Question

Which integrated equation is correct for the following $1^{st}$ order reaction started with only $A\left(g\right)$ in a closed rigid vessel?
$A\left(g\right)\to B\left(g\right)+C\left(g\right)+D\left(g\right)$

where, $P_i=$ initial pressure $;P_t=$ total pressure at time $t$

• A. $K=\frac{2.303}{t}{\log }_{10}\left[\frac{P_i}{P_t}\right]$
• B. $K=\frac{2.303}{t}{\log }_{10}\left[\frac{P_t}{P_i}\right]$
• C. $K=\frac{2.303}{t}{\log }_{10}\left[\frac{2P_i}{3P_i-P_t}\right]$
• D. $K=\frac{2.303}{t}{\log }_{10}\left[\frac{3P_i}{2P_i-3P_t}\right]$

Answer 1

Answer: (C)

$K=\frac{2.303}{t}{\log }_{10}\left[\frac{2P_i}{3P_i-P_t}\right]$

$A\left(g\right)⟶B\left(g\right)+C\left(g\right)+D\left(g\right)$

Initial $Pi$ $0$ $0$ $0$

at time $t$, $Pi-p$ $p$ $p$ $p$

$\therefore$ total pressure at $t$, $P_t=P_i-p+p+p+p=P_i+2p$

Since the reaction is $1^{st}$ order reaction.

$\therefore$ rate constant, $K=\frac{2.303}{t}log\left(\frac{P_i}{P_i-p}\right)$

$\Rightarrow K=\frac{2.303}{t}log\frac{P_i}{(P_i-\frac{P_t-P_i}{2})}$

$\Rightarrow K=\frac{2.303}{t}log\left(\frac{2P_i}{{3P}_i-P_t}\right)$ $\therefore$ Correct answer is C.