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. (b)
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, $\quad \frac{r _1}{r _2}=\frac{\left(a-x _1\right)^{m}}{\left(a-x _2\right)^{m}}$ or $\frac{r _1}{r _2}=\left[\frac{a-x _1}{a-x _2}\right]^{m}$
Now putting the given values
$$ \frac{1}{0.5}=\left(\frac{0.95}{0.67}\right)^{m} \Rightarrow 2=(1.41)^{m} \text { or } m=2 $$
