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

1. 1st order chemical reaction.

2.
`k=0.00067 s^(-1)`.

Step-by-step 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:

`(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)} {(1)/(C_C) } \, dC_C=-k\int\limits^t_0 {} \, dt \\ ln((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,

Answer 2

The reaction is given as follows:

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)`.