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.
$$9.1 \times {10^{ - 2}}\,L\,mo{l^{ - 1}}\,{s^{ - 1}}$$
B.
$$9.1 \times {10^{ - 6}}\,L\,mo{l^{ - 1}}\,{s^{ - 1}}$$
C.
$$3 \times {10^{ - 4}}\,L\,mo{l^{ - 1}}\,{s^{ - 1}}$$
D.
$$27.3 \times {10^{ - 2}}\,L\,mo{l^{ - 1}}\,{s^{ - 1}}$$
Answer :
$$9.1 \times {10^{ - 2}}\,L\,mo{l^{ - 1}}\,{s^{ - 1}}$$
Solution :
$$\eqalign{
& 2A \to {A_2} \cr
& {\text{Rate of formation of dimer}} = k{\left[ A \right]^2} \cr
& k = \frac{{{\text{Rate of}}\,{\text{formation of dimer}}}}{{{{\left[ A \right]}^2}}} \cr
& k = \frac{{9.1 \times {{10}^{ - 6}}\,mol\,{L^{ - 1}}\,{s^{ - 1}}}}{{{{\left( {0.01\,mol\,{L^{ - 1}}} \right)}^2}}} \cr
& \,\,\,\, = 9.1 \times {10^{ - 2}}\,L\,mo{l^{ - 1}}\,{s^{ - 1}} \cr} $$