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?

The reaction is given as follows:

The plot of ln[cyclopropane] verses t is linear with slope .

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:

Here, k is rate constant, t is time of the reaction, is concentration of reactant at time t and is initial concentration of reactant.

Taking ln both sides,

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

Thus, slope =-k

Or, .

Therefore, order of the reaction is first and rate constant is .

particle   p is an electron and particle q is a neutron