Q52 / 180 · COMEDK 2022 · Chemistry
For the reaction $2 \mathrm{~N}_2 \mathrm{O}_5 \longrightarrow 4 \mathrm{NO}_2+\mathrm{O}_2$. If inital pressure is $100 \mathrm{~atm}$ and rate constant $k$ is $3.38 \times 10^{-5} \mathrm{~s}^{-1}$. After $20 \mathrm{~min}$ the final pressure of $\mathrm{N}_2 \mathrm{O}_5$ will be
- Correct answerA$96 \mathrm{~atm}$
- B$50 \mathrm{~atm}$
- C$70 \mathrm{~atm}$
- D$60 \mathrm{~atm}$
Solution
The reaction $2 \mathrm{N}_2 \mathrm{O}_5 \rightarrow 4 \mathrm{NO}_2 + \mathrm{O}_2$ is a first-order reaction. The rate equation for a first-order reaction is given by $P_t = P_0 e^{-kt}$, where $P_0$ is the initial pressure and $P_t$ is the pressure at time $t$.
Given values are:
$P_0 = 100 \text{ atm}$
$k = 3.38 \times 10^{-5} \text{ s}^{-1}$
$t = 20 \text{ min} = 20 \times 60 \text{ s} = 1200 \text{ s}$
Substituting these values into the equation:
$P_t = 100 \times e^{-(3.38 \times 10^{-5}) \times 1200}$
Calculate the exponent:
$-(3.38 \times 10^{-5}) \times 1200 = -0.04056$
Now calculate $P_t$:
$P_t = 100 \times e^{-0.04056}$
Using the approximation $e^{-x} \approx 1 - x$ for small $x$:
$P_t \approx 100 \times (1 - 0.04056) = 100 \times 0.95944 = 95.944 \text{ atm}$
Rounding to the nearest integer, we get $96 \text{ atm}$.
Answer: $96 \mathrm{~atm}$
Given values are:
$P_0 = 100 \text{ atm}$
$k = 3.38 \times 10^{-5} \text{ s}^{-1}$
$t = 20 \text{ min} = 20 \times 60 \text{ s} = 1200 \text{ s}$
Substituting these values into the equation:
$P_t = 100 \times e^{-(3.38 \times 10^{-5}) \times 1200}$
Calculate the exponent:
$-(3.38 \times 10^{-5}) \times 1200 = -0.04056$
Now calculate $P_t$:
$P_t = 100 \times e^{-0.04056}$
Using the approximation $e^{-x} \approx 1 - x$ for small $x$:
$P_t \approx 100 \times (1 - 0.04056) = 100 \times 0.95944 = 95.944 \text{ atm}$
Rounding to the nearest integer, we get $96 \text{ atm}$.
Answer: $96 \mathrm{~atm}$
Hence, the correct answer is option: 'A'.