Question

A reaction $2A\to Products$, has a rate constant $3.5\times {10}^{-4}mo{l}^{-1}litr{e}^{-1}se{c}^{-1}$. If the time required for the concentration of A to change from 0.260 M to 0.011 M $x\times {10}^{5}sec$ then $x$ is___________.(to the nearest integer)

Answer 1

$k=3.5\times {10}^{4}Lit/mol/s$, $a=0.260M,a-x=0.011M,x=0.249M,t=x\times {10}^{5}s$
The units of the rate constant indicates that the reaction is second order reaction. For second order reaction, $t=\frac{x}{ka(a-x)}=\frac{0.249}{3.5\times {10}^{-4}\times 0.260\times 0.011}=2.49\times {10}^{5}s=x\times {10}^{5}s$.
Hence, $x=2.49\approx 2$.