Answer to Question #142346 in Chemistry for Aman

Question

If 5.0 mol of N2O4 is placed in a 1.0L closed container at 150C, what will be the concentrations of N2O4 and NO2 at equilibrium? K=4.5

Answer 1

Solution:

The initial concentration of N₂O₄:

C_o(N₂O₄) = n(N₂O₄) / V(N₂O₄) = 5.0 mol / 1.0 L = 5.0 M

C_o(N₂O₄) = 5.0 M

N₂O₄(g) = 2NO₂(g)

The equlibrium constant expression for the reaction:

K = [NO₂]² / [N₂O₄] = 4.5

........................N₂O₄(g) = 2NO₂(g)

I (initial)..............5.0..............--.......

C (change)..........-x.............+2x......

E (equlibrium)....5.0-x..........2x.......

K = [NO₂]² / [N₂O₄] = (2x)² / (5.0-x) = 4.5

4x² = 4.5 × (5.0-x)

4x² = 22.5 - 4.5x

4x² + 4.5x - 22.5 = 0

x = 1.875

[N₂O₄] = 5.0 - x = 5.0 - 1.875 = 3.125 M

[N₂O₄] = 3.125 M

[NO₂] = 2x = (2 × 1.875) = 3.75 M

[NO₂] = 3.75 M

Answer: at equilibrium, the concentrations of NO₂ and N₂O₄ will be 3.75 M and 3.125 M, respectively (if K = 4.5).