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For the non-equillbrium process,
$A + B ⟶ products$ the rate is first-order w.r.t. A and second-order w.r.t. B. If $1.0$ mole each of A and B introduced into a $1.0 L$ vessel and the initial rate was $1.0 \times 10^{- 2}$ mol, $L^{- 1} g^{- 1}$ rate when half reactants have been turned into products is:

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$1.25 \times 10^{- 3} mol L^{- 1} s^{- 1}$

$1.0 \times 10^{- 2} mol L^{- 1} g^{- 1}$

$2.50 \times 10^{- 3} mol L^{- 1} s^{- 1}$

$2.0 \times 10^{- 2} mol L^{- 1} B^{- 1}$

answer is A.

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Detailed Solution

$R a t e = k [A] [B]^{2}$

The number of moles of the product when half of the reactants are transformed to $A = 0.5 mol and B = 0.5 mol$

When half of reactions have produced products, the rate is,

$Rate = k [0.5 \times A] [0.5 \times B]^{2}$

$Rate = \frac{1 \times 10^{- 2}}{8}$

$= 1.25 \times 10^{- 3} {molL}^{- 1} s^{- 1}$

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