The half-life of a first-order reaction is 6. | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) For a first-order reaction,the rate constant $k$ is given by: $k = \frac{0.693}{t_{1/2}} = \frac{0.693}{6.0 \ h} = 0.1155 \ h^{-1}$ Given initial

Vedclass · Accepted Answer

$10.42$

Vedclass · Answer

(B) For a first-order reaction,the rate constant $k$ is given by: $k = \frac{0.693}{t_{1/2}} = \frac{0.693}{6.0 \ h} = 0.1155 \ h^{-1}$ Given initial concentration $[A]_0 = 0.4 \ M$ and final concentration $[A]_t = 0.12 \ M$. Using the integrated rate equation for a first-order reaction: $t = \frac{2.303}{k} \log_{10} \frac{[A]_0}{[A]_t}$ $t = \frac{2.303}{0.1155 \ h^{-1}} 	imes \log_{10} \left( \frac{0.4}{0.12} ight)$ $t = \frac{2.303}{0.1155} 	imes \log_{10} (3.333)$ $t = 19.939 	imes 0.5228 \approx 10.42 \ h$

The half-life of a first-order reaction is $6.0 \ h$. How long will it take for the concentration of the reactant to decrease from $0.4 \ M$ to $0.12 \ M$ (in $h$)?

Similar Questions

If the rate law is $r = K[A]$,then the concentration of the reactant remaining after time $t = 1/k$ is: ($[A]_0$ is the concentration of the reactant at $t = 0$)

The integrated rate equation is $Rt = \log C_0 - \log C_t$. The straight line graph is obtained by plotting

For the reaction $2A + B \to \text{Product}$,the rate law is given as $\frac{-d[A]}{dt} = K[A]$. At a time when $t = \frac{1}{K}$,the concentration of the reactant $A$ is ($Co =$ initial concentration).

For a first order reaction,the half-life period is

If $50 \%$ of the reactant is converted into a product in a first order reaction in $25 \ min$,how much of it would react in $100 \ min$ (in $\%$)?

Vedclass Test Series

Exam Paper Generator

Online Exam Module