For a process $A + B ⟶$ product, the rate of reaction is second order with respect to A and zero order with respect to B. When 1 mole each of A and B are taken in 1 litre vessel the initial rate is $1 \times 10^{- 2} mol / L - \sec$ . The rate of reaction when 50% of the reactant have been converted to product would be

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$5 \times 10^{- 2} mol / L - \sec$

$2.5 \times 10^{- 3} mol / L - \sec$

$1 \times 10^{- 2} mol / L - \sec$

$5 \times 10^{- 3} mol / L - \sec$

answer is B.

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Detailed Solution

The rate law formula is $r = k [A]^{2} [B]^{0}$

The starting price is $1 \times 10^{- 2}$

Hence, $1 \times 10^{- 2} = k [1]^{2} [1]^{0}$

$k = 1 \times 10^{- 2}$

Reactant quantity equals 0.5 when 50% of the reactants are transformed into products.

Thus, $r = k [A]^{2} [B]^{0} = 1 \times 10^{- 2} [0.5]^{2} [0.5]^{0} = 2.5 \times 10^{- 3}$

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