Question

The decomposition of $N_2{O}_5$ according to the equation, $N_2{O}_5\left(g\right)\to 4N{O}_2\left(g\right)+O_2\left(g\right)$
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 $280mmHg$ and on completion, the total pressure is $580mmHg.$ The rate constant of the reaction is $x\times {10}^{-3}mi{n}^{-1}.$ Find the value of x. Take: log (1.16)=0.065

Answer 1

$2N_2{O}_5\to 4N{O}_2+O_2$
On decomposition of 2 moles of $N_2{O}_5,$ 4 moles of $N{O}_2$ and 1 mole of $O_2$ are produced.

$2N_2{O}_5\to 4N{O}_2+O_2$
$Att=0:P_{0}00$
$Att=30:P_0-2P4PP$
$Att=\infty :02P_0{P}_0/2$
At $t=\infty$
$\text{Total pressure}=2P_0+P_0/2=\frac{5}{2}{P}_0$
$\text{Initial pressure of}{N}_2{O}_5,P_0=\frac{2}{5}\times 580=232mmHg$

At $t=30min$
$\text{Total pressure}=P_0-2P+4P+P$
$P_0+3P=280$
$3P=280-232=48mmHg$
$P=\frac{48}{3}=16mmHg$
Pressure of $N_2{O}_5$ after 30 minutes $=232-(2\times 16)$
$=200mmHg$
$k=\frac{2.303}{30}lo{g}_{10}\frac{232}{200}=5\times {10}^{-3}mi{n}^{-1}$