Chemical Kinetics - Result Question 15
18. At $518^{\circ} C$, the 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 the reaction is :
(2018 Main)
(a) $2$
(b) $3$
(c) $1$
(d) $0$
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Answer:
Correct Answer: 18. (a)
Solution:
- For the reaction,
$ CH _3 CHO(g) \xrightarrow{\text { Decomposes }} CH _4+CO $
Let order of reaction with respect to $CH _3 CHO$ is $m$.
Its given, $r _1=1$ torr $/ sec$. when $CH _3 CHO$ is $5 %$ reacted i.e. $95 %$ unreacted. Similarly, $r _2=0.5$ torr $/ sec$ when $CH _3 CHO$ is $33 %$ reacted i.e., $67 %$ unreacted.
Use the formula, $r \propto(a-x)^{m}$
where $(a-x)=$ amount unreacted
so, $\quad \dfrac{r _1}{r _2}=\dfrac{\left(a-x _1\right)^{m}}{\left(a-x _2\right)^{m}}$ or $\dfrac{r _1}{r _2}=\left[\dfrac{a-x _1}{a-x _2}\right]^{m}$
Now putting the given values
$ \dfrac{1}{0.5}=\left(\dfrac{0.95}{0.67}\right)^{m} \Rightarrow 2=(1.41)^{m} \text { or } m=2 $
BETA
