A flask contains a mixture of compounds A and | vedclass

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

Question

(D) For a first-order reaction,the concentration at time $t$ is given by $[C]_t = [C]_0 e^{-kt},$ where $k = \frac{\ln 2}{t_{1/2}}.$Given $[A]_0 = [B]

Vedclass · Accepted Answer

$900$

Vedclass · Answer

(D) For a first-order reaction,the concentration at time $t$ is given by $[C]_t = [C]_0 e^{-kt},$ where $k = \frac{\ln 2}{t_{1/2}}.$Given $[A]_0 = [B]_0,$ we want to find $t$ such that $[A]_t = 4[B]_t.$Substituting the expressions: $[A]_0 e^{-(\ln 2 / 300)t} = 4[B]_0 e^{-(\ln 2 / 180)t}.$Since $[A]_0 = [B]_0,$ we have $e^{-(\ln 2 / 300)t} = 4 e^{-(\ln 2 / 180)t}.$Rearranging gives $e^{(\frac{\ln 2}{180} - \frac{\ln 2}{300})t} = 4.$Taking the natural logarithm on both sides: $(\frac{\ln 2}{180} - \frac{\ln 2}{300})t = \ln 4 = 2 \ln 2.$Dividing by $\ln 2$: $(\frac{1}{180} - \frac{1}{300})t = 2.$Solving for $t$: $(\frac{300 - 180}{180 	imes 300})t = 2 \Rightarrow \frac{120}{54000}t = 2.$$t = \frac{2 	imes 54000}{120} = \frac{108000}{120} = 900 \ s.$

$t/min$	Pressure of system at time $t/mm \ Hg$
$10$	$160$
$\infty$	$240$

Similar Questions

$A$ molecule undergoes two independent first order reactions whose respective half-lives are $12 \ min$ and $3 \ min$. If both the reactions are occurring simultaneously,the time taken for the $50 \%$ consumption of the reactant is $.......... \ min$ (Nearest integer).

The reaction $X \rightarrow$ products is a first order reaction. In $40 \ min$,the concentration of $X$ changes from $1.0 \ M$ to $0.25 \ M$. What is the rate of reaction when $[X] = 0.1 \ M$? $(\log 4 = 0.60)$

Reaction $A_{(g)} \rightarrow 2B_{(g)} + C_{(g)}$ is a first order reaction. It was started with pure $A$.
$t/min$ Pressure of system at time $t/mm \ Hg$
$10$ $160$
$\infty$ $240$
Which of the following option is incorrect?

An organic compound undergoes first-order decomposition. The time taken for its decomposition to $1/8$ and $1/10$ of its initial concentration are $t_{1/8}$ and $t_{1/10}$ respectively. What is the value of $\frac{t_{1/8}}{t_{1/10}} \times 10$? (Given: $\log_{10} 2 = 0.3$)

What is the half-life of a first-order reaction if the time required to decrease the concentration of the reactant from $0.4 \ M$ to $0.1 \ M$ is $x \ \text{hours}$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module