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].

• A. Both Statement I and Statement II are true
• B. Both Statement I and Statement II are false
• C. Statement I is true but Statement II is false
• D. Statement I is false but Statement II is true

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Statement I requires checking the Arrhenius equation to see if the specified activation energy precisely causes a doubling of the rate constant over a specific temperature shift. Statement II requires knowledge of how half-life scales with initial concentration for different reaction orders.
Step 2: Key Formula or Approach:
1. Arrhenius equation: $\ln\left(\frac{k_2}{k_1}\right) = \frac{E_a}{R} \left( \frac{1}{T_1} - \frac{1}{T_2} \right) = \frac{E_a}{R} \left( \frac{T_2 - T_1}{T_1 T_2} \right)$.
2. Half-life for first-order reaction: $t_{1/2} = \frac{\ln 2}{k} = \frac{0.693}{k}$.
Step 3: Detailed Explanation:
Evaluate Statement I:
We need to verify if $\frac{k_2}{k_1} = 2$.
Given $E_a = 12.6\text{ kcal/mol}$. Convert this to Joules per mole:
$E_a = 12.6 \times 10^3\text{ cal/mol} \times 4.2\text{ J/cal} = 52920\text{ J/mol}$.
Temperatures are $T_1 = 298\text{ K}$ and $T_2 = 308\text{ K}$.
Substitute into the Arrhenius equation:
$\ln\left(\frac{k_2}{k_1}\right) = \frac{52920}{8.314} \times \left( \frac{308 - 298}{298 \times 308} \right)$.
$\ln\left(\frac{k_2}{k_1}\right) = \frac{52920}{8.314} \times \left( \frac{10}{91784} \right)$.
Calculate the numerical value:
$\ln\left(\frac{k_2}{k_1}\right) = \frac{529200}{763092.176} \approx 0.6935$.
Since $\ln 2 \approx 0.693$, we can confidently say that $\frac{k_2}{k_1} \approx 2$.
Therefore, the rate constant doubles. Statement I is true.
Evaluate Statement II:
For a first-order reaction, the half-life is $t_{1/2} = \frac{0.693}{k}$.
This formula clearly shows that the half-life is a constant value and depends exclusively on the rate constant $k$, being entirely independent of the initial concentration $[A]_0$.
A graph of $t_{1/2}$ versus $[A]_0$ would be a perfectly horizontal line.
The statement describes a graph that is a straight line passing through the origin ($t_{1/2} \propto [A]_0$), which is actually characteristic of a zero-order reaction.
Therefore, Statement II is false.
Step 4: Final Answer:
Statement I is true but Statement II is false.