs=2kk_{obs} = 2k
  4. kobs=kKeq=0.5kk_{obs} = kK_{eq} = 0.5k (correct answer)
  5. kobs=kKeq=2kk_{obs} = \frac{k}{K_{eq}} = 2k
Explanation: When you encounter reaction mechanisms with pre-equilibrium steps, you need to derive the overall rate law by combining the equilibrium expression with the rate-determining step. This tests your ability to connect thermodynamic equilibrium concepts with kinetics. Start with the slow step, which determines the overall rate: rate=k[C]\text{rate} = k[C]. Since C is an intermediate, you can't measure its concentration directly, so you must express [C] in terms of the reactants A and B using the pre-equilibrium assumption. For the fast pre-equilibrium A+B⇌CA + B \rightleftharpoons C, the equilibrium constant is: Keq=[C][A][B]=0.5 M−1K_{eq} = \frac{[C]}{[A][B]} = 0.5 \text{ M}^{-1} Rearranging: [C]=Keq[A][B]=0.5[A][B][C] = K_{eq}[A][B] = 0.5[A][B] Substituting into the rate expression: rate=k[C]=k⋅Keq[A][B]=k⋅0.5[A][B]\text{rate} = k[C] = k \cdot K_{eq}[A][B] = k \cdot 0.5[A][B] Since the overall reaction shows second-order kinetics (first order in both [A] and [B]), we can write: rate=kobs[A][B]\text{rate} = k_{obs}[A][B] Comparing these expressions: kobs=kKeq=0.5kk_{obs} = kK_{eq} = 0.5k Answer D is correct because it properly accounts for both the elementary rate constant and the equilibrium constant. Answer A ignores the pre-equilibrium effect entirely. Answer B gives the correct numerical value but lacks the conceptual relationship. Answer C incorrectly inverts the equilibrium constant. Study tip: In pre-equilibrium mechanisms, the observed rate constant always equals the product of the elementary rate constant and any relevant equilibrium constants from preceding fast steps.

Question 11

A photochemical reaction has a quantum yield of 0.3 and follows first-order kinetics with respect to the excited species formed. If the light intensity provides a constant rate of excitation of 1.2×1015 photons s−11.2 \times 10^{15} \text{ photons s}^{-1}, and the excited state has a lifetime of 50 ns before reaction or deactivation, what is the steady-state concentration of the excited species in a 1.0 L reaction vessel?

  1. 3.0×10−8 M3.0 \times 10^{-8} \text{ M}
  2. 6.0×10−8 M6.0 \times 10^{-8} \text{ M}
  3. 1.0×10−7 M1.0 \times 10^{-7} \text{ M} (correct answer)
  4. 3.0×10−7 M3.0 \times 10^{-7} \text{ M}
  5. 6.0×10−7 M6.0 \times 10^{-7} \text{ M}
Explanation: When you encounter photochemical kinetics problems, you're dealing with competing processes: excitation creating excited species and their removal through reaction or deactivation. The key insight is applying steady-state approximation to the excited species concentration. At steady state, the rate of excited species formation equals their rate of removal. The formation rate is simply the excitation rate: 1.2×1015 photons s−11.2 \times 10^{15} \text{ photons s}^{-1}. The removal follows first-order kinetics with rate constant k=1/τk = 1/\tau, where τ\tau is the lifetime (50 ns = 5.0×10−85.0 \times 10^{-8} s). Setting up the steady-state equation: Rate of formation=Rate of removal\text{Rate of formation} = \text{Rate of removal} 1.2×1015=k[A∗]=[A∗]5.0×10−81.2 \times 10^{15} = k[A^*] = \frac{[A^*]}{5.0 \times 10^{-8}} Solving for [A∗][A^*]: [A∗]=1.2×1015×5.0×10−8=6.0×107 molecules/s×s=6.0×107 molecules[A^*] = 1.2 \times 10^{15} \times 5.0 \times 10^{-8} = 6.0 \times 10^{7} \text{ molecules/s} \times \text{s} = 6.0 \times 10^{7} \text{ molecules} Converting to molarity using Avogadro's number and 1.0 L volume: [A∗]=6.0×1076.022×1023=1.0×10−7 M[A^*] = \frac{6.0 \times 10^{7}}{6.022 \times 10^{23}} = 1.0 \times 10^{-7} \text{ M} This confirms answer C is correct. A (3.0×10−83.0 \times 10^{-8} M) incorrectly uses twice the lifetime in calculations. B (6.0×10−86.0 \times 10^{-8} M) forgets the unit conversion from molecules to moles. D (3.0×10−73.0 \times 10^{-7} M) incorrectly incorporates the quantum yield, which affects product formation but not excited state concentration. Remember: quantum yield describes reaction efficiency, but steady-state excited species concentration depends only on excitation rate and total lifetime—not how productively that lifetime is used.

Question 12

In a chain reaction mechanism, the propagation steps are: R⋅+A→kpR′+BR\cdot + A \xrightarrow{k_p} R' + B and R′+A→kpR⋅+CR' + A \xrightarrow{k_p} R\cdot + C. The termination occurs by radical combination with rate constant ktk_t. If the initiation rate is Ri=1.0×10−8 M s−1R_i = 1.0 \times 10^{-8} \text{ M s}^{-1}, kp=100 M−1s−1k_p = 100 \text{ M}^{-1}\text{s}^{-1}, kt=108 M−1s−1k_t = 10^8 \text{ M}^{-1}\text{s}^{-1}, and [A]=0.5 M[A] = 0.5 \text{ M}, what is the steady-state concentration of total radicals?

  1. 1.0×10−8 M1.0 \times 10^{-8} \text{ M} (correct answer)
  2. 3.2×10−8 M3.2 \times 10^{-8} \text{ M}
  3. 1.0×10−7 M1.0 \times 10^{-7} \text{ M}
  4. 3.2×10−7 M3.2 \times 10^{-7} \text{ M}
  5. 1.0×10−6 M1.0 \times 10^{-6} \text{ M}
Explanation: Chain reaction mechanisms involve three key phases: initiation (radical formation), propagation (radical reactions that continue the chain), and termination (radical destruction). When analyzing these systems, you need to apply the steady-state approximation, where the rate of radical formation equals the rate of radical destruction. In this mechanism, radicals are created at rate RiR_i and destroyed through termination. The propagation steps don't change the total radical concentration—they just interconvert R⋅R\cdot and R′R'. For termination by radical combination, two radicals combine to form a stable product, so the termination rate is kt[Rtotal]2k_t[R_{total}]^2, where [Rtotal]=[R⋅]+[R′][R_{total}] = [R\cdot] + [R']. Setting formation rate equal to destruction rate: Ri=kt[Rtotal]2R_i = k_t[R_{total}]^2 Solving for total radical concentration: [Rtotal]=Rikt=1.0×10−8108=1.0×10−16=1.0×10−8 M[R_{total}] = \sqrt{\frac{R_i}{k_t}} = \sqrt{\frac{1.0 \times 10^{-8}}{10^8}} = \sqrt{1.0 \times 10^{-16}} = 1.0 \times 10^{-8} \text{ M} This confirms answer A is correct. Answer B (3.2×10−83.2 \times 10^{-8} M) results from incorrectly using πRi/kt\sqrt{\pi R_i/k_t}. Answer C (1.0×10−71.0 \times 10^{-7} M) comes from the common error of using Ri/ktR_i/k_t instead of taking the square root. Answer D (3.2×10−73.2 \times 10^{-7} M) combines this square root error with an additional factor. Remember: In radical chain reactions, always check whether termination involves one radical (linear in concentration) or two radicals (quadratic in concentration) to set up the steady-state equation correctly.

Question 13

A reaction network has two competing pathways from reactant A: A→k1B→k3DA \xrightarrow{k_1} B \xrightarrow{k_3} D and A→k2C→k4DA \xrightarrow{k_2} C \xrightarrow{k_4} D, where all steps are first-order. Given k1=0.1 s−1k_1 = 0.1 \text{ s}^{-1}, k2=0.3 s−1k_2 = 0.3 \text{ s}^{-1}, k3=0.5 s−1k_3 = 0.5 \text{ s}^{-1}, and k4=0.2 s−1k_4 = 0.2 \text{ s}^{-1}, what is the ratio of D produced via pathway 1 to D produced via pathway 2 at long times?

  1. 0.33 (correct answer)
  2. 0.50
  3. 0.67
  4. 1.00
  5. 1.50
Explanation: When analyzing competing reaction pathways, you need to consider both the branching ratio at the initial step and how efficiently each pathway converts intermediates to the final product. First, determine how A splits between the two pathways. Since both reactions from A are first-order and competing, the branching ratio depends on the relative rate constants: k1:(k1+k2)=0.1:(0.1+0.3)=0.25k_1:(k_1+k_2) = 0.1:(0.1+0.3) = 0.25 goes to pathway 1, while k2:(k1+k2)=0.3:(0.1+0.3)=0.75k_2:(k_1+k_2) = 0.3:(0.1+0.3) = 0.75 goes to pathway 2. At long times, all intermediates B and C will have converted to D (since k3k_3 and k4k_4 are both positive). Therefore, the ratio of D produced via pathway 1 to pathway 2 simply equals the initial branching ratio: 0.250.75=13=0.33\frac{0.25}{0.75} = \frac{1}{3} = 0.33 Looking at the wrong answers: (B) 0.50 might come from incorrectly using k1/k2=0.1/0.3k_1/k_2 = 0.1/0.3, ignoring that this isn't how branching ratios work. (C) 0.67 could result from inverting the calculation or using k2/(k1+k2)k_2/(k_1+k_2) instead of the ratio. (D) 1.00 would suggest equal production through both pathways, perhaps from mistakenly thinking the different values of k3k_3 and k4k_4 somehow balance out the initial branching. Study tip: For competing pathways leading to the same final product, focus on the initial branching step. The relative rates of subsequent steps don't matter for the final product ratio at long times—only how the reactant initially splits between pathways.

Question 14

A research group studies the decomposition of compound X under different conditions. They find that the reaction follows different kinetic orders depending on the concentration regime. The integrated rate law transitions from one form to another as the concentration changes.

Based on the passage, if the integrated rate law changes from [X]0−[X]=kt[X]_0 - [X] = kt to ln⁡([X]0/[X])=kt\ln([X]_0/[X]) = kt as the concentration decreases, what is the most likely explanation for this kinetic transition?

  1. The reaction mechanism changes from unimolecular to bimolecular elementary steps at lower concentrations due to solvent effects
  2. The reaction shifts from zero-order to first-order kinetics, likely due to catalyst saturation at high concentrations (correct answer)
  3. The temperature dependence of the rate constant changes significantly in the lower concentration regime
  4. The reaction exhibits mixed-order kinetics with a transition point where substrate inhibition becomes negligible
  5. The reaction follows Michaelis-Menten kinetics with KmK_m approximately equal to the transition concentration
Explanation: When you encounter changing integrated rate laws, you're dealing with a shift in reaction order, which often reveals important mechanistic information about the system. The key insight here is recognizing what each integrated rate law represents. The equation [X]0−[X]=kt[X]_0 - [X] = kt is the integrated rate law for zero-order kinetics, where the reaction rate is independent of reactant concentration. The equation ln⁡([X]0/[X])=kt\ln([X]_0/[X]) = kt represents first-order kinetics, where rate depends linearly on concentration. Answer B correctly identifies this transition from zero-order to first-order kinetics. At high concentrations, when a catalyst becomes saturated, all active sites are occupied and the reaction rate becomes independent of substrate concentration—classic zero-order behavior. As concentration decreases, the catalyst is no longer saturated, and the reaction rate becomes proportional to the available substrate concentration, shifting to first-order kinetics. Answer A incorrectly suggests a mechanism change from unimolecular to bimolecular steps, but this wouldn't produce the specific integrated rate law transition observed. Answer C focuses on temperature effects, which wouldn't cause the systematic order change described—temperature affects the rate constant kk, not the fundamental kinetic order. Answer D mentions mixed-order kinetics and substrate inhibition, but substrate inhibition typically occurs at high concentrations and wouldn't explain the specific zero-to-first-order transition pattern. Remember: when you see kinetic order changes with concentration, think about catalyst saturation or enzyme kinetics. This is a common mechanism for zero-to-first-order transitions in chemical systems.

Question 15

For a gas-phase reaction 2A(g)→B(g)+C(g)2A(g) \rightarrow B(g) + C(g) following second-order kinetics, the half-life at 1.0 atm partial pressure of A is 25 minutes. Using the ideal gas law, what will be the half-life when the initial partial pressure of A is 0.4 atm at the same temperature?

  1. 10.0 minutes
  2. 15.6 minutes
  3. 25.0 minutes
  4. 39.1 minutes
  5. 62.5 minutes (correct answer)
Explanation: When you encounter gas-phase kinetics problems, remember that the relationship between concentration (or pressure) and half-life depends on the reaction order. For second-order reactions, this relationship is inverse and linear. For a second-order reaction, the half-life equation is: t1/2=1k[A]0t_{1/2} = \frac{1}{k[A]_0} Since we're dealing with gases, we can use partial pressure instead of concentration (they're proportional via the ideal gas law). The key insight is that half-life is inversely proportional to initial pressure for second-order kinetics. Setting up the relationship: t1/2,1t1/2,2=PA,2PA,1\frac{t_{1/2,1}}{t_{1/2,2}} = \frac{P_{A,2}}{P_{A,1}} Substituting our values: 25 mint1/2,2=0.4 atm1.0 atm\frac{25 \text{ min}}{t_{1/2,2}} = \frac{0.4 \text{ atm}}{1.0 \text{ atm}} Solving: t1/2,2=25 min0.4=62.5 minutest_{1/2,2} = \frac{25 \text{ min}}{0.4} = 62.5 \text{ minutes} The distractors represent common misconceptions: A) assumes direct proportionality (first-order thinking), B) appears to use an incorrect square root relationship, C) incorrectly assumes half-life is independent of concentration (zero-order thinking), and D) uses the wrong direction for the inverse relationship. Since 62.5 minutes isn't among the listed options A-D, the correct answer must be E (not shown but implied to exist). Study tip: Memorize that for second-order reactions, when initial concentration decreases, half-life increases proportionally. This inverse relationship is a frequent exam topic that distinguishes second-order from first-order kinetics.

Question 16

Two parallel first-order reactions compete for the same reactant A: A→k1BA \xrightarrow{k_1} B and A→k2CA \xrightarrow{k_2} C, where k1=0.03 s−1k_1 = 0.03 \text{ s}^{-1} and k2=0.07 s−1k_2 = 0.07 \text{ s}^{-1}. What fraction of the original A remains after exactly two effective half-lives of the overall process?

  1. 0.125
  2. 0.25 (correct answer)
  3. 0.33
  4. 0.50
  5. 0.67
Explanation: When you encounter competing parallel reactions, you're dealing with a system where multiple pathways consume the same reactant simultaneously. The key insight is that the overall disappearance of reactant A follows the sum of all rate constants. For this system, the overall rate constant is ktotal=k1+k2=0.03+0.07=0.10 s−1k_{total} = k_1 + k_2 = 0.03 + 0.07 = 0.10 \text{ s}^{-1}. The effective half-life of the overall process is t1/2=ln⁡(2)ktotal=0.6930.10=6.93 st_{1/2} = \frac{\ln(2)}{k_{total}} = \frac{0.693}{0.10} = 6.93 \text{ s}. After exactly two effective half-lives (13.86 s), you can use the first-order decay equation: [A]=[A]0e−ktotalt[A] = [A]_0 e^{-k_{total}t}. Substituting: [A]=[A]0e−0.10×13.86=[A]0e−1.386=[A]0×0.25[A] = [A]_0 e^{-0.10 \times 13.86} = [A]_0 e^{-1.386} = [A]_0 \times 0.25. Therefore, 25% or 0.25 of the original A remains, making (B) correct. (A) 0.125 represents what would remain after three half-lives (123=0.125\frac{1}{2^3} = 0.125), not two. (C) 0.33 might result from incorrectly using only one of the rate constants instead of their sum. (D) 0.50 is what remains after exactly one half-life, not two. Remember that in competing reactions, always sum the rate constants to find the overall disappearance rate of the reactant. The individual products form according to their respective rate constants, but the reactant disappears according to the total rate constant.

Question 17

For a reaction that follows the integrated rate law [A]−2−[A0]−2=2kt[A]^{-2} - [A_0]^{-2} = 2kt, what is the half-life expression when the initial concentration is [A0][A_0]?

  1. t1/2=32k[A0]t_{1/2} = \frac{3}{2k[A_0]} (correct answer)
  2. t1/2=ln⁡(2)kt_{1/2} = \frac{\ln(2)}{k}
  3. t1/2=1k[A0]t_{1/2} = \frac{1}{k[A_0]}
  4. t1/2=34k[A0]t_{1/2} = \frac{3}{4k[A_0]}
  5. t1/2=12k[A0]t_{1/2} = \frac{1}{2k[A_0]}
Explanation: When you encounter an integrated rate law, you're dealing with a specific reaction order that determines how concentration changes over time. The given equation [A]−2−[A0]−2=2kt[A]^{-2} - [A_0]^{-2} = 2kt is the integrated rate law for a third-order reaction. To find the half-life, you need to determine when the concentration drops to half its initial value, so [A]=[A0]2[A] = \frac{[A_0]}{2}. Substituting this into the integrated rate law: ([A0]2)−2−[A0]−2=2kt1/2\left(\frac{[A_0]}{2}\right)^{-2} - [A_0]^{-2} = 2kt_{1/2} 4[A0]2−1[A0]2=2kt1/2\frac{4}{[A_0]^2} - \frac{1}{[A_0]^2} = 2kt_{1/2} 3[A0]2=2kt1/2\frac{3}{[A_0]^2} = 2kt_{1/2} Solving for t1/2t_{1/2}: t1/2=32k[A0]t_{1/2} = \frac{3}{2k[A_0]} This confirms answer A is correct. Answer B (t1/2=ln⁡(2)kt_{1/2} = \frac{\ln(2)}{k}) is the half-life expression for a first-order reaction, which has the integrated rate law ln⁡[A]−ln⁡[A0]=−kt\ln[A] - \ln[A_0] = -kt. Answer C (t1/2=1k[A0]t_{1/2} = \frac{1}{k[A_0]}) represents the half-life for a second-order reaction with integrated rate law [A]−1−[A0]−1=kt[A]^{-1} - [A_0]^{-1} = kt. Answer D gives an incorrect coefficient—it appears to be a mathematical error in the derivation. Study tip: Always identify the reaction order first by examining the integrated rate law format. The power of concentration terms ([A]−2[A]^{-2} here) tells you the order, and each order has its own specific half-life expression that depends differently on initial concentration.

Question 18

The half-life of a second-order reaction with initial concentration 0.80 M is 45 minutes. What will be the half-life when the concentration has decreased to 0.20 M?

  1. 180 minutes (correct answer)
  2. 90 minutes
  3. 45 minutes
  4. 22.5 minutes
Explanation: For second-order reactions, t1/2=1/(k[A]0)t_{1/2} = 1/(k[A]_0). Initially, 45=1/(k×0.80)45 = 1/(k \times 0.80), so k=1/(45×0.80)=0.0278k = 1/(45 \times 0.80) = 0.0278 M⁻¹min⁻¹. When concentration is 0.20 M, the new half-life is t1/2=1/(0.0278×0.20)=180t_{1/2} = 1/(0.0278 \times 0.20) = 180 minutes. Choice B doubles the original half-life incorrectly. Choice C assumes half-life is independent of concentration (first-order thinking). Choice D halves the original half-life incorrectly.

Question 19

Two parallel first-order reactions compete for the same reactant A: A→BA \rightarrow B (k1=0.020 s−1k_1 = 0.020 \text{ s}^{-1}) and A→CA \rightarrow C (k2=0.035 s−1k_2 = 0.035 \text{ s}^{-1}). What is the effective half-life for the consumption of A?

  1. 12.6 s (correct answer)
  2. 20.0 s
  3. 34.7 s
  4. 55.0 s
Explanation: For parallel first-order reactions, the effective rate constant is keff=k1+k2=0.020+0.035=0.055 s−1k_{eff} = k_1 + k_2 = 0.020 + 0.035 = 0.055 \text{ s}^{-1}. The half-life is t1/2=ln⁡(2)/keff=0.693/0.055=12.6t_{1/2} = \ln(2)/k_{eff} = 0.693/0.055 = 12.6 s. Choice B uses only k1k_1. Choice C uses only k2k_2. Choice D incorrectly adds the individual half-lives instead of adding rate constants.

Question 20

A radioactive isotope undergoes first-order decay with a half-life of 8.5 days. If a sample initially contains 3.2×10153.2 \times 10^{15} atoms, how many atoms will remain after exactly 3 half-lives?

  1. 4.0×10144.0 \times 10^{14} atoms (correct answer)
  2. 8.0×10148.0 \times 10^{14} atoms
  3. 1.6×10151.6 \times 10^{15} atoms
  4. 2.4×10142.4 \times 10^{14} atoms
Explanation: After nn half-lives, the remaining amount is N=N0×(1/2)nN = N_0 \times (1/2)^n. After 3 half-lives: N=3.2×1015×(1/2)3=3.2×1015×1/8=4.0×1014N = 3.2 \times 10^{15} \times (1/2)^3 = 3.2 \times 10^{15} \times 1/8 = 4.0 \times 10^{14} atoms. Choice B uses 2 half-lives instead of 3. Choice C uses 1 half-life instead of 3. Choice D uses incorrect exponential calculation.