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The rate of decomposition of a gaseous acetaldehyde (CH³CHO) to gaseous methane and carbon monoxide is found to increase by a factor of 283,When the initial concentration of acetaldehyde is doubled. What is the order of this reaction?

Grace Koranteng , 4 Years ago
Grade 11
anser 1 Answers
Arun

Last Activity: 4 Years ago

As we learned
 
nth order reaction -
 
The rates of the reaction is proportional to nth power of reactant
 
- wherein
 
Differential rate law
 
=\frac{dx}{dt}=k(a-x)^{n}
 
Integrated rate laws,
 
=\frac{1}{n-1}[(a-x)^{1-n} -a^{(1-n)}]=k_nt
 
a= initial, concentration of reactant at t=0 sec
 
x= concentration of product formed at t= tsec
 
t_\frac{1}{2}=\frac{1}{(n-1)(a^{n-1})(k_n)}[2^{n-1}-1]
 
Formulae for all the order except n=1

Let rate law be:
 
r=K\left [ acetaldehyde \right ]^{n}
 
\therefore 1= K\left ( 0.95\: P_{o} \right )^{n}
 
0.5=K\left ( 0.67\: P_{o})^{n} \right
 
So\: 2= \left ( \frac{0.95}{0.67} \right )\Rightarrow 2= \left ( 1.41 \right )^{n}\therefore n;2

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