The rate constant k = 2.303 min^{-1} for a p | vedclass

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(B) The reaction is a $1^{st}$ order reaction because the unit of the rate constant is $min^{-1}$.For a $1^{st}$ order reaction,the integrated rate la

Vedclass · Accepted Answer

$0.2303 \ M \ min^{-1}$

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(B) The reaction is a $1^{st}$ order reaction because the unit of the rate constant is $min^{-1}$.For a $1^{st}$ order reaction,the integrated rate law is $k = \frac{2.303}{t} \log \frac{[A_0]}{[A]}$.Given: $k = 2.303 \ min^{-1}$,$[A_0] = 1 \ mol/L$,and $t = 1 \ min$.Substituting the values: $2.303 = \frac{2.303}{1} \log \frac{1}{[A]}$.This simplifies to $1 = \log \frac{1}{[A]}$,which means $\frac{1}{[A]} = 10^1 = 10$.Therefore,$[A] = 0.1 \ mol/L$.The rate of reaction is given by $Rate = k 	imes [A]$.$Rate = 2.303 \ min^{-1} 	imes 0.1 \ M = 0.2303 \ M \ min^{-1}$.

The rate constant $k = 2.303 \ min^{-1}$ for a particular reaction,and the initial concentration of the reactant is $1 \ mol/L$. What is the rate of reaction after $1 \ minute$?

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