Question

In a homogeneous reaction $A⟶B+C+D$ the initial pressure was $P_0$ and after time $t$ it was $P$. Expression for rate constant $k$ in terms of $P_0,$ $P$ and $t$ will be:

• A. $k=\frac{2.303}{t}\log \frac{2P_0}{{3P}_0-P}$
• B. $k=\frac{2.303}{t}\log \frac{2P_0}{P_0-P}$
• C. $k=\frac{2.303}{t}\log \frac{{3P}_0-P}{2P_0}$
• D. $k=\frac{2.303}{t}\log \frac{2P_0}{{3P}_0-2P}$

Answer 1

The correct option is A $k=\frac{2.303}{t}\log \frac{2P_0}{{3P}_0-P}$

$A⟶B+C+D$
Initial $a$ $0$ $0$ $0$
After time $t$ $a-x$ $x$ $x$ $x$
It is given time $a=P_0...................\left(i\right)$
$a-x+x+x+x=P.....\left(ii\right)$
From $\left(i\right)$
$P_0+2x=P$ or $x=\frac{P-P_0}{2}$
From rate equation
$k=\frac{2.303}{t}\log \frac{a}{a-x}$
$k=\frac{2.303}{t}\log \frac{2P_0}{{3P}_0-P}$