Calculate the equilibrium concentration of [N2][N2] if the initial concentrations are 2.25 MM N2N2 and 0.55 MM O2O2. (This N2/O2N2/O2 concentration ratio is the ratio found in air.)

The equilibrium concentration of N2 is 2.22765 M

Explanation:

The air pollutant NO is produced in automobile engines from the high-temperature reaction:

N2(g)+O2(g) <--> 2NO(g)  Kc=1.7 * 10^-3

Part A: Calculate the equilibrium concentration of N2 if the initial concentrations are 2.25 M N2 and 0.55 M O2. (This N2/O2 concentration ratio is the ratio found in air.)

Step 1: Data given

Initial concentration of N2 = 2.25 M

Initial concentration of O2 = 0.55M

Kc = 1.7 * 10^-3

Step 2: The initial concentrations

[N2] = 2.25 M

[O2] = 0.55 M

[NO] = 0M

Step 3: The concentrations at equilibrium

[N2] = 2.25 - x M

[O2] = 0.55 - x M

[NO] = 2x M

Step 4: Kc

Kc = 1.7 *10^-3 = (2x)² / (2.25 -x)(0.55 -x )

1.7 * 10^-3 = 4x² / (1.2375 - 2.80x  +x² )

0.0017x² - 0.00476x + 0.002104 = 4x²

3.9983x² + 0.00476x - 0.002104 = 0

x = 0.02235

[N2] = 2.25 - x M = 2.25 - 0.02235 = 2.22765 M

[O2] = 0.55 - x M = 0.55 - 0.02235 = 0.52765 M

[NO] = 2x M = 2* 0.02235 = 0.0447 M

The equilibrium concentration of N2 is 2.22765 M

b. th3 n4zc4 pl4t3 15 c0nv3rg1ng w1th the s0uth 4m3ric4n pl4t3.

the positions which points up in upward carbon and down in downward carbon are axial

and vice versa holds true for equatorial positions

so the correct answer with labelled diagram will be