Statement-I: When temperature is increased from 298 K to 308 K, having \( E_a = 12.72 \, \text{Kcal mol}^{-1} \), Rate constant (K) is doubled. Statement-II: For a first (1st) order reaction \( A \rightarrow B \), the following graph is valid.
Statement I is correct but Statement II is incorrect.
Statement I is incorrect but Statement II is correct.
Both Statement I and Statement II are incorrect.
Show Solution
The Correct Option isB
Solution and Explanation
Step 1: Understanding the Question:
This question evaluates the effect of temperature on the rate constant (Arrhenius equation) and the graphical representation of a first-order reaction's half-life. Step 2: Key Formula or Approach:
Arrhenius equation: \[ \ln\left(\frac{K_2}{K_1}\right) = \frac{E_a}{R} \left[ \frac{T_2 - T_1}{T_1 T_2} \right] \]
For 1st order reaction: \[ t_{1/2} = \frac{\ln 2}{K} \] Step 3: Detailed Explanation: Checking Statement-I:
Given: $T_1 = 298$ K, $T_2 = 308$ K, $E_a = 12.72$ Kcal/mol = $12.72 \times 10^3$ cal/mol.
$R \approx 2$ cal mol$^{-1}$K$^{-1}$.
\[ \ln\left(\frac{K_2}{K_1}\right) = \frac{12.72 \times 10^3}{2} \left[ \frac{308 - 298}{298 \times 308} \right] \]
\[ \ln\left(\frac{K_2}{K_1}\right) = 6360 \times \left[ \frac{10}{91784} \right] \approx 0.693 \]
Since $\ln(2) = 0.693$, then $\frac{K_2}{K_1} = 2$. Statement-I is correct.
Checking Statement-II:
For a first-order reaction, the half-life ($t_{1/2}$) is independent of the initial concentration $[A]$.
The graph shows $t_{1/2} \propto [A]$, which is characteristic of a zero-order reaction ($t_{1/2} = \frac{[A]_0}{2K}$).
Therefore, the graph is incorrect for a first-order reaction. Statement-II is incorrect. Step 4: Final Answer:
Statement I is correct but Statement II is incorrect.
