Question:

Given below are two statements:
Statement I: When \(E_a = 12.6\) kcal/mol, the room temperature rate constant is doubled by a 10 °C increase in temperature (298 K to 308 K).
Statement II: For a first order reactions \(A \to B\), [the graph of rate vs [A] is a straight line through origin].

Updated On: Apr 13, 2026
  • Both Statement I and Statement II are true
  • Both Statement I and Statement II are false
  • Statement I is true but Statement II is false
  • Statement I is false but Statement II is true
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The Correct Option is A

Solution and Explanation

Step 1: Understanding the Concept:
Statement I involves the Arrhenius equation relating rate constants at two temperatures. Statement II involves the rate law for first-order kinetics.
Step 2: Detailed Explanation:
Statement I: Using \(\log \frac{k_2}{k_1} = \frac{E_a}{2.303R} \left( \frac{T_2 - T_1}{T_1 T_2} \right)\). Given \(E_a = 12.6 \text{ kcal/mol} = 12600 \text{ cal/mol}\), \(R \approx 2 \text{ cal/mol}\cdot\text{K}\), \(T_1 = 298\), \(T_2 = 308\). \[ \log \frac{k_2}{k_1} = \frac{12600}{2.303 \times 2} \left( \frac{10}{298 \times 308} \right) \approx 0.301 \] Since \(\text{antilog}(0.301) = 2\), the rate constant doubles. (True)
Statement II: For a first-order reaction, Rate \(= k[A]^1\). This is an equation of the form \(y = mx\), which represents a straight line passing through the origin. (True)
Step 3: Final Answer:
Both statements are true.
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