JEE Main On 08 April 2018 Question 12
Question: At $ \text{518 }{}^\circ\text{ C} $ , die rate of decomposition of a sample of gaseous acetaldehyde, initially at a pressure of 363 Torr, was 1.00 Torr $ {s^{-1}} $ when 5% had reacted and, 0.5 Torr $ {s^{-1}} $ when 33% had reacted. The order of me reaction is: [JEE Main Online 08-04-2018]
Options:
A) 1
B) 0
C) 2
D) 3
Show Answer
Answer:
Correct Answer: C
Solution:
rate $ \text{=-}\frac{dP}{dt}\text{=k}( {P _{C{H_3}CHO}} ) $
$ \frac{r _1}{r _2}=\frac{1}{0.5}=\frac{{{( \frac{363\times 95}{100} )}^{x}}}{{{( 363\times \frac{67}{100} )}^{x}}} $
$ 2={{(1.41)}^{x}} $
$ 2={{(\sqrt{2})}^{x}} $ $ $
