For the reaction equilibrium N_2 O_4(~g) leftharpoons 2 NO_2(~g) the concentrations of N_2 O_4 and NO_2 at equilibrium are 4.8 × 10^ -2 and 1.2 × 10^ -2 ~mol ~L^ -1 respectively. The value of K_ c for the reaction is

Correct answer: B. 3 × 10^ -3 ~mol ~L^ -1 K_ C = [NO_2]^2 [~N_2 O_4] = [1.2 × 10^ -2 ]^2 [4.8 × 10^ -2 ] =3 × 10^ -3 ~mol / L

JEE Main 2003

JEE Main 2003 · Chemistry · Chemical Equilibrium

For the reaction equilibrium N_2 O_4(~g) leftharpoons 2 NO_2(~g)...

For the reaction equilibrium

\mathrm{N}_2 \mathrm{O}_4(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_2(\mathrm{~g})

the concentrations of

\mathrm{N}_2 \mathrm{O}_4

and

\mathrm{NO}_2

at equilibrium are

4.8 \times 10^{-2}

and

1.2 \times 10^{-2} \mathrm{~mol} \mathrm{~L}^{-1}

respectively. The value of

\mathrm{K}_{\mathrm{c}}

for the reaction is

B. $3 \times 10^{-3} \mathrm{~mol} \mathrm{~L}^{-1}$

\mathrm{K}_{\mathrm{C}}=\frac{\left[\mathrm{NO}_2\right]^2}{\left[\mathrm{~N}_2 \mathrm{O}_4\right]}=\frac{\left[1.2 \times 10^{-2}\right]^2}{\left[4.8 \times 10^{-2}\right]}=3 \times 10^{-3} \mathrm{~mol} / \mathrm{L}