Question

For the reaction cyclopropane(g) → propene(g) at 500◦c, a plot of ln[cyclopropane] vs t gives a straight line with a slope of −0.00067 s−1 . What is the order of this reaction and what is the rate constant?

Answer 1

The reaction is given as follows:

$cyclopropane (g)\rightarrow propene (g)$

The plot of ln[cyclopropane] verses t is linear with slope $-0.00067 s^{-1}$.

The plot for ln[cyclopropane] verses t is linear for first order reaction (as in the diagram attached). The integrated rate law equation is as follows:

$[A]=[A_{0}]e^{-kt}$

Here, k is rate constant, t is time of the reaction, $[A]$ is concentration of reactant at time t and $[A_{0}]$ is initial concentration of reactant.

Taking ln both sides,

$ln[A]=ln[A_{0}]-kt$

Comparing with equation for linear graph, y=mx+c

Thus, slope =-k

Or, $k=0.00067 s^{-1}$.

Therefore, order of the reaction is first and rate constant is $0.00067 s^{-1}$.

Answer 2

Answer:

1. 1st order chemical reaction.

2. $k=0.00067 s^{-1}$.

Explanation:

Hello,

In this case, as long as the information states that when graphing the ln[cyclopropane] vs time a straight line is obtained, one infers that it is about a first order chemical reaction, this could be substantiated via the integration of the rate law:

$\frac{dC_C}{dt}=-k C_C^n$

If we set n=1, we are going to obtain the aforesaid logarithm as shown below:

$\int\limits^{C_{C}}_{C_{C_0}} {\frac{1}{C_C} } \, dC_C=-k\int\limits^t_0 {} \, dt \\ ln(\frac{C_{C}}{C_{C_0}} )=-kt$

In such a way, the straight line accounts for the rate constant including the negative sign (cyclopropane's consumption), thus, the rate constant is:

$k=-(-0.00067 s^{-1})\\k=0.00067 s^{-1}$

Which matches with a 1st order constant rate because of its units.

Best regards,