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Question: For a reaction 2A + B \(\rightarrow\) product, rate law is –\(\frac{d\lbrack A\rbrack}{dt}\) = k[A]...

For a reaction 2A + B →\rightarrow product, rate law is –d[A]dt\frac{d\lbrack A\rbrack}{dt}

= k[A]. At a time when t = 1k\frac{1}{k}, concentration of the reactant is : (C0 = initial concentration)

A

C0e\frac{C_{0}}{e}

B

C0e

C

C0e2\frac{C_{0}}{e^{2}}

D

1C0\frac{1}{C_{0}}

Answer

C0e\frac{C_{0}}{e}

Explanation

Solution

2A+B→product2A + B\overset{\quad\quad}{\rightarrow}product $- \frac{d\lbrack A\rbrack}{dt} = K\lbrack A\rbrack

}{C_{t} = C_{0}e^{- Kt} }{C_{t} = C_{0}e^{- Kx\frac{1}{K}} }{C_{t} = C_{0}e^{- 1} }{C_{t} = \frac{C_{0}}{e}}$$