Question

Assume ideal gas behaviour. For the reaction, $N_2{O}_5\left(g\right)\rightleftharpoons 2N{O}_2\left(g\right)+0.5O_2\left(g\right)$, the initial pressure is 600 mmHg and the pressure at any time is 960 mmHg. The fraction of $N_2{O}_5\left(g\right)$ decomposed at constant volume and temperature is:

• A. 0.4
• B. 0.5
• C. 0.1
• D. 0.3

Answer 1

The correct option is A

0.4

This is a beautifully worded problem. The initial pressure is 600 mmHg while the "pressure at any time” is 960 mmHg. We can make sense of it by assuming that only ($N_2{O}_5$g) remains initially. It is given that the decomposition takes place at constant volume and constant temperature.

It is a roundabout way of saying that the number of moles of $N_2{O}_5$(g) at t=0 corresponds to 600 mmHg. Or we can say that the number of moles of $N_2{O}_5$(g) at t= 0 exert a pressure of 600 mmHg.
After the system attains chemical equilibrium, at "any time” the equilibrium pressure is 960 mmHg
Volume and temperature are constant. So we can say that at t = 0 and at t = any time after equilibrium, Pressure is proportional to the number of moles.
So we can either speak in terms of concentrations, or in no of moles or in partial pressures.
$\begin{array}{cccc}& N_2{O}_5\left(g\right)\leftrightharpoons 2N{O}_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)& & \\ t=0& 600mmHg& 0mmHg& 0mmHg\\ atanytime& 600mmHg-p& 2p& p/2\end{array}$
Pressure at any time = $(600mmHg-p)+2p+\frac{p}{2}=600mmHg+\frac{3}{2}p=960mmHg$
Solving, p = 244 mmHg.

We need the fraction of $N_2{O}_5$ that dissociates = $\frac{p}{\left(600mmHg\right)}$ = $\frac{\left(244mmHg\right)}{\left(600mmHg\right)}=\mathrm{0.407.}$