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Reaction: $\ce{2 C5H6 -> C10H12}$

For this reaction it is stated that $\pu{0.609 M}$ cyclopentadiene was formed in an ethanol solution. When the solution was heated $(\pu{40 ^\circ C})$ and dicyclopentadiene was formed. The amount of dicyclopentadiene after time/h was:

$$ \begin{array}{lr} \hline \text{time}/\pu{h} & c/\pu{M} \\ \hline 6 & \pu{0.017} \\ 24 & \pu{0.059} \\ 48 & \pu{0.099} \\ 78 & \pu{0.135} \\ 143 & \pu{0.18} \\ 167 & \pu{0.191} \\ 192 & \pu{0.201} \\ \hline \end{array} $$

When given data like this, i.e. the concentration for dicyclopentadiene and not cyclopentadiene, do I have to convert the concentration to the concentration of cyclopentadiene? And in that case, will the concentration of cyclopentadiene at each of given time become:

$$ [\ce{C5H6}] = 0.609 - 2x$$

where $x$ is the concentration of dicyclopentadiene at each of corresponding time?

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  • $\begingroup$ Since the concentrations of product with time is given, you can find the order of the reaction by plotting time versus product concentrarion accordingly to find out it was 1st or second order. $\endgroup$ Apr 10 at 20:30

1 Answer 1

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Yes. You are right. The concentration of cyclopentadiene is equal to $c=0.609−2x$. And afterwards, you should plot the natural logarithm of this concentration c vs. time. If the points are well alined, the order is $1$. If the alinement is not good, plot the inverse $1/c$ versus time. If the points are well alined, the order is $2$. In both cases the slope of the line is the rate constant $k_1$ or $k_2$

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  • $\begingroup$ Thank you! Perhaps you might help me with a follow up question. By doing this I got that the reaction order is $2$ and k = $0.0166$ M$^{-1}$h$^{-1}$. If I then want to calculate how long it will take for $10$% of cyclopentadiene to react at $40$°C if the initial concentration of cyclopentadiene is $0.609$ M and $1$ mM resp., do I simply use the equation $\frac{1}{a-x}-\frac{1}{a}$= kt, where the a-x is the $10$% of the initial value and a is the initial value? Thus, getting t$_{0.609} = 890$ h and t$_{0.001} = 542168$ h $\endgroup$
    – katara
    Apr 11 at 10:15
  • $\begingroup$ Your $k$ value is OK. But your $t$ value is not. $a-x$ is not $0.1 a$. No ! It is $0.9 a$. Using $a - x = 0.9 a = 0.9·0.609 = 0.5481$, you must get $\pu{ t = 10.96 h}$ $\endgroup$
    – Maurice
    Apr 11 at 14:48
  • $\begingroup$ Oh yeah, I missed that! Thank you very much! :) $\endgroup$
    – katara
    Apr 11 at 17:07

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