For a chemical reaction,2A + 2B to C + D,the | vedclass

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(B) The rate law for the reaction is given by: $	ext{rate} = k[A]^1[B]^1 = k[A][B]$.Initial rate: $R_1 = k[A]_0[B]_0 = 4 	imes 10^{-2} \ mol \ L^{-1

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$1 	imes 10^{-2} \ mol \ L^{-1} \ s^{-1}$

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(B) The rate law for the reaction is given by: $	ext{rate} = k[A]^1[B]^1 = k[A][B]$.Initial rate: $R_1 = k[A]_0[B]_0 = 4 	imes 10^{-2} \ mol \ L^{-1} \ s^{-1}$.When $50\%$ of the reactants are converted into products,the concentration of each reactant becomes half of its initial concentration:$[A] = \frac{[A]_0}{2}$ and $[B] = \frac{[B]_0}{2}$.New rate: $R_2 = k \left( \frac{[A]_0}{2} ight) \left( \frac{[B]_0}{2} ight) = \frac{1}{4} k[A]_0[B]_0$.Substituting the initial rate value:$R_2 = \frac{1}{4} 	imes (4 	imes 10^{-2} \ mol \ L^{-1} \ s^{-1}) = 1 	imes 10^{-2} \ mol \ L^{-1} \ s^{-1}$.Therefore,the correct option is $B$.

$Expt.$	$[X]_0 / mol \ L^{-1}$	$[Y]_0 / mol \ L^{-1}$	$Rate / mol \ L^{-1} s^{-1}$
$1$	$0.25$	$0.25$	$1.0 \times 10^{-6}$
$2$	$0.50$	$0.25$	$4.0 \times 10^{-6}$
$3$	$0.25$	$0.50$	$8.0 \times 10^{-6}$

For a chemical reaction,$2A + 2B \to C + D$,the order of reaction is $1$ with respect to $A$ and $1$ with respect to $B$. The initial rate of the reaction is $4 \times 10^{-2} \ mol \ L^{-1} \ s^{-1}$. When $50\%$ of the reactants are converted into products,the rate of the reaction would become:

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