For the equilibrium N2O4(g)rightleftha…

For the equilibrium N2O4(g)rightleftharpoons 2NO2(g), Kp=0.15 atm at 298 K and Δ H^°=+57.2 kJ mol^-1 (assumed independent of temperature). Take…

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For the equilibrium

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

K_p=0.15\ \mathrm{atm}

298\ \mathrm{K}

and

\Delta H^\circ=+57.2\ \mathrm{kJ\,mol^{-1}}

(assumed independent of temperature). Take

R=8.314\ \mathrm{J\,K^{-1}\,mol^{-1}}

. Pure

\mathrm{N_2O_4}

is introduced into a vessel that is maintained at

350\ \mathrm{K}

and at a constant total pressure of

2.0\ \mathrm{atm}

until equilibrium is established. The degree of dissociation of

\mathrm{N_2O_4}

is closest to

Show answer and solution

Answer: (B) $0.61$

First obtain

K_p

350\ \mathrm{K}

from the van't Hoff equation:

\ln\dfrac{K_2}{K_1}=-\dfrac{\Delta H^\circ}{R}\left(\dfrac1{T_2}-\dfrac1{T_1}\right)=-\dfrac{57200}{8.314}\left(\dfrac1{350}-\dfrac1{298}\right)=3.43

, so

K_2\approx0.15\times e^{3.43}\approx4.63\ \mathrm{atm}

.
For

\mathrm{N_2O_4}

with degree of dissociation

\alpha

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

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

, hence

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

and

\alpha=\sqrt{\dfrac{K_p}{K_p+4P}}=\sqrt{\dfrac{4.63}{4.63+8}}=0.61

.
Distractors:

0.14

uses

K_p

298\ \mathrm{K}

(no temperature correction);

0.73

uses

P=1\ \mathrm{atm}

;

0.02

reverses the sign in the van't Hoff equation (as if the reaction were exothermic). Answer: (B).

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