Q11 / 180 · COMEDK 2025 (10 May Shift 2) · Chemistry
The rate constant for a zero order reaction $\mathrm{A} \rightarrow \mathrm{B}+\mathrm{C}$ is $6.0 \times 10^{-3} \mathrm{molL}^{-1} \mathrm{~s}^{-1}$.
What would be the time taken for the initial concentration of A to decrease from 0.2 M to 0.024 M ?
- A37.34 s
- Correct answerB29.33 s
- C15.83 s
- D31.90 s
Solution
For a zero order reaction, the integrated rate equation is given by $[A]_t = [A]_0 - kt$, where $[A]_0$ is the initial concentration, $[A]_t$ is the concentration at time $t$, and $k$ is the rate constant.
Given values are $[A]_0 = 0.2 \text{ M}$, $[A]_t = 0.024 \text{ M}$, and $k = 6.0 \times 10^{-3} \text{ mol L}^{-1} \text{ s}^{-1}$.
Rearranging the equation to solve for $t$:
$t = \dfrac{[A]_0 - [A]_t}{k}$
Substituting the values:
$t = \dfrac{0.2 - 0.024}{6.0 \times 10^{-3}}$
$t = \dfrac{0.176}{6.0 \times 10^{-3}}$
$t = \dfrac{176}{6} \text{ s}$
$t = 29.333... \text{ s}$
Rounding to two decimal places, $t = 29.33 \text{ s}$.
Answer: 29.33 s
Given values are $[A]_0 = 0.2 \text{ M}$, $[A]_t = 0.024 \text{ M}$, and $k = 6.0 \times 10^{-3} \text{ mol L}^{-1} \text{ s}^{-1}$.
Rearranging the equation to solve for $t$:
$t = \dfrac{[A]_0 - [A]_t}{k}$
Substituting the values:
$t = \dfrac{0.2 - 0.024}{6.0 \times 10^{-3}}$
$t = \dfrac{0.176}{6.0 \times 10^{-3}}$
$t = \dfrac{176}{6} \text{ s}$
$t = 29.333... \text{ s}$
Rounding to two decimal places, $t = 29.33 \text{ s}$.
Answer: 29.33 s
Hence, the correct answer is option: 'B'.