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Question: The rate of formation of a dimer in a second order dimerisation reaction is \(9.1 \times 10^{- 6}\)m...

The rate of formation of a dimer in a second order dimerisation reaction is 9.1×10−69.1 \times 10^{- 6}mol L−1s−1L^{- 1}s^{- 1} at

0.01molL−10.01molL^{- 1} monometer concentration . what will be the rate constant for the reaction?

A

9.1×10−2Lmol−1s−19.1 \times 10^{- 2}Lmol^{- 1}s^{- 1}

B

9.1×10−6Lmol−1s−19.1 \times 10^{- 6}Lmol^{- 1}s^{- 1}

C

3×10−4Lmol−1s−13 \times 10^{- 4}Lmol^{- 1}s^{- 1}

D

27.3×10−2Lmol−1s−127.3 \times 10^{- 2}Lmo{l^{-}}^{1}s^{- 1}

Answer

9.1×10−2Lmol−1s−19.1 \times 10^{- 2}Lmol^{- 1}s^{- 1}

Explanation

Solution

2A→A22A \rightarrow A_{2}

Rate of formation of dimer =k[A]2= k\lbrack A\rbrack^{2}

k=Rateofformationofdim⁡er[A]2k = \frac{Rateofformationof\dim er}{\lbrack A\rbrack^{2}}

k=9.1×10−6molL−1s−1(0.01molL−1)2=9.1×10−2Lmol−1s−1k = \frac{9.1 \times 10^{- 6}molL^{- 1}s^{- 1}}{(0.01molL^{- 1})^{2}} = 9.1 \times 10^{- 2}Lmol^{- 1}s^{- 1}