Question

The rate of formation of a dimer in a second order dimerisation reaction is $9.1 \times 10^{-6}\, mol\, L^{-1}s^{-1}$ at $0.01\, mol\, L^{-1}$ monomer concentration. What will be the rate constant for the reaction?

• A. $9.1 \times 10^{-2} \,L \,mol^{-1} s^{-1}$
• B. $9.1 \times 10^{-6} L \,mol^{-1} s^{-1}$
• C. $3\times 10^{-4}\,L\,mol^{-1} s^{-1}$
• D. $27.3 \times 10^{-2} \,L\, mol^{-1} s^{-1}$

Answer 1

Answer: (A)

$2A \rightarrow A_{2} $
Rate of formation of dimer $= k\left[A\right]^{2}$
$k=\frac{Rate \,of \,formation \,of\, dimer}{\left[A\right]^{2}}$
$k=\frac{9.1\times10^{-6} mol L^{-1}s^{-1}}{\left(0.01 mol L^{-1}\right)^{2}}$
$=9.1\times10^{-2} L\,mol^{-1}\, s^{-1}$