JEE Main 2023 · Chemistry · Chemical Equilibrium · medium

N2O4(g)rightleftharpoons 2NO2(g) is 20% dissociated at a total pressure of 1 atm at a fixed temperature. At what total pressure (same temperature)…

Single correct option

\mathrm{N_2O_4(g)\rightleftharpoons 2NO_2(g)}

20\%

dissociated at a total pressure of

1\ \mathrm{atm}

at a fixed temperature. At what total pressure (same temperature) will it be

50\%

dissociated?

Show answer and solution

Answer: (A) $0.125\ \mathrm{atm}$

With one mole initially, total moles

=1+\alpha

, so

p_{\mathrm{NO_2}}=\frac{2\alpha}{1+\alpha}P

and

p_{\mathrm{N_2O_4}}=\frac{1-\alpha}{1+\alpha}P

. Hence

K_p=\frac{4\alpha^2P}{1-\alpha^2}

. At

\alpha=0.2

P=1

K_p=\frac{4(0.04)}{0.96}=\frac16

. At

\alpha=0.5

\frac16=\frac{4(0.25)P}{0.75}=\frac{4P}{3}

, so

P=\frac18=0.125\ \mathrm{atm}

. A common error is to use

1-\alpha

instead of

1-\alpha^2

in the denominator.

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