Question

The following data were obtained during the first order thermal decomposition of $N_2{O}_5\left(g\right)$ at constant volume :
$2N_2{O}_5\left(g\right)\to 2N_2{O}_4\left(g\right)+O_2\left(g\right)$

$\begin{array}{ccc}\text{S.NO}& \text{Time/s}& \text{Total Pressure/(atm)}\\ 1& 0& 0.5\\ 2& 100& 0.512\end{array}$

Calculate the rate constant.

Answer 1

Total pressure

Let the pressure of $N_2{O}_5\left(g\right)$decrease by
$2x$ atm

As two moles of$N_2{O}_5$ decompose to give
one mole of $O_2\left(g\right)$ and two moles of
$N_2{O}_5\left(g\right)$, the pressure of $N_2{O}_5\left(g\right)$ increases
by 2x atm and that of $O_2\left(g\right)$ increases by $x$
atm.

It can be written as -

$2N_2{O}_5\left(g\right)\to 2n_2{O}_4\left(g\right)+O_2\left(g\right)Att=00.5atm0atm0atmAttimet(0.5-2x)atm2xatmxatm$

At time t, total pressure $p_t$ is-
$p_t=p_{N_2{O}_5}+p_{N_2{O}_4}+p_{O_2}=(0.5-2x)+2x+x=0.5+x\Rightarrow x=p_t-0.5So,p_{N_2{O}_5}=0.5-2x$

$=0.5-2(p_t-0.5)=1.5-2p_t$

At $t=100s;p_t=0.512atm$

$\therefore p_{N_2{O}_5}=1.5-2\times 0.512=0.476atm$

Rate constant

We know that,
$k=\frac{2.303}{t}\log \frac{p_t}{p_A}$

Putting values in above equation,

$k=\frac{2.303}{100s}\log \frac{0.5atm}{0.476atm}$

$=\frac{2.303}{100s}\times 0.0216$

$=4.98\times {10}^{-4}{s}^{-1}$

Final answer: The rate constant of reaction is $=4.98\times {10}^{-4}{s}^{-1}$