For the following equilibrium, N2O42NO2;Kc=0.67. If we start with 3 moles of NO2 and 1 mole of N2O4 in 1L flask, then NO2 present at equilibrium is

Question

For the following equilibrium, N2O42NO2;Kc=0.67. If we start with 3 moles of NO2 and 1 mole of N2O4 in 1L flask, then NO2 present at equilibrium is (A) 1.5 mol (B) 2.0 mol (C) 0.5 mol (D) 1.0 mol

Tardigrade Admin · Accepted Answer

N2O4(g)leftharpoons 2NO2(g) Q=([NO2]2/[N2O4])=([3]2/[1])=9>Kc Thus, equilibrium is displaced in backward side. underset beginsmallmatrix 3 (3-2x) endsmallmatrix mathop2NO2 leftharpoons underset beginsmallmatrix 1 (1+x) endsmallmatrix mathopN2O4 Kc=(1/Kc)=(1/0.67)=(1+x/(3-2x)2)=1.5 Only permissible value of x that satisfies above equation is x=0.5 Thus, NO2 at equilibrium =3-2x=2 mol.

Q. For the following equilibrium, N2​O4​2NO2​;Kc​=0.67. If we start with 3 moles of NO2​ and 1 mole of N2​O4​ in 1L flask, then NO2​ present at equilibrium is

1.5 mol

2.0 mol

0.5 mol

1.0 mol

Q. For the following equilibrium, $N_{2} O_{4} 2 N O_{2}; K_{c} = 0.67.$ If we start with 3 moles of $N O_{2}$ and 1 mole of $N_{2} O_{4}$ in 1L flask, then $N O_{2}$ present at equilibrium is