For a first-order reaction at 100,^oC,the rat | vedclass

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

Question

(D) For a first-order reaction,the integrated rate equation is: $\ln([A]_t / [A]_0) = -Kt$ Given: $K = 1.5 	imes 10^{-2} \, s^{-1}$ $t = 10 \, min =

Vedclass · Accepted Answer

$10^{-2} \, mol \, L^{-1}$

Vedclass · Answer

(D) For a first-order reaction,the integrated rate equation is: $\ln([A]_t / [A]_0) = -Kt$ Given: $K = 1.5 	imes 10^{-2} \, s^{-1}$ $t = 10 \, min = 600 \, s$ $[A]_0 = 100 \, mol \, L^{-1}$ Substituting the values: $\ln([A]_t / 100) = -(1.5 	imes 10^{-2} \, s^{-1}) 	imes (600 \, s)$ $\ln([A]_t / 100) = -9$ $[A]_t / 100 = e^{-9}$ $[A]_t = 100 	imes e^{-9} \approx 100 	imes 1.23 	imes 10^{-4} \approx 1.23 	imes 10^{-2} \, mol \, L^{-1}$ Since the calculated value is approximately $10^{-2} \, mol \, L^{-1}$,the correct option is $D$.

For a first-order reaction at $100\,^oC$,the rate constant is $K = 1.5 \times 10^{-2} \, s^{-1}$. If the initial concentration of the reactant is $100 \, mol \, L^{-1}$,what will be the concentration of the reactant after $10 \, min$?

Similar Questions

For the reaction $A \rightarrow B$,the following graph was obtained. The time required (in seconds) for the concentration of $A$ to reduce to $2.5 \ g \ L^{-1}$ (if the initial concentration of $A$ was $50 \ g \ L^{-1}$) is $........$ (Nearest integer). Given: $\log 2 = 0.3010$.

$A$ first-order reaction takes $30$ minutes for $75\%$ decomposition. Calculate its rate constant. (in $\text{ min}^{-1}$)

For a first-order reaction,the time taken for the concentration of the reactant to reach $3/4$ of its initial value is $t_{1/4}$. If the rate constant for the reaction is $K$,then $t_{1/4}$ can be expressed as: (in $/K$)

$A$ first order reaction is $75 \%$ completed in $60 \ min$. The time required for its $50 \%$ completion is: (in $min$)

The half-life period of a first-order chemical reaction is $6.93 \, \text{min}$. The time required for the completion of $99 \%$ of the chemical reaction will be ........ $\text{min}$. $(\log 2 = 0.301)$

Vedclass Test Series

Exam Paper Generator

Online Exam Module