The reaction can be represented by the following equation:
Cyclopropane → Propane
The rate law for this reaction can be written as:
Rate = k[Cyclopropane]
Where k is the rate constant and [Cyclopropane] is the concentration of cyclopropane.
Given that the rate constant k = 2.74×10-3 s-1, the initial concentration of cyclopropane [Cyclopropane]0 = 0.290 M, and the time t = 100 s, we can use the first-order integrated rate law equation to calculate the concentration of cyclopropane after 100 seconds:
ln([Cyclopropane]/[Cyclopropane]0) = -kt
where [Cyclopropane] is the concentration of cyclopropane at time t.
Substituting the given values, we get:
ln([Cyclopropane]/0.290) = -(2.74×10-3 s-1) × (100 s)
ln([Cyclopropane]/0.290) = -0.274
[Cyclopropane]/0.290 = e-0.274
[Cyclopropane] = 0.290 × e-0.274
[Cyclopropane] = 0.251 M
Therefore, the concentration of cyclopropane after 100 seconds is 0.251 M.
Cyclopropane rearreanges to form propane CH2CH2CH2--> CH2=CHCH3 by first order kinetics .the rate constant is k=2.74×10-3 s-1.the initial concentration of cyclopropane is 0.290 M.what will be the concentration of cyclopropabe after 100 seconds
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