37.8 g N2O5 was taken in a 1 L reaction vessel and allowed to undergo the following reaction at 500 K

Question

\(37.8 \mathrm{~g} \mathrm{~N}_{2} \mathrm{O}_{5}\) was taken in a 1 L reaction vessel and allowed to undergo the following reaction at 500 K

\(2 \mathrm{N}_{2} \mathrm{O}_{5(\mathrm{~g})} \rightarrow 2 \mathrm{N}_{2} \mathrm{O}_{4(\mathrm{~g})}+\mathrm{O}_{2(\mathrm{g})}\)

The total pressure at equilibrium was found to be 18.65 bar.

Then, \(\mathrm{Kp}=\) _____ \(\times 10^{-2}\) [nearest integer]

Assume \(\mathrm{N}_{2} \mathrm{O}_{5}\) to behave ideally under these conditions

Given : \(\mathrm{R}=0.082\) bar \(\mathrm{L} \mathrm{mol}^{-1} \mathrm{~K}^{-1}\)

Answer 1

Answer is : 962

Initial pressure of \(\mathrm{N}_{2} \mathrm{O}_{5}\) 

\( =\frac{\frac{37.8}{108} \times 0.082 \times 500}{1}=14.35 \mathrm{bar} \) 

 \( 2 \mathrm{~N}_{2} \mathrm{O}_{5} \rightleftharpoons 2 \mathrm{N}_{2} \mathrm{O}_{4}+\mathrm{O}_{2} \) 

\(\mathrm{t}=0\quad 14.35\) 

 \(t = eq\quad 14.35 - 2P\quad 2P\quad P\) 

\(\mathrm{P}_{\text {Total }}\) at \(\mathrm{eqb}=14.35+\mathrm{P}=18.65\) 

\( \mathrm{P}=4.3\) 

\( \mathrm{P}_{\mathrm{N}_{2} \mathrm{O}_{5}}=5.75 \mathrm{bar}\)

\( \mathrm{P}_{\mathrm{N}_{2} \mathrm{O}_{4}}=8.6 \mathrm{bar}\) 

\( \mathrm{P}_{\mathrm{O}_{2}}=4.3 \mathrm{bar}\)  

\(\mathrm{k}_{\mathrm{p}}=\frac{(8.6)^{2} \times(4.3)}{(5.75)^{2}}=9.619=\mathrm{x} \times 10^{-2}\) 

\( \mathrm{x}=961.9 \approx 962\)