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

The rate for a first order reaction is $0.6932 \times 10^{-2} \, mol \, L^{-1} \, min^{-1}$ and the initial concentration of the reactants is $1 \, M$. The half-life $T_{1/2}$ is equal to ........ $min$.

$A$ first order reaction is given as $A \rightarrow \text{products}$. Its integrated rate equation is:

If the rate constant for a first order reaction is $k$,the time $(t)$ required for the completion of $99 \%$ of the reaction is given by

For the reaction $2N_2O_{5(g)} \rightarrow 4NO_{2(g)} + O_{2(g)}$,which is a first-order reaction with respect to $N_2O_5$,which of the following plots gives a straight line?

For a first-order reaction,it takes $40 \ min$ for $90\%$ completion. The half-life of the reaction is ..... (in $min$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module