Question

The decomposition of $N_2{O}_5$ according to the equation $2N_2{O}_5\left(g\right)\to 4N{O}_2\left(g\right)+O_2\left(g\right)$ is a first order reaction. After 30 min, from the start of the decomposition in a closed vessel, the total pressure developed is found to be 284.5 mm Hg. On complete decomposition, the total pressure is 584.5 mm Hg. The rate constant of the reaction is:

• A. $5.21\times {10}^{-3}$ min${}^{-1}$
• B. $5.12\times {10}^{-3}$ min${}^{-1}$
• C. $5.89\times {10}^{-3}$ min${}^{-1}$
• D. None of these

Answer 1

The correct option is A $5.21\times {10}^{-3}$ min${}^{-1}$

$\begin{array}{cccc}& N_2{O}_5\left(g\right)\to & 2N{O}_2\left(g\right)& +\frac{1}{2}{O}_2\left(g\right)\\ Initialpressure& a& 0& 0\\ Attimet& a-x& 2x& \frac{x}{2}\\ Atcomplete& & & \\ dissociation& 0& 2a& \frac{a}{2}\end{array}$

$2N_2{O}_5\left(g\right)\to 4N{O}_2\left(g\right)+O_2\left(g\right)$
OR
Total number of moles at time $t=a-x+2x+\frac{x}{2}$
$=a+\frac{3}{2}x$
Total number of moles at complete dissociation $=2a+\frac{a}{2}=\frac{5}{2}a$
$\therefore \frac{5}{2}a\propto 584.5mmHg$ ----- (1)
and $a+\frac{3}{2}x\propto 284.5mmHg$ -------(2)

Solving 1 equation, we get $a\propto$ 233.8 mm Hg and on solving equation 2,we get x $\propto$ 33.8 mm Hg. Substituting the value in first order equation:

$k=\frac{2.303}{t}log\frac{a}{a-x}=\frac{2.303}{30min}log\frac{233.8}{(233.8-33.8)}$
$=5.21\times {10}^{-3}mi{n}^{-1}$

Hence, the correct option is $\text{A}$