For N2O4(g) 2NO2(g), K_p = 3.0 atm at a given temperature. A gas mixture containing 1.0 mol N2O4 and 3.0 mol NO2 is held at a total pressure of 2.0 atm at this temperature. The mixture will

shift backward, since Q_p = 4.5 atm is greater than K_p

shift forward, since Q_p = 2.25 is less than K_p

shift forward, since p_NO2 > p_N2O4

not shift, since the total pressure is constant

shift backward, since Q_p = 4.5 atm is greater than K_p

JEE Main levelSingle correct optionDifficulty 2 of 3ce123

For $N X_{2} O X_{4} (g) 2 NO X_{2} (g)$ , $K_{p} = 3.0$ atm at a given temperature. A gas mixture containing $1.0$ mol $N X_{2} O X_{4}$ and $3.0$ mol $NO X_{2}$ is held at a total pressure of $2.0$ atm at this temperature. The mixture will

Partial pressures:

p_{N X_{2} O X_{4}} = 2.0 \times 1/4 = 0.5

atm and

p_{NO X_{2}} = 2.0 \times 3/4 = 1.5

atm.

Q_{p} = p_{NO X_{2}}^{2} / p_{N X_{2} O X_{4}} = (1.5)^{2} /0.5 = 4.5

atm, which exceeds

K_{p} = 3.0

atm, so the reaction goes backward. Using mole fractions in place of pressures gives 2.25 and the wrong direction.