Chemistry Daily Challenge 17-July-2015

We note that $N_2O_4 \rightleftharpoons 2NO_2$ , it means that $1$ mol of $N_2O_4$ gives $2$ mol of $NO_2$ . From Ideal Gas Law $pV = nRT$ and with constant volume $V$ , we have:

$\begin{aligned} \frac{p_2}{n_2T_2} & = \frac{p_1}{n_1T_1} \\ \frac{p_2}{(0.8+2\times 0.2) (600)} & = \frac{1}{(1)(300)} \\ p_2 & = \frac{1.2\times 600}{300} = \boxed{2.4} \end{aligned}$

After heating moles of $N{O}_{2}$ =0.4

After heating moles of ${N}_{2}{O}_{4}$ =0.8

Therefore, total moles after heating = 1.2

$\frac{{n}_{1}{T}_{1}}{{P}_{1}}=\frac{{n}_{2}{T}_{2}}{{P}_{2}}$

On substituting the values we get, ${P}_{2}$ =2.4 atm.

Keep on posting problems on chemistry!