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(D) The rate constant for a second-order reaction with equal initial concentrations is given by $k = \frac{1}{t} \cdot \frac{x}{a(a-x)}$.Let the initi

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$500$

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(D) The rate constant for a second-order reaction with equal initial concentrations is given by $k = \frac{1}{t} \cdot \frac{x}{a(a-x)}$.Let the initial concentration $a = 1$.For $60\%$ completion,$x = 0.6$ and $t = 3000 \ s$.$k = \frac{1}{3000} \cdot \frac{0.6}{1(1-0.6)} = \frac{1}{3000} \cdot \frac{0.6}{0.4} = \frac{1}{3000} \cdot 1.5 = \frac{1}{2000} \ s^{-1}$.For $20\%$ completion,$x = 0.2$ and $a = 1$.$t = \frac{1}{k} \cdot \frac{x}{a(a-x)} = 2000 \cdot \frac{0.2}{1(1-0.2)} = 2000 \cdot \frac{0.2}{0.8} = 2000 \cdot 0.25 = 500 \ s$.

$Expt. \ No.$	$[A]$	$[B]$	$Initial \ Rate$
$1$	$0.012$	$0.035$	$0.10$
$2$	$0.024$	$0.070$	$0.80$
$3$	$0.024$	$0.035$	$0.10$
$4$	$0.012$	$0.070$	$0.80$

For a second-order reaction where the initial concentrations of both reactants are equal,it takes $3000 \ s$ for the reaction to be $60\%$ complete. How much time (in $s$) will it take for the reaction to be $20\%$ complete?

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$Expt. \ No.$ $[A]$ $[B]$ $Initial \ Rate$

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