In the equilibrium N2O4(g) 2NO2(g) at constant temperature, the equilibrium partial pressures are p(N2O4) = 1.00 atm and p(NO2) = 0.60 atm. The volume is halved isothermally and a new equilibrium is established. Calculate the new p(NO2) in atm, to two decimal places.

JEE Main levelNumerical answerDifficulty 2 of 3ce030

In the equilibrium $N X_{2} O X_{4} (g) 2 NO X_{2} (g)$ at constant temperature, the equilibrium partial pressures are $p (N X_{2} O X_{4}) = 1.00$ atm and $p (NO X_{2}) = 0.60$ atm. The volume is halved isothermally and a new equilibrium is established. Calculate the new $p (NO X_{2})$ in atm, to two decimal places.

Numerical answer type, answer in atm, tolerance 0.005.

K_{p} = (0.60)^{2} /1.00 = 0.36

atm. Halving the volume doubles the partial pressures to 2.00 and 1.20 atm. If

y

atm of

N X_{2} O X_{4}

forms from

2 y

atm of

NO X_{2}

(1.20 - 2 y)^{2} = 0.36 (2.00 + y)

, giving

y = 0.1592

. New

p (NO X_{2}) = 1.20 - 2 y = 0.882

atm.