 # Calculate the rate of reaction after [A] is reduced to 0.06

For the reaction, 2A + B > \$A _{ 2 }B\$;
: The rate = k [A] [\${{B}^{2}}\$] with
k= 2.0x \${{10}^{-6}}\$\${{mol}^{-2}}\$ \${{L}^{2}}\$ \${{s}^{-1}}\$. Calculate the initial rate of reaction when [A] = 0.1mol\${{L}^{-1}}\$
[B] =0.2 mol \${{L}^{-1}}\$. Calculate the rate of reaction after [A] is reduced to 0.06 mol \${{L}^{-1}}\$.

(i) Case I Initial rate = k [A] \${{B}^{2}}\$
= 2.0 x \${{10}^{-6}}\$ \${{mol}^{-2}}\$\${{L}^{2}}\$\${{s}^{-1}}\$) x (0.1 mol \${{L}^{-1}}\$) x (0.2 x mol \${{L}^{-1}}\$)
=8.0 x \${{10}^{-9}}\$mol \${{L}^{-1}}\$\${{s}^{-1}}\$
(ii) Case II Concentration of A at a particular time = 0.06 mol \${{L}^{-1}}\$
Amount of A reacted = (0.1 - 0.06) = 0.04 mol \${{L}^{-1}}\$
Amount of B reacted = 1/2 x 0.04 mol \${{L}^{-1}}\$ = 0.02 mol \${{L}^{-1}}\$
Concentration of B at a particular time = (0.2 - 0.02) mol \${{L}^{-1}}\$= 0.18 mol \${{L}^{-1}}\$
Rate = k [A] [\${{B}^{2}}\$] =
(2.0 x \${{10}^{-6}}\$ \${{mol}^{-2}}\$\${{L}^{2}}\$\${{s}^{-1}}\$) x
(0.06 mol \${{L}^{-1}}\$) x
(0.18mol\${{L}^{-1}}\$=
3.89x\${{10}^{-9}}\$mol\${{L}^{-1}}\$\${{s}^{-1}}\$