Question

The decomposition of $C{l}_2{O}_7$ at $400$ K in the gas phase to $C{l}_2$ and $O_2$ is of $1^{st}$ order. After $55$ sec at 400 K, the pressure of $C{l}_2{O}_7$ falls from $0.062$ to $0.044$ atm.

The rate constant is :

• A. $6.24\times {10}^{-3}se{c}^{-1}$
• B. $5.20\times {10}^{-3}se{c}^{-1}$
• C. $7.34\times {10}^{-3}se{c}^{-1}$
• D. none of the above

Answer 1

The correct option is A $6.24\times {10}^{-3}se{c}^{-1}$

The given reaction is:

$C{l}_2{O}_7\to C{l}_2+\frac{7}{2}{O}_2$

Mole at $t=0$ sec	$a$	0	0
Mole at $t=55$ sec	$a-x$	$x$	$7x/2$

Since, mole at any time $\propto$ pressure at that time (at constant V and T)

$\therefore a\propto 0.062$

and $(a-x)\propto 0.044$ at $t=55sec$

$\because K=\frac{2.303}{t}log\frac{a}{(a-x)}=\frac{2.303}{55}log\frac{0.062}{0.044}$

$=6.24\times {10}^{-3}se{c}^{-1}$

Hence, the correct option is $A$