Search any question & find its solution
Question:
Answered & Verified by Expert
At $300 \mathrm{~K}$, the half-life period of a gaseous reaction at an initial pressure of $40 \mathrm{kPa}$ is 350 s. When pressure is $20 \mathrm{kPa}$, the half-life period is $175 \mathrm{~s}$. What is the order of the reaction?
Options:
Solution:
2308 Upvotes
Verified Answer
The correct answer is:
Zero
We know that
$\frac{\left(t_{1 / 2}\right)_1}{\left(t_{1 / 2}\right)_2}=\left(\frac{p_2}{p_1}\right)^{n-1}$
where, $n$ is the order of reaction and $t_{1 / 2}$ is half-life.
$\begin{gathered}
\frac{350}{175}=\left(\frac{20}{40}\right)^{n-1} \\
2=\left(\frac{1}{2}\right)^{n-1} \\
\Rightarrow n-1=-1 \\
n=0
\end{gathered}$
$\therefore$ It is a zero order reaction.
$\frac{\left(t_{1 / 2}\right)_1}{\left(t_{1 / 2}\right)_2}=\left(\frac{p_2}{p_1}\right)^{n-1}$
where, $n$ is the order of reaction and $t_{1 / 2}$ is half-life.
$\begin{gathered}
\frac{350}{175}=\left(\frac{20}{40}\right)^{n-1} \\
2=\left(\frac{1}{2}\right)^{n-1} \\
\Rightarrow n-1=-1 \\
n=0
\end{gathered}$
$\therefore$ It is a zero order reaction.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.