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What is the rate constant of a first-order reaction that takes 7.30 minutes for the reactant concentration to drop to half of its initial value?

User Jayne
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1 Answer

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First-order reaction => r = - dC / dt k * C^m

Where: m is the order of the reaction, which is 1, r is the rate of reaction, k is the rate constant, C is the concentration, and dC / dt is the derivative of the concentration respect time.

=> - dC/dt = K * C

Solving: dC / C = k dt

=> - [ln(C) - ln Co] = kt, where Co is the initial concentration (at t = 0)

=> - ln (C / Co) = kt

Concentration dropo to half => C / Co = 1/2

=> - ln (1/2) = k * 7.30 min

=> k = ln(2) / 7.30 min = 0.095 M / min

Answer: k = 0.095 M / min
User Foysal Osmany
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