Consider a reaction that is first order in bo | vedclass

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

Question

(C) For a reversible first-order reaction $A \underset{K_b}{\stackrel{K_f}{\rightleftharpoons}} B$,the rate law is given by: $-\frac{d[A]}{dt} = K_f[A

Vedclass · Accepted Answer

$t = \frac{\ln 2}{(K_{f} + K_{b})}$

Vedclass · Answer

(C) For a reversible first-order reaction $A \underset{K_b}{\stackrel{K_f}{ightleftharpoons}} B$,the rate law is given by: $-\frac{d[A]}{dt} = K_f[A] - K_b[B]$.At equilibrium,$-\frac{d[A]}{dt} = 0$,so $K_f[A_{	ext{eq}}] = K_b[B_{	ext{eq}}]$. Since $[B_{	ext{eq}}] = A_0 - A_{	ext{eq}}$,we have $K_b = \frac{K_f A_{	ext{eq}}}{A_0 - A_{	ext{eq}}}$.The integrated rate equation for this system is $(K_f + K_b)t = \ln \left( \frac{A_0 - A_{	ext{eq}}}{A_t - A_{	ext{eq}}} ight)$.Given $A_t = \frac{A_0 + A_{	ext{eq}}}{2}$,we substitute this into the expression:$A_t - A_{	ext{eq}} = \frac{A_0 + A_{	ext{eq}}}{2} - A_{	ext{eq}} = \frac{A_0 - A_{	ext{eq}}}{2}$.Substituting this into the integrated rate equation:$(K_f + K_b)t = \ln \left( \frac{A_0 - A_{	ext{eq}}}{(A_0 - A_{	ext{eq}})/2} ight) = \ln(2)$.Therefore,$t = \frac{\ln 2}{K_f + K_b}$.

Similar Questions

The equilibrium constant at $298 \ K$ for a reaction $A + B \rightleftharpoons C + D$ is $100$. If the initial concentration of all the four species were $1 \ M$ each,then the equilibrium concentration of $D$ (in $mol \ L^{-1}$) will be:

For the reaction $N_2 + 3H_2 \rightleftharpoons 2NH_3$,what is the relationship between $\Delta H$ and $\Delta E$?

In a closed vessel,the decomposition of solid $Mg(HCO_3)_2$ is given as follows:
$Mg(HCO_3)_{2(s)} \rightleftharpoons MgCO_{3(s)} + CO_{2(g)} + H_2O_{(g)}$
Given $K_p = 64 \ bar^2$ and the mole fraction of $CO_2$ is $0.8$,calculate the total pressure at equilibrium. (in $bar$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module