Question
The rate of a reaction increases four-fold when the concentration of reactant is increased 16 times. If the rate of reaction is $$4 \times {10^{ - 6}}mol\,{L^{ - 1}}{s^{ - 1}}$$ when the concentration of the reactant is $$4 \times {10^{ - 4}}\,mol\,{L^{ - 1}}.$$ The rate constant of the reaction will be
A.
$$2 \times {10^{ - 4}}\,mo{l^{\frac{1}{2}}}{L^{ - \frac{1}{2}}}{s^{ - 1}}$$
B.
$$1 \times {10^{ - 2}}{s^{ - 1}}$$
C.
$$2 \times {10^{ - 4}}mo{l^{ - \frac{1}{2}}}{L^{\frac{1}{2}}}{s^{ - 1}}$$
D.
\[25\,mo{{l}^{-1}}L\,{{\min }^{-1}}\]
Answer :
$$2 \times {10^{ - 4}}\,mo{l^{\frac{1}{2}}}{L^{ - \frac{1}{2}}}{s^{ - 1}}$$
Solution :
$$\eqalign{
& {\text{Rate}} \propto \sqrt {{\text{Concentration}}} = k\sqrt {{\text{Concentration}}} \cr
& k = \frac{{{\text{Rate}}}}{{{{\left( {{\text{Concentration}}} \right)}^{\frac{1}{2}}}}} \cr
& = \frac{{4 \times {{10}^{ - 6}}}}{{{{\left( {4 \times {{10}^{ - 4}}} \right)}^{\frac{1}{2}}}}} \cr
& = \frac{{4 \times {{10}^{ - 6}}}}{{2 \times {{10}^{ - 2}}}} \cr
& = 2 \times {10^{ - 4}}mo{l^{\frac{1}{2}}}{L^{ - \frac{1}{2}}}{s^{ - 1}} \cr} $$