A 2.0 L vessel contains an equilibrium mixture of 0.080 mol N2O4 and 0.040 mol NO2 for N2O4(g) 2NO2(g) at fixed temperature. Then 0.060 mol of NO2 is injected and, at the same moment, the volume is reduced to 1.0 L (temperature unchanged). Immediately afterwards the system has

Q_c = 12.5 K_c and shifts backward

Q_c = 6.25 K_c and shifts backward

Q_c = 12.5 K_c and shifts backward

Q_c = 0.08 K_c and shifts forward

Q_c = 25.0 K_c and shifts backward

JEE Advanced levelSingle correct optionDifficulty 3 of 3ce144

A $2.0 L$ vessel contains an equilibrium mixture of $0.080$ mol $N X_{2} O X_{4}$ and $0.040$ mol $NO X_{2}$ for $N X_{2} O X_{4} (g) 2 NO X_{2} (g)$ at fixed temperature. Then $0.060$ mol of $NO X_{2}$ is injected and, at the same moment, the volume is reduced to $1.0 L$ (temperature unchanged). Immediately afterwards the system has

Initially

K_{c} = (0.020)^{2} /0.040 = 0.0100 mol L^{- 1}

. After the changes there are

0.100

mol

NO X_{2}

and

0.080

mol

N X_{2} O X_{4}

1.0 L

, so

Q_{c} = (0.100)^{2} /0.080 = 0.125

, which is

12.5 K_{c}

. Hence the system shifts backward. Ignoring the volume reduction would give only 6.25

K_{c}