4.2     For the reaction:
           2A+B\rightarrow A_{2}B
           the rate =k\left [ A \right ]\left [ B \right ]^{2} with k=2.0\times 10^{-6}mol^{-2}L^{2}s^{-1}. Calculate the initial rate of the reaction when\left [ A \right ]=0.1 mol L^{-1}\: \: ,\left [ B \right ]=0.2 mol L^{-1}. Calculate the rate of reaction after \left [ A \right ] is reduced to 0.06molL ^{-1}.

Answers (1)
M manish

The initial rate of reaction =

 rate = k[A][B]^{2}

substitute the given values of [A], [B] and k,

rate  = 2\times 10^{-6}\times 0.1\times (0.2)^{2}

        =8\times 10^{-9}mol^{-2}\ L^{2}\ s^{-1}

When [A] is reduced from 0.1 mol/L to 0.06 mol/L [A^{'}=0.06]

So, conc. of A reacted = 0.1-0.06 = 0.04 mol/L

and conc. of B reacted = 1/2(0.04) = 0.02mol/L

conc. of B left = (0.2-0.02) = 0.18 mol/L    [B^{'}=0.18]

Now,  the rate of the reaction is (R) = k[A^{'}][B^{'}]

                                                         =2\times 10^{-6}\times 0.06\times (0.18)^{2}

                                                        =3.89\times 10^{-6} mol L^{-1}s^{-1}

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