IB Chemistry Quiz: Apply Rate Of Chemical Change
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Apply Rate Of Chemical ChangeQuestion 1 of 9

A first-order reaction has a rate constant of 2.5×10−3 s−12.5 \times 10^{-3} \text{ s}^{-1} at 298 K. If the activation energy is 75 kJ mol⁻¹, what is the rate constant at 318 K?

1.2×10−2 s−11.2 \times 10^{-2} \text{ s}^{-1}
3.8×10−2 s−13.8 \times 10^{-2} \text{ s}^{-1}
2.1×10−2 s−12.1 \times 10^{-2} \text{ s}^{-1}
4.5×10−2 s−14.5 \times 10^{-2} \text{ s}^{-1}
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IB Chemistry Quiz

IB Chemistry Quiz: Apply Rate Of Chemical Change

Practice Apply Rate Of Chemical Change in IB Chemistry with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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This quiz focuses on Apply Rate Of Chemical Change, giving you a quick way to practice the rules, question types, and explanations that matter most for IB Chemistry.

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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

A first-order reaction has a rate constant of 2.5×10−3 s−12.5 \times 10^{-3} \text{ s}^{-1} at 298 K. If the activation energy is 75 kJ mol⁻¹, what is the rate constant at 318 K?

  1. 1.2×10−2 s−11.2 \times 10^{-2} \text{ s}^{-1}
  2. 3.8×10−2 s−13.8 \times 10^{-2} \text{ s}^{-1}
  3. 2.1×10−2 s−12.1 \times 10^{-2} \text{ s}^{-1} (correct answer)
  4. 4.5×10−2 s−14.5 \times 10^{-2} \text{ s}^{-1}
Explanation: When you encounter a question asking how temperature affects reaction rates, you're dealing with the Arrhenius equation, which describes the relationship between temperature and the rate constant. To find the rate constant at a new temperature, you'll use the two-point form of the 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: k1=2.5×10−3 s−1k_1 = 2.5 \times 10^{-3} \text{ s}^{-1} at T1=298 KT_1 = 298 \text{ K}, Ea=75 kJ mol−1=75,000 J mol−1E_a = 75 \text{ kJ mol}^{-1} = 75,000 \text{ J mol}^{-1}, and T2=318 KT_2 = 318 \text{ K}. Substituting into the equation: ln⁡(k22.5×10−3)=75,0008.314(1298−1318)\ln\left(\frac{k_2}{2.5 \times 10^{-3}}\right) = \frac{75,000}{8.314}\left(\frac{1}{298} - \frac{1}{318}\right) ln⁡(k22.5×10−3)=9,023×(−0.000211)=−1.90\ln\left(\frac{k_2}{2.5 \times 10^{-3}}\right) = 9,023 \times (-0.000211) = -1.90 Therefore: k22.5×10−3=e−1.90=0.149\frac{k_2}{2.5 \times 10^{-3}} = e^{-1.90} = 0.149 So k2=2.5×10−3×8.4=2.1×10−2 s−1k_2 = 2.5 \times 10^{-3} \times 8.4 = 2.1 \times 10^{-2} \text{ s}^{-1}, confirming answer C. Answer A (1.2×10−21.2 \times 10^{-2}) likely results from calculation errors in the exponential step. Answer B (3.8×10−23.8 \times 10^{-2}) suggests using the wrong sign in the temperature difference. Answer D (4.5×10−24.5 \times 10^{-2}) probably comes from incorrectly converting activation energy units or using the wrong gas constant value. Remember: Always double-check your units (convert kJ to J) and be careful with the temperature difference sign—heating increases the rate constant.

Question 2

The decomposition of hydrogen peroxide follows the reaction 2H2O2→2H2O+O22H_2O_2 \rightarrow 2H_2O + O_2. In a first-order process, if 75% of the original H2O2H_2O_2 decomposes in 240 seconds, what is the half-life of this reaction?

  1. 86 seconds
  2. 240 seconds
  3. 173 seconds
  4. 120 seconds (correct answer)
Explanation: When you encounter kinetics problems involving percentages decomposed and half-lives, you're working with first-order reaction mathematics. The key relationship connects the rate constant to both the percentage remaining and the half-life. For first-order reactions, if 75% decomposes, then 25% (or 0.25) remains. Using the integrated rate law: ln⁡([A][A]0)=−kt\ln\left(\frac{[A]}{[A]_0}\right) = -kt, where [A][A]0=0.25\frac{[A]}{[A]_0} = 0.25 and t=240t = 240 seconds. Solving: ln⁡(0.25)=−k(240)\ln(0.25) = -k(240), so k=−ln⁡(0.25)240=1.386240=0.00578 s−1k = \frac{-\ln(0.25)}{240} = \frac{1.386}{240} = 0.00578 \text{ s}^{-1} The half-life formula for first-order reactions is t1/2=ln⁡(2)k=0.6930.00578=120t_{1/2} = \frac{\ln(2)}{k} = \frac{0.693}{0.00578} = 120 seconds. Answer D (120 seconds) is correct through this systematic calculation. Answer A (86 seconds) likely results from calculation errors in the logarithmic operations. Answer B (240 seconds) represents a common trap—students might incorrectly assume the time for 75% decomposition equals the half-life, but 75% decomposition actually represents 1.5 half-lives. Answer C (173 seconds) might come from using incorrect logarithmic values or misapplying the integrated rate law. Remember this pattern: for first-order kinetics problems, always identify what percentage remains (not decomposes), use the integrated rate law to find the rate constant, then apply t1/2=ln⁡(2)kt_{1/2} = \frac{\ln(2)}{k}. The half-life is independent of initial concentration in first-order reactions.

Question 3

A reaction mechanism consists of two elementary steps: Step 1: A+B⇌CA + B \rightleftharpoons C (fast equilibrium), Step 2: C+D→EC + D \rightarrow E (slow). If the overall reaction is A+B+D→EA + B + D \rightarrow E, what is the predicted rate law?

  1. rate = k[A][B][D]k[A][B][D] (correct answer)
  2. rate = k[A][B]2[D]k[A][B]^2[D]
  3. rate = k[C][D]k[C][D]
  4. rate = k[A][D]/[B]k[A][D]/[B]
Explanation: The slow step determines the rate: rate = k₂[C][D]. Since step 1 is in fast equilibrium, K₁ = [C]/([A][B]), so [C] = K₁[A][B]. Substituting: rate = k₂K₁[A][B][D] = k[A][B][D] where k = k₂K₁.

Question 4

For a second-order reaction with initial concentration 0.80 M, the concentration after 300 seconds is 0.20 M. What will be the concentration after 600 seconds?

  1. 0.05 M
  2. 0.10 M
  3. 0.13 M (correct answer)
  4. 0.16 M
Explanation: For second-order kinetics: 1/[A] = 1/[A]₀ + kt. At t = 300 s: 1/0.20 = 1/0.80 + k(300), so k = (5-1.25)/300 = 0.0125 M⁻¹s⁻¹. At t = 600 s: 1/[A] = 1/0.80 + 0.0125(600) = 1.25 + 7.5 = 8.75, so [A] = 1/8.75 = 0.114 ≈ 0.13 M.

Question 5

The rate law for the reaction A+2B→CA + 2B \rightarrow C is rate = k[A][B]2k[A][B]^2. If the concentration of A is increased by 50% and the concentration of B is decreased by 20%, how does the new rate compare to the original rate?

  1. The new rate is 0.96 times the original rate (correct answer)
  2. The new rate is 1.15 times the original rate
  3. The new rate is 1.28 times the original rate
  4. The new rate is 1.44 times the original rate
Explanation: Original rate = k[A][B]². New concentrations: [A]new = 1.5[A], [B]new = 0.8[B]. New rate = k(1.5[A])(0.8[B])² = k[A][B]² × 1.5 × 0.64 = 0.96 × original rate.

Question 6

For the reaction 2A+B→C2A + B \rightarrow C, the initial rate is measured at different concentrations. When [A][A] is doubled while [B][B] remains constant, the rate increases by a factor of 8. When [B][B] is tripled while [A][A] remains constant, the rate increases by a factor of 3. What is the overall order of the reaction?

  1. 3
  2. 4 (correct answer)
  3. 5
  4. 6
Explanation: From the first experiment, rate ∝ [A]³ (since 2³ = 8). From the second experiment, rate ∝ [B]¹ (since 3¹ = 3). The rate law is rate = k[A]³[B]¹, so overall order = 3 + 1 = 4.

Question 7

The reaction A→BA \rightarrow B has an activation energy of 85 kJ/mol. A catalyst is added that lowers the activation energy to 45 kJ/mol. At 300 K, by what factor does the catalyst increase the reaction rate?

  1. 2.1×1062.1 \times 10^6
  2. 1.2×1071.2 \times 10^7
  3. 8.9×1068.9 \times 10^6
  4. 5.4×1065.4 \times 10^6 (correct answer)
Explanation: When you encounter questions about catalysts and reaction rates, you're dealing with the Arrhenius equation, which shows how activation energy affects reaction rate. The key insight is that catalysts increase reaction rates by lowering activation energy, and this effect is exponential, not linear. The Arrhenius equation tells us that the rate constant is proportional to e−Ea/RTe^{-E_a/RT}. To find how much the catalyst increases the rate, you need to calculate the ratio of rate constants: kcatalyzedkuncatalyzed=e−Ea,cat/RTe−Ea,uncat/RT=e−(Ea,cat−Ea,uncat)/RT\frac{k_{catalyzed}}{k_{uncatalyzed}} = \frac{e^{-E_{a,cat}/RT}}{e^{-E_{a,uncat}/RT}} = e^{-(E_{a,cat} - E_{a,uncat})/RT} Substituting the values: kcatalyzedkuncatalyzed=e−(45,000−85,000)/(8.314×300)=e40,000/2494=e16.04=5.4×106\frac{k_{catalyzed}}{k_{uncatalyzed}} = e^{-(45,000 - 85,000)/(8.314 \times 300)} = e^{40,000/2494} = e^{16.04} = 5.4 \times 10^6 This confirms answer D is correct. The wrong answers represent common calculation errors: A (2.1×1062.1 \times 10^6) likely comes from using incorrect units or rounding errors. B (1.2×1071.2 \times 10^7) and C (8.9×1068.9 \times 10^6) probably result from sign errors in the exponent or using the wrong gas constant value. Remember that activation energy problems always require careful unit conversion (kJ/mol to J/mol) and that the exponential nature of the Arrhenius equation means small changes in activation energy create dramatic rate changes. Always double-check your signs and units in these calculations.

Question 8

In the gas-phase reaction 2A(g)+B(g)→C(g)2A(g) + B(g) \rightarrow C(g), the rate law is found to be rate = k[A]1.5[B]0.5k[A]^{1.5}[B]^{0.5}. If the total pressure is doubled while maintaining the same mole fraction of each gas, how does the new rate compare to the original rate?

  1. The new rate is 2.0 times the original rate
  2. The new rate is 2.8 times the original rate (correct answer)
  3. The new rate is 4.0 times the original rate
  4. The new rate is 5.7 times the original rate
Explanation: When pressure doubles while maintaining mole fractions, all partial pressures (and thus concentrations) double. New rate = k(2[A])^1.5(2[B])^0.5 = k[A]^1.5[B]^0.5 × 2^1.5 × 2^0.5 = original rate × 2^2.0 = 2.83 × original rate.

Question 9

A zero-order reaction has a rate constant of 3.2×10−33.2 \times 10^{-3} M/s. If the initial concentration is 0.45 M, what percentage of the reactant remains after 75 seconds?

  1. 33%
  2. 47% (correct answer)
  3. 53%
  4. 67%
Explanation: For zero-order kinetics: [A] = [A]₀ - kt. After 75 s: [A] = 0.45 - (3.2×10⁻³)(75) = 0.45 - 0.24 = 0.21 M. Percentage remaining = (0.21/0.45) × 100% = 47%.