By Fatskills Exam Guides Team — the exam nerds behind 28,500+ quizzes and 2.1M practice questions across 500+ global exams.
Intermediate — requires understanding of mathematical forms, graph interpretation, and conceptual clarity between order, molecularity, and temperature effects.
Question: What is the unit of the rate constant for a first-order reaction? A) mol L?¹ s?¹ B) L mol?¹ s?¹ C) s?¹ D) mol² L?² s?¹ Answer: C Explanation: For first-order reactions, rate = ( k[R] ), so unit of ( k ) is s?¹. Why others fail: Option A is for zero-order, commonly misremembered.
Question: The half-life of a first-order reaction is 30 minutes. What is its rate constant? A) 0.0231 min?¹ B) 0.0462 min?¹ C) 0.0154 min?¹ D) 0.0300 min?¹ Answer: A Explanation: ( k = \frac{0.693}{t_{1/2}} = \frac{0.693}{30} = 0.0231 ) min?¹. Why others fail: Option B is double the correct value, a common calculation error.
Question: For a reaction, the rate constant increases from ( 2 \times 10^{-3} ) s?¹ at 300 K to ( 4 \times 10^{-3} ) s?¹ at 310 K. What is the activation energy? (Use ( R = 8.314 ) J/mol·K) A) 53.6 kJ/mol B) 42.1 kJ/mol C) 27.6 kJ/mol D) 38.3 kJ/mol Answer: A Explanation: Using ( \log \frac{k_2}{k_1} = \frac{E_a}{2.303R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) ), ( E_a \approx 53.6 ) kJ/mol. Why others fail: Option B results from using natural log instead of log, a frequent mistake.
Question: Which of the following is correct for a zero-order reaction? A) Half-life is independent of initial concentration B) Rate constant has units of L mol?¹ s?¹ C) Plot of [R] vs time is linear with negative slope D) Rate doubles if concentration is doubled Answer: C Explanation: For zero-order, [R] = [R]? – kt, so [R] vs t is linear with slope = –k. Why others fail: Option D is true for first-order, often confused.
Question: The activation energy of a reaction is 98 kJ/mol. By what factor does the rate increase when temperature rises from 300 K to 310 K? A) 1.8 B) 2.5 C) 3.2 D) 4.0 Answer: B Explanation: Using ( \log \frac{k_2}{k_1} = \frac{98000}{2.303 \times 8.314} \left( \frac{1}{300} - \frac{1}{310} \right) \approx 0.40 ), so ( \frac{k_2}{k_1} \approx 10^{0.4} = 2.5 ). Why others fail: Option C comes from incorrect temperature difference or arithmetic.
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