The reaction, $A + B \to$ products, is first order with respect of $A$ and second order with respect to $B$ . When $1.0$ mole each of $A$ and $B$ were taken in one litre flask, the initial rate of the reaction is $1.0 \times 10^{- 2} m o l L^{- 1} s^{- 1}$ . The rate of the reaction when $50$ % of the reactants have been converted into products is:

1.00 \times 10^{- 3} m o l L^{- 1} s^{- 1}

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0.05 \times 10^{- 2} m o l L^{- 1} s^{- 1}

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1.25 \times 10^{- 3} m o l L^{- 1} s^{- 1}

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4.00 \times 10^{- 2} m o l L^{- 1} s^{- 1}

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2.00 \times 10^{- 3} m o l L^{- 1} s^{- 1}

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Solution

The correct option is C $1.25 \times 10^{- 3} m o l L^{- 1} s^{- 1}$
$A + B \to P r o d u c t (C)$
Initially no. of moles $110$
At time= t $\frac{1}{2} \frac{1}{2} 1$

$∴$ Rate of reaction, $r = k [A] [B]^{2}$

$= 1 \times 10^{- 2} (\frac{1}{2}) {(\frac{1}{2})}^{2}$

$= 1.25 \times 10^{- 3} m o l L^{- 1} s^{- 1}$ .

Hence, option C is correct.

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The correct option is C 1.25×10−3mol L−1s−1 A+B→Product(C)Initially no. of moles 1 1 0At time= t 12 12 1∴ Rate of reaction, r=k[A][B]2 =1×10−2(12)(12)2 =1.25×10−3 mol L−1s−1.Hence, option C is correct.