In the reaction of: 2O4 (g) ⇌ 2O2(g) If the equilibrium constant is Kc = 0.513 and the initial concentration of 2O4 = 0.0500 . What are the equilibrium concentrations of 2O4 and O2 under these conditions?

[N₂O₄] = 0.00411 M

[NO₂] = 0.0918 M

Explanation:

In the reaction of: N₂O₄(g) ⇌ 2 NO₂(g) If the equilibrium constant is Kc = 0.513 and the initial concentration of N₂O₄ = 0.0500 M. What are the equilibrium concentrations of N₂O₄ and NO₂ under these conditions?

Step 1: Given data

Initial concentration of N₂O₄: 0.0500 M

Equilibrium constant (Kc): 0.513

Step 2: Make an ICE chart

        N₂O₄(g) ⇌ 2 NO₂(g)

I        0.0500           0

C           -x              +2x

E      0.0500-x        2x

Step 3: Find the value of "x"

The equilibrium constant is:

Kc = [NO₂]² / [N₂O₄]

0.513 = (2x)² / (0.0500-x)

4x² + 0.513 x -0.02565

We solve for x and we get x₁ = 0.0458942 and x₂ = −0.558894 (neglected).

Step 4: Calculate the concentrations at equilibrium

[N₂O₄] = 0.0500-x = 0.00411 M

[NO₂] = 2x = 0.0918 M

(newtons first law)

conservation of momentum law states :

momentum before collision = momentum after collision

momentum p (kg-m/s)= mass * velocity

say moving car=1 000 kg, velocity = 10 m/s then p = 1 000*10= 10 000 kg-m/s

say still car = 2 000 kg, velocity = 0 m/s then p = 2 000* 0 = 0 kg-m/s

total momentum prior to collision = 10 000 + 0 = 10 000 kg-m/s

momentum after collision = 10 000 kg-m/s

mass now = 1 000 kg+2 000 kg = 3 000 kg

10 000 = 3 000 kg* velocity m/s

10 000/3 000 = velocity

3.33 m/s = velocity after collision

an impulse is a force applied over time , momentum is exclusive of external forces and a perfect collision is implied