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(C) For a first order reaction,the rate constant $K$ is given by:$K = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$Given:Initial concentration $[A]_0 = \f

Vedclass · Accepted Answer

$4.606 	imes 10^{-3} \; \sec^{-1}$

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(C) For a first order reaction,the rate constant $K$ is given by:$K = \frac{2.303}{t} \log \frac{[A]_0}{[A]_t}$Given:Initial concentration $[A]_0 = \frac{1}{10} \; M = 0.1 \; M$Final concentration $[A]_t = \frac{1}{100} \; M = 0.01 \; M$Time $t = 8 \; 	ext{minutes} + 20 \; 	ext{seconds} = (8 	imes 60) + 20 = 500 \; \sec$Substituting the values:$K = \frac{2.303}{500} \log \frac{0.1}{0.01}$$K = \frac{2.303}{500} \log 10$Since $\log 10 = 1$:$K = \frac{2.303}{500} = 0.004606 \; \sec^{-1}$$K = 4.606 	imes 10^{-3} \; \sec^{-1}$

$t \ (sec)$	$P \ (mm \ of \ Hg)$
$0$	$35.0$
$360$	$54.0$
$720$	$63.0$

$A$ first order reaction was started with a decimolar solution of the reactant. $8$ minutes and $20$ seconds later,its concentration was found to be $M/100$. The rate constant of the reaction is:

Similar Questions

If benzene diazonium chloride undergoes first order decomposition at $T \ K$ with a rate constant of $6.93 \times 10^{-2} \ min^{-1}$,the time for completion of $90 \%$ of the reaction (in $min$) is (nearest integer) $(\log 2 = 0.30, \log 3 = 0.477)$.

$A$ first-order reaction decomposes such that the time taken for $1/8$ and $1/10$ of the initial concentration to decompose are $t_{1/8}$ and $t_{1/10}$ respectively. Find the value of $\frac{t_{1/8}}{t_{1/10}} \times 10$. (Given: $\log_{10} 2 = 0.3$)

If the time required for $90\%$ completion of a first-order reaction is '$t$', what is the time required for $99.9\%$ completion of the reaction at the same temperature?

For the decomposition of azoisopropane to hexane and nitrogen at $543 \ K,$ the following data are obtained.

$t \ (sec)$ $P \ (mm \ of \ Hg)$

$0$ $35.0$

$360$ $54.0$

$720$ $63.0$

Calculate the rate constant.

$A$ first order reaction requires $30 \ min$ for $50\%$ completion. The time required to complete the reaction by $75\%$ will be .......... $\min.$

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