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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)

best_answer

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) = $k[A^{'}][B^{'}]$

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

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

Posted by

manish

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