Question

For a first order reaction, A(g) $\rightarrow$ B(g) at 35 $^\circ$C, the volume of "A" left in the reaction vessel at various times are given below. [Given data: $\log(5/4) = 0.0969$]

What is the value of rate constant?

• A. $0.02231 \text{ min}^{-1}$
• B. $0.04231 \text{ min}^{-1}$
• C. $0.06231 \text{ min}^{-1}$
• D. $0.08231 \text{ min}^{-1}$
• E. $0.1231 \text{ min}^{-1}$

Hint:

For first-order reactions, the half-life is independent of the initial concentration/volume. You can check $k$ at different intervals ($t=20, 30$) to verify consistency.

Answer 1

The Correct Option is A

Concept: For a first-order reaction, the rate constant $k$ is calculated using the integrated rate law formula: $k = \frac{2.303}{t} \log \left( \frac{V_0}{V_t} \right)$, where $V_0$ is the initial volume and $V_t$ is the volume at time $t$.

Step 1: {Extract data from the table.} At $t = 0$, $V_0 = 25 \text{ mL}$. At $t = 10$, $V_{10} = 20 \text{ mL}$.

Step 2: {Substitute values into the rate constant formula.} $$k = \frac{2.303}{10} \log \left( \frac{25}{20} \right)$$ $$k = 0.2303 \times \log \left( \frac{5}{4} \right)$$

Step 3: {Solve for $k$ using the given log value.} Using the provided data $\log(5/4) = 0.0969$: $$k = 0.2303 \times 0.0969$$ $$k = 0.022316 \dots$$ Rounding gives $0.02231 \text{ min}^{-1}$.