Question
The rate of the reaction : $$C{H_3}COO{C_2}{H_5} + NaOH \to $$ $$C{H_3}COONa + {C_2}{H_5}OH$$ is given by the equation, $${\text{rate}} = k\left[ {C{H_3}COO{C_2}{H_5}} \right]\left[ {NaOH} \right]$$
If concentration is expressed in $$mol/L,$$ the units of $$k$$ are
A.
$$mo{l^{ - 2}}\,{L^2}\,{s^{ - 1}}$$
B.
$$mol\,{L^{ - 1}}\,{s^{ - 1}}$$
C.
$$L\,mo{l^{ - 1}}\,{s^{ - 1}}$$
D.
$${s^{ - 1}}$$
Answer :
$$L\,mo{l^{ - 1}}\,{s^{ - 1}}$$
Solution :
For a second order reaction, $$\frac{{dx}}{{dt}} = k{\left[ A \right]^2}$$
\[\frac{\text{conc}\text{.}}{\text{time}}=k{{\left[ \text{conc}\text{.} \right]}^{2}}\]
$$\frac{{mol\,{L^{ - 1}}}}{2} = k\,mol\,{L^{ - 1}} \times mol\,{L^{ - 1}};$$ $$k = L\,mo{l^{ - 1}}\,{s^{ - 1}}$$