Question

The conversion of trypsinogen (A) into trypsin (B) is an autocatalytic reaction, $A\to B$, where, B catalyses the reaction. The rate equation is $\frac{-dx}{dt}=K\cdot x\cdot y$, where x and y are concentration of tripsinogen at time t. Integrate this equation for initial concentration of $x_0$ and $y_0$ for A and B.what is the value of $Kt$ ?

• A. $Kt=\frac{2.303}{x_0+y_0}log\frac{y\cdot x_0}{x\cdot y_0}$
• B. $Kt=\frac{23.03}{x_0+y_0}log\frac{x\cdot y_0}{x\cdot y_0}$
• C. $Kt=\frac{2.303}{x_0-y_0}log\frac{y\cdot x_0}{x\cdot y_0}$
• D. None of these

Answer 1

The correct option is A $Kt=\frac{2.303}{x_0+y_0}log\frac{y\cdot x_0}{x\cdot y_0}$

$\underset{x_0}{A}\to \underset{y_0}{B}$
$\underset{=x}{(x_0-a)}\underset{=y}{(y_0+a)}$
$\therefore x+y=x_0-a+y_0+a=x_0+y_0$
$-\frac{dx}{dt}=kx\cdot y$
Put $x_0-a=x$
$\therefore -\frac{da}{dt}=\frac{dx}{dt}$
$\frac{da}{ddt}=K(x_0-a)(y_0+a)$
$=K[x_0{y}_0+a{x}_0-a{y}_0-a^2]$
$=K[x_0{y}_0+(x_0-y_0)a-a^2]$
$=K\left[\right(x_0{y}_0)-\{a^2-2\left(\frac{x_0-y_0}{2}\right)\cdot a-{\left(\frac{x_0-y_0}{2}\right)}^2+{\left(\frac{x_0-y_0}{2}\right)}^2\}]$
$=K[x_0{y}_0-{(a-\frac{x_0-y_0}{2})}^2+{\left(\frac{x_0-y_0}{2}\right)}^2]$
$=K[{\left(\frac{x_0+y_0}{2}\right)}^2-{(a-\left(\frac{x_0-y_0}{2}\right))}^2]$
On integrating,
$Kt+C=\frac{1}{\frac{2(x_0+y_0)}{2}}log\frac{\frac{x_0+y_0}{2}+[a-\frac{x_0-y_0}{2}]}{\frac{x_0+y_0}{2}-[a-\frac{x_0-y_0}{2}]}$
$Kt+C=\frac{1}{x_0+y_0}log\frac{a+y_0}{(x_0-a)}$
at $t=0,a=0$
$C=\frac{1}{x_0+y_0}log\frac{y_0}{x_0}$
$\therefore Kt=\frac{1}{x_0+y_0}log\frac{a+y_0}{(x_0-a)}\cdot \frac{x_0}{y_0}=\frac{1}{x_0+y_0}log\frac{y\cdot x_0}{x\cdot y_0}$
$=\frac{2.303}{(x_0+y_0)}log\frac{y\cdot x_0}{x\cdot y_0}$