Question

The decomposition of $N_2{O}_5$ according to 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 $30$ minutes from the start of the decomposition in a closed vessel, the total pressure developed is found to be $284.5$ mm of $Hg$ and on complete decomposition, the total pressure is $584.5$ mm of $Hg$.

The rate constant of a reaction is:

• A. $5.206\times {10}^{-3}mi{n}^{-1}$
• B. $4.261\times {10}^{-3}mi{n}^{-1}$
• C. $3.316\times {10}^{-4}mi{n}^{-1}$
• D. None of these

Answer 1

The correct option is A $5.206\times {10}^{-3}mi{n}^{-1}$

For the given reaction:

$2N_2{O}_5\left(g\right)\to 4N{O}_2\left(g\right)+O_2\left(g\right)$

$a$ $0$ $0$ -------- initial time $(t=0)$
$a-X$ $2X$ $X/2$ ------- after time $(t=30min)$

$0$ $2a$ $a/2$ ----- after complete reaction

$\because$ No. of mole at any time $\propto$ pressure developed at that time

$\therefore a\propto P_0$ at $t=0$ ------- 1

$a+\left(3X/2\right)\propto 284.5$ at $t=30$ ------ 2

$\left(5a/2\right)\propto 584.5$ at $t=completelyreacted$ ------ 3

From 1 and 3 equation, we get

$\therefore a\propto 233.8$ ----------------------- 4

Substituting 4 in 2 equation we get

$X\propto 33.8$ --------------------------------- 5

Now,

$\therefore K=\frac{2.303}{t}log\frac{a}{(a-X)}$

$\therefore K=\frac{2.303}{30}log\frac{233.8}{200}$

$=5.206\times {10}^{-3}mi{n}^{-1}$
Option A is correct.