Question

For a first-order homogeneous gaseous reaction, $A⟶2B+C$.
If the total pressure after time $t$ was $P_t$ and after long time $(t\to \infty )$ was $P_{\infty }$ then $k$ in terms of $P_t,P_{\infty }$ and $t$ is :

• A. $k=\frac{2.303}{t}\log \left(\frac{2P_{\infty }}{P_{\infty }-P_t}\right)$
• B. $k=\frac{2.303}{t}\log \left(\frac{2P_{\infty }}{2(P_{\infty }-P_t)}\right)$
• C. $k=\frac{2.303}{t}\log \left(\frac{2P_{\infty }}{3(P_{\infty }-P_t)}\right)$
• D. none of these

Answer 1

The correct option is C $k=\frac{2.303}{t}\log \left(\frac{2P_{\infty }}{3(P_{\infty }-P_t)}\right)$

$A\to 2B+C$
$t=0$ $P_i$ $0$ $0$
$t$ $P_t-x$ $2x$ $x$

$t\to \infty$ $0$ $2P_i$ $P_i$

$P_{\infty }=3P_i$ or $P_i=\frac{P_{\infty }}{3};P_i+2x=P_t$
$x=\frac{P_t-P_i}{2}$

As we know $k=\frac{2.303}{t}log\left(\frac{P_i}{P_i-x}\right)$

so $k=\frac{2.303}{t}log\left(\frac{2P_{\infty }}{3(P_{\infty }-P_t)}\right)$

Hence optoin (C) is correct.