The slope of a graph log [A]_t versus 't' for | vedclass

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(C) For a first-order reaction, the integrated rate equation is $\ln [A]_t = -kt + \ln [A]_0$. Converting this to base $10$ logarithm: $\log [A]_t = -

Vedclass · Accepted Answer

$5.757 	imes 10^{-3} \,s^{-1}$

Vedclass · Answer

(C) For a first-order reaction, the integrated rate equation is $\ln [A]_t = -kt + \ln [A]_0$. Converting this to base $10$ logarithm: $\log [A]_t = -\frac{k}{2.303}t + \log [A]_0$. The slope of the graph $\log [A]_t$ versus '$t$' is given by $m = -\frac{k}{2.303}$. Given slope $m = -2.5 	imes 10^{-3} \,s^{-1}$. Therefore, $-\frac{k}{2.303} = -2.5 	imes 10^{-3} \,s^{-1}$. $k = 2.5 	imes 10^{-3} 	imes 2.303 \,s^{-1} = 5.7575 	imes 10^{-3} \,s^{-1}$. Thus, the rate constant is $5.757 	imes 10^{-3} \,s^{-1}$.

$S$.No.	Time	Total pressure in Pascal
$1$.	At the end of $10 \ min$	$300$
$2$.	After completion	$200$

The slope of a graph $\log [A]_t$ versus '$t$' for a first-order reaction is $-2.5 \times 10^{-3} \,s^{-1}$. Find the rate constant of the reaction.

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