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Tema [17]
3 years ago
7

Integrated rate law for second order unimolecular irreversible

Chemistry
1 answer:
kirill115 [55]3 years ago
5 0

Answer:

The rate law for second order unimolecular irreversible reaction is

\frac{1}{[A]} = k.t + \frac{1}{[A]_{0} }

Explanation:

A second order unimolecular irreversible reaction is

2A → B

Thus the rate of the reaction is

v = -\frac{1}{2}.\frac{d[A]}{dt} = k.[A]^{2}

rearranging the ecuation

-\frac{1}{2}.\frac{k}{dt} = \frac{[A]^{2}}{d[A]}

Integrating between times 0 to <em>t </em>and between the concentrations of [A]_{0} to <em>[A].</em>

\int\limits^0_t -\frac{1}{2}.\frac{k}{dt} =\int\limits^A_{0} _A\frac{[A]^{2}}{d[A]}

Solving the integral

\frac{1}{[A]} = k.t + \frac{1}{[A]_{0} }

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Explanation:

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Molecular formula = [CH₂.₇O]₃

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