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algol13
3 years ago
9

The reaction is first order in cyclopropane and has a measured rate constant of k=3.36×10−5 s−1 at 720 k. if the initial cyclopr

opane concentration is 0.0445 m, what will the cyclopropane concentration be after 235.0 min
Chemistry
1 answer:
sergey [27]3 years ago
6 0

Answer:

0.0277 M.

Explanation:

The integral rate law of a first order reaction:

<em>Kt = ln ([A₀]/[A]),</em>

where, k is the rate constant of the reaction <em>(k = 3.36 × 10⁻⁵ s⁻¹)</em>,

t is the time of the reaction <em>(t = 235.0 min = 14100 s)</em>,

[A₀] is the initial concentration of cyclopropane <em>([A₀] = 0.0445 M)</em>

<em>∵ Kt = ln ([A₀]/[A]),</em>

∴ (3.36 × 10⁻⁵ s⁻¹)(14100 s) = ln (0.0445 M)/[A]

Taking the exponential of both sides:

1.6 = (0.0445 M)/[A]

<em>∴ [A] = (0.0445 M)/1.6 = 0.0277 M.</em>

<em />

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